TYPICAL QUESTIONS & ANSWERS

PART - I

OBJECTIVE TYPE QUESTIONS

Each Question carries 2 marks.

Choose correct or the best alternative in the following:

Q.1 The value of limit is

(A) 0 (B) 1

(C) 2 (D) does not exist

Ans: D

Q.2 If , then equals

(A) 0 (B) u

(C) 2u (D) 3u

Ans: A

Q.3 Let . Then the value of is

(A) (B) 0

(C) (D)

Ans: B

Q.4 The value of is

(A) 1 (B)

(C) (D) 3

Ans: A

Q.5 The solution of is

(A) (B)

(C) (D)

Ans: C

Q.6 The solution of is

(A) sin x (B) cos x

(C) x sin x (D) x cos x

Ans: D

Q.7 . Let and be elements of . The set of vectors is

(A) linearly independent (B) linearly dependent

(C) null (D) none of these

Ans: A

Q.8 The eigen values of the matrix are

(A) and 1 (B) 0, 1 and 2

(C) –1, –2 and 4 (D) 1, 1 and –1

Ans: A

Q.9 Let , , be the Legendre polynomials of order 0, 1, and 2, respectively. Which of the following statement is correct?

(A) (B)

(C) (D)

Ans: B

Q.10 Let be the Bessel function of order n. Then is equal to

(A) (B)

(C) (D)

Ans: D

Q.11 The value of limit

(A) 0 (B)

(C) (D) does not exist

Ans: D

Q.12 Let a function f(x, y) be continuous and possess first and second order partial derivatives at a point (a, b). If is a critical point and , , then the point P is a point of relative maximum if

(A) (B)

(C) (D)

Ans: B

Q.13 The triple integral gives

(A) volume of region T (B) surface area of region T

(C) area of region T (D) density of region T

Ans: A

Q.14 If then matrix A is called

(A) Idempotent Matrix (B) Null Matrix

(C) Transpose Matrix (D) Identity Matrix

Ans: A

Q.15 Let be an eigenvalue of matrix A then , the transpose of A, has an eigenvalue as

(A) (B)

(C) (D)

Ans: C

Q.16 The system of equations is said to be inconsistent, if it has

(A) unique solution (B) infinitely many solutions

(C) no solution (D) identity solution

Ans: C

Q.17 The differential equation is an exact differential equation if

(A) (B)

(C) (D)

Ans: B

Q.18 The integrating factor of the differential equation is

(A) (B)

(C) xy (D)

Ans: D

Q.19 The functions defined on an interval I, are always

(A) linearly dependent (B) homogeneous

(C) identically zero or one (D) linearly independent

Ans: D

Q.20 The value of , the second derivative of Bessel function in terms of and is

(A) (B)

(C) (D)

Ans: C

Q.21 The value of limit is

(A) 0 (B) 1

(C) -1 (D) does not exist

Ans: A

Q.22 If , the total differential of the function at the point (1, 2) is

(A) e (dx + dy) (B) e²(dx + dy)

(C) e⁴(4dx + dy) (D) 4e⁴(dx + dy)

Ans: D

Q.23 Let , x > 0, y > 0 then

equals

(A) 0 (B) 2u

(C) u (D) 3u

Ans: B

Q.24 The value of the integral over the domain E bounded by planes x = 0, y = 0, z = 0, x + y + z = 1 is

(A) (B)

(C) (D)

Ans: C

Q.25 The value of α so that is an integrating factor of the differential equation is

(A) -1 (B) 1

(C) (D)

Ans: C

Q.26 The complementary function for the solution of the differential equation is obtained as

(A) Ax + Bx ^-3/2 (B) Ax + Bx ^3/2

(C) Ax² + Bx (D) Ax ^-3/2 + Bx ^3/2

Ans: A

Q.27 Let be elements of R³. The set of vectors is

(A) linearly independent (B) linearly dependent

(C) null (D) none of these

Ans: A

Q.28 The value of µ for which the rank of the matrix is equal to 3 is

(A) 0 (B) 1

(C) 4 (D) -1

Ans: B

Q.29 Using the recurrence relation, for Legendre’s polynomial (n + 1) , the value of P₂ (1.5) equals to

(A) 1.5 (B) 2.8

(C) 2.875 (D) 2.5

Ans: C

Q.30 The value of Bessel function J₂(x) in terms of J₁(x) and J₀(x) is

(A) 2J₁(x) – x J₀(x) (B)

(C) (D)

Ans: D

Q.31 The value of the integral where C is the contour is

(A) . (B) .

(C) 0. (D) .

Ans: C

Because z = 1 is a pole for given function f and it lies outside the circle

|z| = ½ . Therefore, by Cauchy’s Theorem

Q.32 If X has a Poisson distribution such that then the variance of the distribution is

(A) 1. (B) -1.

(C) 2. (D) 0.

Ans: A

Because P (x = 2) = 9 P (x = 4) + 90 P (x = 6)

=>

=>

Because m¹0, Therefore, 3m² + m⁴ – 4 = 0

=> m = 1

Q.33 The vector field function is called solenoidal if

(A) curl =0. (B) div =0.

(C) grad =0. (D) grad div =0.

Ans: B

A vector field is solenoidal if div = 0

Q.34 The number of distinct real roots of in the interval is

(A) 0. (B) 2.

(C) 3. (D) 1.

Ans: D

=> (cos x – sin x)² (sin x + 2 cos x) = 0

Its only root which lies in .

Q.35 The solution of : is

(A) . (B) .

(C) . (D) .

Ans: A

2y sin x + cos 2x = a

I.F.

Therefore, the solution is given as

=> 2y sin x + cos 2x = a

Q.36 If then is equal to

(A) . (B) .

(C) . (D) .

Ans: A

where u and v are homogeneous functions of order 6 and 0 respectively. Using Euler’s theorem = 6 u + 0 v = 6 u.

Q.37 The value of Legendre’s Polynomial, is

(A) 1. (B) -1.

(C) . (D) 0.

Ans: D

By Rodrigue’s formula,

P_n(0) = 0 if n is odd.

Q.38 The value of integral over the region bounded by the line y = x and the curve is

(A) . (B) .

(C) . (D) .

Ans: C

=

Q.39 The value of the integral where C is the semi-circular arc above the real axis is

(A) . (B) .

(C) . (D) .

Ans: A

Let z = e^iq then

= i p

Q.40 Residue at z = 0 of the function is

(A) . (B) .

(C) . (D) .

Ans: B

Let

Residue = coefficient of

Q.41 In solving any problem, odds against A are 4 to 3 and odds in favour of B in solving the same problem are 7 to 5. The probability that the problem will be solved is

(A) . (B) .

(C) . (D) .

Ans: B

P(A) = , P(B) = . Probability problem will be solved i.e. P(AÈB)

P(AÈB) = P(A) + P(B) – P(AB)

Because A & B are independent, So P(AB) = P(A) P(B)

P(AÈB) =

Q.42 The value of the integral over the area in the first quadrant by the curve is

(A) . (B) .

(C) . (D) .

Ans: D

over x² – 2ax + y² = 0

=

Q.43 The surface will be orthogonal to the surface at the point for values of a and b given by

(A) a = 0.25, b = 1. (B) a = 1, b = 2.5.

(C) a = 1.5, b = 2. (D) .

Ans: A

a= 0.25, b = 1

Let F = ax² – byz – (a + 2) = 0

G = 4x²y + z³ – 4 = 0

These surfaces will be orthogonal if

Also since (1, -1, 2) lies on F

\ a + 2b – a – 2 = 0 b = 1 , thus a =

Q.44 If and and if z = u + v then equals

(A) 4 v. (B) 4 u.

(C) 2 u. (D) 4 u + v.

Ans: C

z = u + v

i.e. u is homogeneous function of degree 2 and v is homogeneous function of degree 0. By Euler’s Theorem,

Q.45 The series equals

(A) . (B) .

(C) . (D) .

Ans: C

Q.46 The value of integral , where is a Legendre polynomial of degree 3, equals

(A) . (B) 0.

(C) . (D) .

Ans: D

As

Q.47 For what values of x, the matrix is singular?

(A) 0, 3 (B) 3, 1

(C) 1, 0 (D) 1, 4

Ans: A

The matrix is singular if its determinant is zero. Solving determinant, we get equations

x(x-3)²=0.

Q.48 If then

(A) 3 ab (B) 2 abz

(C) abz (D) 3 abz

Ans: B

Because

Q.49 The value of the integral is

(A) . (B) 2.

(C) -2. (D) 0.

Ans: D

Since it is an odd function.

Q.50 If and then div

(A) 5 (B) 5u

(C) (D) 0

Ans: B

Q.51 The solution of the differential equation is given as

(A) (B)

(C) (D)

Ans: A

Dividing by z, we get

,

Let 1/(log z)=u, then above differential equation becomes

Q.52 The value of the integral where C is the circle is given as

(A) (B)

(C) 0 (D)

Ans: C

The given function has a pole at z=1, which lies outside the circle C. So by Cauchy’s theorem integral is zero.

Q.53 The value of the Legendre’s polynomial if

(A) (B)

(C) (D)

Ans: D

By orthogonal property of Legendre’s polynomial.

Q.54 Two persons A and B toss an unbiased coin alternately on the understanding that the first who gets the head wins. If A starts the game, then his chances of winning is

(A) (B)

(C) (D)

Ans: C

Probability of getting head=1/2= probability of getting tail.

If A starts the game, then in first chance either A wins the game, in second case A fails, B fails and A won the match and so on, we get an infinite series. Let H_A, H_B, T_A, T_B, denotes the getting of head and tails by A and B respectively.

P(wining of A)=P(H_A)+P(T_AT_BH_A)+ P(T_AT_BT_AT_BH_A)+…….

=

This is an infinite G.P. series with common ratio 1/4. Thus

P(winning of A) = .

Q.55 The value of limit

(A) equals 0. (B) equals .

(C) equals 1. (D) does not exist.

Ans: D

Let y= mx² be equation of curve. As x→0, y also tends to zero.

=

, which depends on m.

Thus it does not exist.

Q.56 If then equals

(A) . (B) .

(C) . (D) .

Ans: A

Eliminating we get

Q.57 The function has

(A) a minimum at (0, 0).

(B) neither minimum nor maximum at (0, 0).

(C) a minimum at (1, 1).

(D) a maximum at (1, 1).

Ans: B

f (x, y) = y² – x³

f_x = - 3x² = 0 , f_y = 2y = 0

gives (0,0) is a critical point.

∆f (x, y) = f(∆x,∆y)= (∆y)² –(∆ x)³

> 0 , if (∆y)² >(∆ x)³

< 0 , if (∆y)² < (∆ x)³

This means in the neighborhood of (0,0) f changes sign. Thus (0,0) is neither a point of maximum nor minimum.

Q.58 The family of orthogonal trajectories to the family , where k is an arbitrary constant, is

(A) . (B) .

(C) . (D) .

Ans: A

y = (x – k)² Diff. w.r.t. x

y₁ = 2(x – k) => y₁ = 2

For orthogonal trajectories y₁ is replaced by -1/y₁.

Therefore, -1/y₁ = 2

=> 2dy + dx = 0

Integrating, we get y^3/2 = ¾ (c-x)

Q.59 Let be two linearly independent solutions of the differential equation . Then , where are constants is a solution of this differential equation for

(A) . (B) .

(C) no value of . (D) all real .

Ans: B

yy” – (y’)² = 0

Because, y₁, y₂ are solutions

Therefore, y₁y₁” – (y₁’)² = 0

y₂y₂” – (y₂’)² = 0

Now (c₁y₁ + c₂y₂) (c₁y₁ + c₂y₂)” – ((c₁y₁ + c₂y₂)’)²

= (c₁y₁ + c₂y₂) (c₁y₁” + c₂y₂”) – (c₁y₁’² + c₂y₂’²) - 2 c₁y₁’c₂y₂’

= c₁²(y₁y₁” – (y₁’)²) + c₂² (y₂y₂” – (y₂’)²) + c₁ c₂ (y₁ y₂”+y₂y₁” - 2y₁’y₂’)

= 0, if c₁c₂ = 0.

Q.60 If A, B are two square matrices of order n such that AB=0, then rank of

(A) at least one of A, B is less than n.

(B) both A and B is less than n.

(C) none of A, B is less than n.

(D) at least one of A, B is zero.

Ans: B

Since A, B are square matrix of order n such that AB = 0, then rank of both A and B is less than n.

Q.61 A real matrix has an eigen value i, then its other two eigen values can be

(A) 0, 1. (B) -1, i.

(C) 2i, -2i. (D) 0, -i.

Ans: D

Because i is one eigen value so another eigen value must be – i.

Q.62 The integral, n>1, where is the Legendre’s polynomial of degree n, equals

(A) 1. (B) .

(C) 0. (D) 2.

Ans: C

Let I =

Let cosq = t. –sinqdq = dt

= 0

Q.63 The value of limit is

(A) 0 (B) 1

(C) limit does not exist (D) -1

Ans.: A

Language of the question is not up to the mark in the sense that its statement does not go with all the alternatives consequently, change is in order.

The suggested change is either satisfies the statement given in the alternative (C) or assumes the value given in one of three remaining alternatives A, B and D.

Q.64 If then the value of is equal to

(A) 0 (B)

(C) (D)

Ans.: B

Since taking log on both sides we get log(u)=y log(x)

Q.65 If , then the value of is

(A) z (B) 2z

(C) tan(z) (D) sin(z)

Ans.: C

If u(x,y) = , is a homogeneous function of degree n, then from Euler’s theorem

= nu.

Here = is a homogeneous function of degree 1.

Therefore u = sin z

From u = sin z; ,

Q.66 The value of integral is equal to

(A) (B)

(C) (D)

Ans.: B

Q.67 The differential equation of a family of circles having the radius r and the centre on the x-axis is given by

(A) (B)

(C) (D)

_{Ans.: A}

Let (h,0) be centre on x-axis. Thus eq. of circle is

Differentiating, w.r. to x, we get

Eliminating h between and

We get .

Q.68 The solution of the differential equation satisfying the initial conditions y(0)=1, y(π/2) = 2 is

(A) y = 2cos(x) + Sin(x) (B) y = cos(x) + 2 sin(x)

(C) y = cos(x) + Sin(x) (D) y = 2cos(x) + 2 sin(x)

Ans.: B

On solving the differential equation _, we get y = Acosx + Bsin x, Since

y(0)=1, Thus, y = cos(x) + 2 sin x

Q.69 If the matrix then

(A) C=Acos(θ) – Bsin(θ) (B) C=Asin(θ) + Bcos(θ)

(C) C=Asin(θ) – Bcos(θ) (D) C=Acos(θ) + Bsin(θ)

Ans.: D

Q.70 The three vectors (1,1,-1,1), (1,-1,2,-1) and (3,1,0,1) are

(A) linearly independent (B) linearly dependent

(C) null vectors (D) none of these.

Ans.: B

Let a,b,c be three constants such that a(1,1,-1,1)+b (1,-1,2,-1) +c(3,1,0,1)=(0,0,0,0).

This yields a + b + 3c = 0, a – b + c = 0, -a + 2b = 0, a – b + c = 0.

On solving, we get a = 2b = -2c → b = - c. Since a, b, c are non-zero, therefore three vectors are linearly dependent.

Q.71 The value of is equal to

(A) 1 (B) 0

(C) (D)

Ans.: B

Q.72 The value of the integral is

(A) (B)

(C) (D)

Ans.: C

. Here v=1.

Q.73 The value of limit is

(A) limit does not exist (B) 0

(C) 1 (D) -1

Ans.: A

Consider the path y = mx² As (x,y)→(0,0), we get x →0. Therefore

which depends on m. Thus limit does not exist.

Q.74 If then the value of is equal to

(A) 0 (B)

(C) (D)

Ans.: B

Since taking log on both sides we get log(u)=y log(x)

Q.75 If , then the value of is

(A) u (B) 2u

(C) 3u (D) 0

Ans.: D

Let

Here and are homogenous functions of degree zero.

Consequently

Or ; similarly .

;

= 0 + 0 = 0.

Q.76 The value of integral is equal to

(A) 22 (B) 26

(C) 5 (D) 25

Ans.: B

Q.77 The solution of the differential equation is given by

(A) (B)

(C) (D)

Ans.: A

Let x + y = t, Differentiating w r to x we get

Or ,

; integrating we get

→

Or or .

Q.78 The solution of the differential equation is

(A) (B)

(C) (D)

Ans.: A

The solution of differential equation is given as C.F.

P.I. =

In writing the C.F. we have used the roots of the auxiliary equation

i.e. m = 1, 2. For writing the P.I we have used ;

Q.79 If 3x+2y+z= 0, x+4y+z=0, 2x+y+4z=0, be a system of equations then

(A) system is inconsistent

(B) it has only trivial solution

(C) it can be reduced to a single equation thus solution does not exist

(D) Determinant of the coefficient matrix is zero.

Ans. B

, then system has only trivial solution.

Q.80 If λ is an eigen value of a non-singular matrix A then the eigen value of A^-1 is

(A) 1/ λ (B) λ

(C) -λ (D) -1/ λ

Ans. A By definition of A^-1.

Q.81 The product of eigen value of the matrix is

(A) 3 (B) 8

(C) 1 (D) -1

Ans.: B

Eigen values are 1,2,4.

Thus product = 8.

Q.82 The value of the integral is

(A) (B)

(C) (D)

Ans.: C . Here v=2.

Q.83 If then

(A) (B)

(C) (D)

Ans.: B

Since is a homogeneous function of degree 0. Thus by Euler’s theorem .

Q.84 If , then the value of is

(A) 1 (B) r

(C) 1/r (D) 0

Ans.: B

Q.85 The value of integral is equal to

(A) -4 (B) 3

(C) 4 (D) -3

Ans.: C

Q.86 The solution of differential equation under condition y(1)=1 is given by

(A) (B)

(C) (D)

Ans.: B

The given differential is a particular case of linear differential equation of first order

. Here

. Multiplying throughout by x, it can be written as

; Integrating w.r. to x we get

; Given y(1) = 1;

or which is alternative B.

Q.87 The particular integral of the differential equation is

(A) (B)

(C) (D)

Ans.: A

P.I. is a case of failure of ;

In such cases .

Q.88 The product of the eigen values of is equal to

(A) 6 (B) -8

(C) 8 (D) -6

Ans.: C

.

The eigenvalues are 1,2,4. Thus product of eigen values = 8.

Q.89 If then matrix A is equal to

(A) (B)

(C) (D)

Ans.: D

Q.90 The value of (m being an integer < n) is equal to

(A) 1 (B) -1

(C) 2 (D) 0

Ans.: D

Using Rodrigue formula can be expressed as

Q.91 The value of the is

(A) (B)

(C) (D)

Ans.: A

.

PART – II

NUMERICALS

Q.1 Consider the function f (x, y) defined by

Find and .

Is differentiable at (0, 0)? Justify your answer. (8)

Ans:

The partial derivatives are

Therefore, df = 0

Let dx = r cosq dy = r sinq

\ f(X,Y) is differentiable.

Q.2 Find the extreme values of subject to the constraints of (x, y, z) = 2x + y =0 and h(x, y, z) = x + y + z = 1 (8)

Ans:

Consider the Auxiliary function

For the extremum, we have the necessary conditions

From (4) we get y = -2x.

Taking y = -2x in (1), we get -2x + 2l₁ + l₂ = 0 -----------(6)

(2) & (6) implies 3l₁ + 2l₂ = 0 -----------(7)

From (5) x + y = 1- z. putting this in (1), we get 2 – 2z + 2l₁ + l₂ = 0 ----(8)

(3) and (8) implies 2l₁ + 2l₂ = -2 -----------(9)

(7) and (9) implies l₁ = 2, l₂ = -3

The point of extremum is

The extremum value is

Q.3 Find all critical points of and determine relative extrema at these critical points. (8)

Ans:

\ The only critical point is (x, y) = (0, 0)

Q.4 Find the second order Taylor expansion of about the point . (4)

Ans:

Second order Taylor expansion of f (x, y) is

Q.5 Change the order of integration in the following double integral and evaluate it : . (4)

Ans:

The region of integration is given by y £ x £ 1 and 0 £ y £ 1.

Hence, it is bounded by the straight lines x= y and x = 1 between y = 0 and y = 1.

y

x = y

y = 1

y = 0 0 x

x = 1

To find the limits of integration in the reverse order, we observe that the region is

Also given by 0 £ y £ x and 0 £ x £ 1

Hence,

Q.6 Solve the differential equation . (4)

Ans:

This is linear equation of 1^st order

I.F. =

Solution is

.

Therefore is the solution.

Q.7 Solve the differential equation . (6)

Ans:

Hence the equations is exact.

\ From (1) and (2) above, we get

Therefore, j(y) = -y + c

Therefore,

Q.8 Find the general solution of the differential equation by method of undetermined coefficients. (6)

Ans:

So the linearly independent solutions of the homogeneous equation are

The particular solution

By the method of undetermined coefficients, the particular solution is in the form y(z) = c₁e^z + c₂z e^z

Substituting in the equation(1), we get

Hence the general solution

y = c₃ Sin2z + c₄ Cos2z

y = c₃ Sin2(logx) + c₄ Cos2(logx) .

Q.9 Find the general solution of the differential equation . (9)

Ans:

This is Cauchy’s homogeneous linear equation.

Putting x = e^t,

Then given equation becomes (D(D – 1)(D -2) – D(D-1) + 2D – 2)y = e^3t.

which is linear equation with constant coefficients.

A.E. is .

=> D = 1, 1, 2.

.

.

= .

Q.10 Show that the eigen values of a Hermitian matrix are real. (7)

Ans:

We have Ax = lx. Premultiplying by , we get

The denominator x is always real and positive. Therefore the behaviour of l is

determined by Ax.

Q.11 Using Frobenius method, find two linearly independent solutions of the differential equation . (10)

Ans:

The point x = 0 is a regular singular point of the equation.

The lowest degree term is the term containing x^r-1.

Setting this coefficient to zero, we get

(1) may be written as

\ For m ³ 0, we get

when r = 0

when r = ½

Q.12 Solve the following system of equations by matrix method: (6)

Ans:

Therefore rank(A|b) = rank(A) = 3

So System admits unique solution.

Now the resultant system is

4z = 3 =>

y + 2z = 1 =>

x + y +2z = 0 => x = -1.

Q.13 Express the polynomial in terms of Legendre polynomials.

(8)

Ans:

.

writing various powers of x in terms of legendre polynomials.

Now,

Q.14 Let be the Bessel function of order . Show . (8)

Ans:

Now we know that

=

=

=

= .

Q.15 If A is a diagonalizable matrix and f (x) is a polynomial, then show that f(A) is also diagonalizable. (7)

Ans:

by induction, we may show that

Since D is a diagonal matrix, f(D) is also diagonal.

Thus f(A) is diagonalizable.

Q.16. Let. Find the matrix P so that is a diagonal matrix. (9)

Ans:

Eigen values are

Therefore l = 2, 2, 4 are Eigen values .

Eigen values corresponding to l = 2 is

The eigen vectors are (-2 1 0) ^T and (-1 0 1) ^T

Eigen vector corresponding to l = 4 is

\ The eigen vector is (1 0 1)^T

Q.17 Show that the function

is continuous at (0, 0) but its partial derivatives and do not exist at (0, 0). (8)

Ans: We have

Hence the given function is continuous at (0, 0)

Now at (0, 0), we have

Hence limit does not exist. Therefore f_x does not exist at (0, 0).

Also at (0, 0), the limit

Hence limit does not exist at (0, 0).

Q.18 Find the linear and the quadratic Taylor series polynomial approximation to the function about the point (1, 2). Obtain the maximum absolute error in the region and for the two approximations. (8)

Ans:

The linear approximation is given by

Maximum absolute error in the linear approximation is given by

The quadratic approximation is given by

The maximum absolute error in the quadratic approximation is given by

Q.19 Find the shortest distance between the line and the ellipse . (8)

Ans:

Let (x, y) be a point on the ellipse and (u, v) be a point on the line and the ellipse is the square root of the minimum value of

Subject to the constraints

Dividing we get 8y=9x. Substituting in ellipse we get

u = 2v – 2.

Substituting in the equation of line 2u + v –10 = 0, we get

Hence an extremum is obtained when (x, y) =

The distance between the two points is .

The distance between these two and pts is .

Hence shortest distance between line and ellipse is .

Q.20 Evaluate the double integral , where R is the region bounded by the x-axis, the line y = 2x and the parabola . (8)

Ans:

The points of intersection of the curves

The region

We evaluate the double integrate as

Q.21 Evaluate the integral where R is the parallelogram with successive vertices at , , and . (8)

Ans:

The region R is given in figure

The equations of the sides AB, BC, CD and DA are respectively

x – y = p, x + y = 3p, x – y = -p, x + y = p.

Let y – x = u, y + x = v. Then -p £ u £ p and p £ v £ 3p

Q.22 Show that where is the Bessel function of order. (8)

Ans:

We know that

Substituting n=0, 1,2, …………..adding we get

Q.23 Show that . (6)

where are the Legendre polynomials of order K.

Ans:

Integrating by parts

Now by orthogenality property, we have

If m = n

= .

=

= .

, m = n.

Q.24 Find the power series solution about x =2, of the initial value problem

. Express the solution in closed form. (10)

Ans:

We have

Putting x = 2, we get

Q.25 Solve the initial value problem y(0) = 0, , . (8)

Ans:

A.E. is

m = 1, 2, 3.

.

y(0) = 0

=>

y' (0) = 1

=>

y" (0) = -1

=> .

Thus

~

Solving .

Thus solution of initial value problem is .

Q.26 Solve . (8)

Ans:

Let x = e^t.

(D(D – 1) + D – 1)y = .

A.E. m² – 1 = 0.

m = 1.

C.F. .

P.I. =

=

=

=

=

.

Q.27 Show that set of functions forms a basis of the differential equation . Obtain a particular solution when . (6)

Ans:

Hence y₁(x) and y₂(x) are solutions of the given equation.

The Wronskian is given by

Thus the set{y₁(x), y₂(x)} forms a basis of the equation.

The general solution is

Q.28 Solve the following differential equations: (25 = 10)

(i)

(ii)

Ans:

(i) ,

M = ,

.

.

Thus, the equation is not exact and M, N are homogenous function of degree 2,

Thus equation is

Now M₁ = .

(ii)

Integrating, we get

Q.29 Let be a linear transformation defined by . Taking as a basis in determine the matrix of linear transformation. (8)

Ans:

The given matrix which maps the elements in R³ into R² is a 2 x 3 matrix.

Solving these equations, we obtain the matrix

Q.30 If then show that , for . Hence find . (8)

Ans:

The characteristic equation of A is given by

Using Cayley Hamilton theorem, we get

Adding we get

.

Q.31 Examine whether matrix A is similar to matrix B, where , . (8)

Ans:

The given matrices are similar if there exists an inevitable matrix P such that

Solving, we get a = 1, b = 1, c = 1, d = 2

Thus which is a non-singular matrix. Hence the matrices A and B

are similar.

Q.32 Discuss the consistency of the following system of equations for various values of

and if consistent, solve it. (8)

Ans:

The augmented matrix is

rank (A:B) = 3 and rank (A) = 2, hence the system is inconsistent.

When l = 7, rank (A:B) = 2 = rank (A), hence consistent.

Q.33 Show that for the function f(x, y) = , partial derivatives and both exist at the origin and have value 0. Also show that these two partial derivatives are continuous except at the origin. (8)

Ans:

Now at (0, 0),

If the function is differentiable at (0, 0), then by definition , where φ and are functions of k and h and tend to zero as (h, k) → (0, 0). Putting h = r cosθ, k = r sinθ and dividing by r, we get . Now for arbitrary θ, r → 0, implies that (h, k) → (0, 0). Taking the limit as r → 0, we get , which is impossible for all arbitrary θ. Hence the function is not differentiable at (0, 0) and consequently the partial derivatives cannot be continuous at (0, 0). For (x, y) ≠ (0, 0).

=

Now as h → 0, we can take x + h > 0, i.e. = x + h, when x > 0 and x + h < 0 or .

Similarly,

which is not continuous at the origin.

Q.34 In a plane triangle ABC, if the sides a, b be kept constant, show that the variations of its angles are given by the relation

(8)

Ans:

By the sine formula we have

Taking differentials on both sides, we get a cos B dB = b cos A dA.

Also, by the projection rule in triangle ABC we have a cos B + b cos A = c, and

A + B + C = p, we have dA + dB + dC = 0 or dA + dB = - dC. Therefore, equation (1), becomes

Q.35 Find the shortest distance from (0, 0) to hyperbola in XY plane. (8)

Ans:

We have to find the minimum value of x² + y² (the square of the distance from the origin to any point in the xy plane) subject to the constraint x² + 8xy + 7y² = 225.

Consider the function F(x, y) = x² + y² + l (x² + 8xy + 7y² – 225), where x, y are independent variables and l is a constant.

Thus dF = (2x + 2xl + 8yl) dx + (2y + 8xl + 14yl) dy

Therefore, (1+l) x + 4ly = 0 and, 4lx + (1 + 7l) y = 0.

Thus l = 1, -1/9. For, l = 1, x = -2y, and substitution in x² + 8xy + 7y² = 225, gives y² = - 45, for which no real solution exists.

For l = -1/9, y = 2x, and substitution in x² + 8xy + 7y² = 225, gives x² = 5, y² = 20 and so x² + y² = 25.

and cannot vanish because (dx, dy) ¹ (0,0). Hence the function x² + y² has a minimum value 25.

Q.36 Express , as a single integral and then evaluate it. (8)

Ans:

Let and

Let R₁ and R₂ be the regions over which I₁ and I₂ are being integrated respectively and are depicted by the shaded portion in fig. I.

As it is clear from Fig.II, R = R₁+ R₂

_{Fig. I Fig. II}

Thus, I = I₁ + I₂ = . For evaluating I we change the order of integration hence the elementary strip has to be taken parallel to x-axis from y = x to

dy

Q.37 Obtain the volume bounded by the surface and a quadrant of the elliptic cylinder , z > 0 and where a, b > 0 (8)

Ans:

In solid geometry represents a cylinder whose axis is along z-axis and guiding curve ellipse. Required volume is given by

Let us use elliptic polar co-ordinates x = a r cosθ, y = b r sinθ, where 0 ≤ r ≤ 1,

dx dy = ab r dr dθ, and , hence

.

Q.38 Solve the following differential equations: (8)

(i)

(ii)

Ans:

(i)The given equation can be written as

It is linear in y, and here P = - cos x and Q = sin x cos x.

Then I.F. = . Hence the solution of (1) is

Y(I.F.) = or

Now, putting –sin x = t, we get

Therefore, the solution is y = - (1 + sin x) +

=> y + 1 + sinx = ce^sinx

This is the required solution where c is an arbitrary constant.

(ii)

The given equation is of the form M dx + N dy = 0 where

Multiplying the given equation throughout by cos y, we get

which is exact.

Therefore, the solution is

This is the required solution where c is an arbitrary constant.

Q.39 Solve the following differential equation by the method of variation of parameters.

(9)

Ans:

First of all we find the solution of the equation .

This is a homogeneous equation. Putting x = e^z and , the equation reduces to

[D(D-1)+D-1] y = 0 , which gives D = 1, -1.

Therefore, ,

Therefore, ,

Writing the given equation in standard form, we get

(1)

Let y = Ax+ B/x be a solution of equation (1), where A, B are functions of x. Then

, Choose A and B such that (2)

Therefore, ,

Substituting the values of y, in equation(1), we get

(3)

Solving equations (2) and (3), we get

.

On integrating, we get

.

Thus, the complete solution is y = Ax + B/x

i.e.

Q.40 Solve (7)

Ans:

The auxiliary equation is m² – 4m + 1 = 0 which gives m = 2 ± Ö3.

Therefore, C.F. = y = .

Further P.I. =

Hence, the general solution of the given equation is

y = .

Q.41 Show that non-trivial solutions of the boundary value problem

, are y(x) = , where D_n are constants. (9)

Ans:

Assume the solution to be of the form y = e^mx . The characteristic equation is given as m⁴ – ω⁴ = 0 or .

The general solution is given by

y(x) =

Substituting the initial conditions, we get y(0) = A + C = 0.

;

.

Solving, the two equations, we get A = 0, C = 0. We also have

Adding, we obtain 2B sinh ωl = 0 or B = 0. Therefore, we obtain

D sin ωl = 0. Since we require non-trivial solutions, we have D ≠0.

Hence, sin ωl = 0 = sin nπ, n = 1,2,3,…..

Therefore, .

The solution of the boundary value problem is

By superposition principle, the solution can be written as

.

Q.42 Show that the matrices A and A^T have the same eigenvalues. Further if λ, μ are two distinct eigenvalues, then show that the eigenvector corresponding to λ for A is orthogonal to eigenvector corresponding to μ for A^T. (7)

Ans:

We have ,

Since A and A^T have the same characteristic equation, they have the same eigenvalues.

Let λ and µ be two distinct eigenvalues of A. Let x be the eigenvector corresponding to the eigenvalue λ for A and y be the eigenvector corresponding to the eigenvalue µ for A^T. We have Ax = λx. Premultiplying by y^T , we get

= (1)

and

Post multiplying by x, we get (2)

Subtracting equation (1) and (2), we obtain

Since λ ≠ μ, we obtain .

Therefore, the vectors x and y are mutually orthogonal.

Q.43 Let T be a linear transformation defined by

Find (7)

Ans:

The matrices , , , are linearly independent and hence form a basis in the space of 2X2 matrices. We write for any scalars not all zero

= .

Comparing the elements and solving the resulting system of equations, we get , , , . Since T is a linear transformation, we get

Q.44 Find the eigen values and eigenvectors of the matrix . (9)

Ans:

The characteristic equation of A is

Thus, the eigen values of A are ₁=0, ₂ = 3, ₃ = 15. The eigenvector X₁ of A corresponding to ₁=0, is the solution of the system of equations

Eliminating from the last two equations gives .

Setting

Therefore, the eigenvector of A corresponding to = 0, is

Similarly for = 3, we get X₂ is

, we get X₃ as .

Further, since A is symmetric matrix, the eigen vectors X₁, X₂, X₃ should be mutually orthogonal. Let us verify that

X₁ * X₂ = (1)(2) + (2)(1) + (2) (-2) = 0.

X₂ * X₃ = (2)(2) + (1)(-2) + (-2) (1) = 0.

X₃ * X₁ = (2)(1) + (-2)(2) + (1) (2) = 0.

Q.45 Solve the following system of equations:

(6)

Ans:

The given system in the matrix equation form is AX = B; where

~

(on operating R₂ → R₂ - 3 R₁ , R₃ → R₃ - 2 R₁ and R₄ → R₄ - R₁ )

~ (on operating R₂ → R₂ - R₃ and R₃ → R₃ – 2 R₄ )

~ (on operating R₂ → - R₂ and R₄ → R₄ + 3 R₂ )

~ (on operating R₄ → R₄ - 2 R₃ )

Therefore, rank (A : B) = rank A = 3 = number of unknowns, hence unique solution. To obtain this unique solution, we have

(A : B) ~ (on operating R₁ → R₁ + R₃ and R₁ → R₁ - 2 R₂ )

Therefore, the unique solution is x = -1, y = 4, z = 4.

Q.46 Find the series solution about the origin of the differential equation

. (10)

Ans:

We find that x = 0 is a regular singular point of the equation. Therefore, Frobenius series solution can be obtained.

Let y(x) = be solution about x = 0

Then given differential equation becomes

.

The lowest degree term is the term containing x^r . Equating coefficient of x^r to zero, we get

The indicial roots are r = -2, -3. Setting the Coefficients of x^r+1 to zero, we get

. For r= -2, a₁ is zero and for r = -3, a₁ is arbitrary. Therefore, the indicial root r = -3, gives the complete solution as the corresponding solution contains two arbitrary constants. The remaining terms are

.

Setting n-2 = t in the first sum and changing the dummy variable t to n, we get

.

Setting the coefficient of x^n+r+2 to zero, we get

, n ≥ 0.

We have , ,……

For r = -3, we get ,……

The solution is given by

=

For r = -2, we get

Therefore, the solution is

We find that the indicial root r = -2, product a linearly dependent solution.

Q.47 Express f(x) = in terms of Legendre polynomials. (8)

Ans:

We know that, various powers of x in terms of Legendre polynomials can be written as

=

Therefore,

.

Q.48 Evaluate , where denotes Bessel function of order n. (8)

Ans:

Using the recurrence relation,

Using the recurrence relation,

Q.49 Find the stationary value of subject to the fulfilment of the condition , given a, b, c are not zero. (7)

Ans:

Let u = a³x² + b³y² + c³z²

Let where l is constant using Lagrange’s multiplier method.

For stationary values,

=> ax = by = cz = k

Then, gives a + b + c = k

Therefore,

Stationary value of u is

.

Q.50 Find the volume enclosed by coordinate planes and portion of the plane lying in the first quadrant. (7)

Ans:

The region of integration R is bounded by x = 0, y = 0, and lx + my = 1

{projection of lx + my + nz = 1 on z = 0}

.

Q.51 If the directional derivative of at (1, 1,1) has maximum magnitude 15 in the direction parallel to line find the value of a, b, c. (7)

Ans:

This is along normal to the surface and is the maximum directional derivative. Thus is || to line .

Therefore,

Therefore, a = 4l, b= -11l, c = 10l and

Therefore, .

Q.52 Verify divergence theorem for the vector field taken over the region bounded by cylinder . (7)

Ans:

By Divergence theorem,

Now,

Let x = 2 sin then dx = 2cos, for x = 0, = 0 and for x = 2, =

Now Surface S consists of three surfaces, the one leaving base S₁ (z = 0), second leaving top S₂ (z = 3), third the curved surface S₃ of cylinder x² + y² = 4 between z = 0, z = 3

On S₃ the outer normal is in the direction of . Therefore a unit vector along normal to the curved surface is given by

, thus

Hence divergence theorem proved.

Q.53 Show that and hence deduce that . (7)

Ans:

We know that

Putting, , Then

Comparing real and imaginary part

Let

.

Therefore, . Let b > 0, then

Hence,

Putting b = 0, we get

Q.54 Solve in the interval subject to the boundary conditions :

(i) (ii)

(iii) (iv) . (7)

Ans:

Let U = X(x) Y(y), then equation becomes

.

Let , then

Therefore, X = A cos lx + B sin lx

Y = Ce^ly + De^-ly

U (x, y) = (A cos lx + B sin lx) (Ce^ly + De^-ly)

(iv) condition gives U (x, y) → 0 as y→ µ => C = 0

\ U = (A cos lx + B sin lx) e^-ly

(i) gives U (0, y) = 0 => A = 0

Hence, U = B sin lx . e^-ly

(iii) gives U (x, 0) = 1 => 1 = b_l sin x

Q.55 Use Cayley - Hamilton theorem to express in terms of A and the identity matrix I, where . (7)

Ans:

The characteristic equation of A is |A-lI| = 0

=> l² – 4l – 5 = 0

=> A² – 4A – 5I = 0, by Cayley Hamilton Theorem.

Thus A⁵ – 4A⁴ – 7A³ + 11A² – A – 10I

= (A² – 4A – 5I) (A³ – 2A + 3I) + A + 5I = A + 5I.

Q.56 Solve . (7)

Ans:

(D² + 5D + 6) y = e^-2x sec²x (1 + 2 tan x)

A.E. : m² + 5m + 6 = 0

m = -2 , -3

C.F. = c₁e^-2x + c₂e^-3x

Thus y = c₁e^-2x + c₂ e^-3x + e^-2x tan x.

Q.57 Find analytic function whose real part is . (7)

Ans:

Let and f(z) = u + iv

since u is an analytic function, thus it must

satisfies C-R equations, thus

Using Milne’s Thomson method, Let x = z , y = 0

, where c is a constant of integration.

Q.58 Evaluate , using contour integration. (7)

Ans:

Consider the function . It has simple pole at z = 0. where C consists of the part of real axis from –R to -r and from r to R, the small semi circle C_r from –r to r with center at origin and radius r, which is small and large semi-circle C_R from R to –R as shown in Fig. f (z) is analytic inside C (z = 0, the only singularity has been deleted by indenting the origin by drawing C_r).

Therefore, by Cauchy’s Theorem,

For C_R, we have z = Re^iq , 0 £ q £ p and C_r , z = re^iq , 0 £ q £ p

Since decreases from 1 to as q increases from 0 to

Putting values in (1) and applying limits r→0, R→¥, we get

Q.59 In a normal distribution 31% of the items are under 45 and 8% are over 64. Find the mean and standard deviation of the distribution.

[Given that ,

where is pdf of standard normal distribution.] (7)

Ans:

Since 31% of items are under 45. Hence 19% of items lies between and 45. Since ,

thus, .Similarly

Solving, we get s = 10, .

Q.60 A can hit a target 3 times in 5 shots, B 2 times in 5 shots and C 3 times in 4 shots. All of them fire one shot each simultaneously at the target. What is the probability that (i) two shots hit (ii) atleast two shots hit? (7)

Ans:

p(A) = Probability of hitting target by A = 3/5

p(B) = Probability of hitting target by B = 2/5

p(C) = Probability of hitting target by C = 3/4

(i) p₁ = Chance A, B hit & C fails

=

p₂ = Chance B, C hit & A fails = 12/100

p₃ = Chance C, A hit & B fails = 27/100

Since all these events are mutually exclusive, therefore,

P(two shots hit the target) = p₁ + p₂ + p₃ = 0.45

(ii) In case atleast two shots may hit target, we must also consider case when all hit the target.

p₄ = Probability A, B, C hit target = 18/100.

Therefore, P(atleast two shot hit the target) = p₁ + p₂ + p₃ + p₄ = 63/100 = 0.63.

Q.61 Diagonalize the matrix . (7)

Ans:

The characteristic equation is |A – lI| = 0

i.e. l³ - 7l² + 36 = 0

=> l = -2 , 3, 6

These are eigen values of given matrix A. For eigen vectors we find X ≠ 0, such that (A - lI) X = 0

For l = -2

3x + y + 3z = 0

x + 7y + z = 0

Therefore, for l = -2, eigen vector is (-1, 0, 1)’

Similarly for l = 3, eigen vector is (1, -1, 1)’

for l = 6, eigen vector is (1, 2, 1)’

The modal matrix P is given by

The diagonal matrix D is given by

Q.62 Investigate the values of and so that equations

have (i) no solution (ii) a unique solution (iii) infinite number of solutions. (7)

Ans:

We have

i.e. AX = B

(i) (A : B) =

R₃→R₃ – R₁, R₂→7R₁ – 2R₂

If l = 5, system will have no solution for those values of m, for which rank A ¹ rank (A : B). If l = 5, m ¹ 9, then rank (A) = 2 and rank (A : B) = 3. Hence no solution

(ii) The system admits unique solution iff coefficient matrix is of rank 3

\

Thus for unique solution l ¹ 5 and m may have any value.

(iii) If l = 5,m = 9, system of equation have infinitely many solution

Q.63 The height h and semi vertical angle of a cone are measured and the total area A of surface of cone including that of base is calculated in terms of h, . If h and are in error by small quantities and respectively, find the corresponding error in the area. Show further that if an error of +1% in h will be approximately compensated by an error of –0.33 degrees in . (7)

Ans:

Let r be base radius and l be slant height of cone.

Total area A = area of base + area of curved surface

= pr² + prl = pr (r + l) = ph² tan a (tan a + sec a)

dA = 2ph (tan² a + tana seca) dh + ph² (2tana sec²a + sec³a + tan²a seca) da

The error in h will be compensated by error in a, when

dA = 0

radians = -0.33^o .

Q.64 If what values of n will make (7)

Ans:

- (A)

Differentiating (A) partially w.r.t. t, we get

Differentiating (A) partially w.r.t. r, we get

equating (1) and (2), we get n = -3/2.

Q.65 A vector field is given by . Show that the field is irrotational and find its scalar potential. Hence evaluate line integralfrom (1, 2) to (2, 1). (7)

Ans:

Curl

=

\ vector field is irrotational then $ a scalar function j s.t.

Integrating, we get

Because, field is irrotational

Q.66 Solve . (7)

Ans:

Therefore, solution is

=> t = 1 + cx

=> (log z)^-1 = 1 + cx or .

Q.67 Express in terms of and . (7)

Ans:

We know

For n = 1, 2 , 3

Q.68 Find Taylor’s expansion of about the point z = i. (7)

Ans:

We have to expand about z = i

Q.69 Examine the following system of equations for consistency :

Reduce the augmented matrix of the above system of equations to Echelon form and find the solution of the above system, if it exists. (7)

Ans:

The system of equations can be written as AX = B

where

Augmented matrix = [A : B]

[A : B] =

[A : B]

[A : B]

This is row Echelon Form of A. Since the number of non-zero rows in the row-echelon form is 3. So,

Hence system of equations has unique solution, and is given by solving

Q.70 Find the eigen values and the corresponding eigen vectors of the matrix A defined by

Obtain the modal matrix and reduce the given matrix to the diagonal matrix. (7)

Ans:

Characteristic equation is | A – lI | = 0

Eigen values of matrix A are 1, 1, 5

For l = 5, eigen vector is obtained by solving the system of equations

i.e. –3x + 2y + z = 0

x – 2y + z = 0

x + 2y –3z = 0

Solving, we get

i.e. eigen vector is (1, 1, 1)’

For l = 1, eigen vector is obtained by solving the system of equations

i.e. x + 2y + z = 0

There are two linearly independent eigen vectors for l = 1 . These are obtained by putting x=0 and y =0 respectively in the equation.

For x = 0, 2y + z = 0

Eigen vector is (0, -1, 2)’

For y = 0, Eigen vector is (-1, 0, 1)’

\ Modal Matrix = P =

\ Diagonal Matrix = D = P^-1AP

Q.71 If and is an analytic function of , find f(z) subject to the condition that at . (7)

Ans:

Then

By C.R. equations,

Subtracting (2) from (1)

Adding (1) and (2)

Using Milne’s Thomson method, putting x = z, y = 0 in (3) and (4), we get

Q.72 Prove that the relation transforms the real axis in the z-plane into a circle in the w-plane. Find the centre and the radius of the circle and the point in the z-plane which is mapped on the centre of the circle. (7)

Ans:

Since,

Thus image of real axis in the z plane (means y = 0) is given by

4(u²+v²) + 7v – 2 = 0

i.e.

which is an equation of circle with centre at and radius

For u = 0, v = , we get x = 0, y = . Thus centre of circle in w plane is image of point in z plane.

Q.73 Solve in series the differential equation :

(8)

Ans:

x = 0 is a regular singular point.

Let

Substituting in the given differential equation

The lowest power of x is x^r-1. Its coefficient equated to zero gives C₀r(r+3) = 0. Because C₀ ¹ 0

=> r = 0, r = –3

The coefficient of x^m+r is equated to zero gives

When r = 0

, for r = 0. Thus one solution is

When r = -3, C₁ = -5C₀ , C₂ = 10C₀ , C₃ = -10C₀ , C₄ = 5C₀ ------Thus is another solution.

is general solution.

Q.74 Prove that

where c is an arbitrary constant. (6)

Ans:

Q.75 Show that the vector field represented by

is irrotational but not solenoidal. Also obtain a scalar function such that grad =. (5)

Ans:

except for x = -2.

Therefore, given vector field is not solenoidal.

= 0

\ Given vector field is irrotational. Thus it can be expressed as , where f is scalar function.

,,

Integrating w.r.t. x, y, z we get

Since these three must be equal

+c

Q.76 If and , show that . Also show that is a possible solution of where A and B are arbitrary constants. (5)

Ans:

We know that

Q.77 Evaluate where and s is the surface of the cylinder included in the first octant between z = 0 and z = 5. (4)

Ans:

Q.78 If , then show that . (7)

Ans:

Given

Adding & subtracting respectively, we get

Let

Similarly

Similarly

Adding, we get

Q.79 The Luminosity L of a star is connected with its mass M by the relation where and a, b being given constants. If p is the percentage error made in the estimate of M, express the resulting percentage error in the calculated luminosity in terms of p and and show that it lies between p and 3p. (7)

Ans:

L = aM (1 – b) , 1 – b = bb⁴M²

Eliminating from (1) and (2), we get

is the required expression of percent change in L. Since 0 < β < 1, we have

Q.80 Solve the following differential equations :

(i) . (5)

(ii) . (5)

Ans:

(i)

Let

I.F.

(ii)

Integrating,

Q.81 Change the order of integration in the following integral :

(4)

Ans:

This integration is first w.r.t. y and then w.r.t. x. On changing the order of integration, we first integrate w.r.t. x and then w.r.t. y. This divided into three regions.

Region I: The strip extends from parabola y² = 2ax i.e. to x = 2a.

Then y = a to y = 2a

Region II: The strip extends from to circle .

Then y = 0 to a

Region III: The strip extends from circle, to x = 2a

Then y = 0 to y = a

Q.82 A string is stretched and fastened to two points at distance l apart. Motion is ensued by displacing the string into the form from which it is released at time t = 0. Find the displacement at any point x and any time t. (5)

Ans:

The vibration of the string is given by

- (i)

As the end points of the strings are fixed, for all time

y (0, t) = 0 = y (L, t) - (ii)

Since initial transverse velocity at any point of the string is zero.

- (iii)

Also, - (iv)

Using method of separation of variables and since the vibration of the string is periodic, therefore, the solution of (i) is of the form

- (v)

By (ii),

This should be true if c₁ = 0

Hence - (vi)

By (iii)

If c₂ = 0, then (vi) will be y (x, t) = 0

\ C₄ = 0

Thus (vi) becomes

Hence (vi) reduces to

From (iv)

\ Solution is

Q.83 The ends A and B of an insulated rod of length , have their temperatures at and until steady state conditions prevail. The temperatures of these ends are changed suddenly to and respectively. Find the temperature distribution in the rod at any time t. (9)

Ans:

Let the equation for conduction of heat be

Prior to temperature change at end B, when t = 0, the heat flow was independent of time (steady state condition), when u depends only on x i.e.

Since u = 20 for x = 0 and u = 80 for x = L

\ b = 20 , a =

Thus initial condition is expressed as

The boundary conditions are

The temperature function u (x, t) can be written as

where u_s(x) is a solution of (i) involving x only and satisfying boundary conditions (iii), which is steady state solution, u_t (x, t) is a transient part of the solution which decreases with increase of t.

Since, u_s(0) = 40 , u_s(L) = 60 \ using (ii), we get

From (iv), we get

General solution of (i) is given as

where

Q.84 Represent the function in Laurent’s series

(i) within

(ii) in the annular region within and

(iii) Exterior to . (8)

Ans:

Q.85 Apply the calculus of residue to evaluate . (6)

Ans:

Let

The pole of ¦(z) enclosed by contour C, which consists of semicircle C_R and real axis segment from –R to R taken in counter clockwise directions, are z = i, z = 3i each of order 1.

Residue at

Residue at

Let R → ¥

, since second integral on the left hand side tends to 0.

Q.86 Show that the stationary values of , where and are the roots of the equation . (7)

Ans:

Let

Let Using Lagrange’s method of multiplier, for extreme values, , f = g = 0.

Multiplying (1), (2), (3) by x, y, z respectively and adding, we get

Substituting in (4), we get

is satisfied by stationary points.

Q.87 Expand in powers of and upto 3^rd degree terms. (7)

Ans:

Q.88 A man takes a step forward with probability 0.4 and backward with probability 0.6. Find the probability that at the end of 11 steps, he is just one step away from the starting point. (7)

Ans:

Let us call a forward step “a success” and a backward step “a failure”.

Let X = no. of forward steps. Then X has binomial distribution with n = 11 and probability of success p = 0.4.

Required probability =P(X=6)+P(X=5)

Q.89 In a certain factory turning out razor blades, there is a small chance of for any blade to be defective. The blades are supplied in packets of 10. Using Poisson’s distribution calculate the approximate number of packets containing

(i) no defective blade (ii) one defective blade (iii) two defectives blades

in a consignment of 10,000 packets . (7)

Ans:

, n = 10

\ m = np = 0.02, e^-0.2 = 0.9802

(i) Probability of no defective blade = P (X=0)

= e^–m = 0.9802 (approximately)

\ Mean number of packets containing no defective blade is

= 10000 x 0.9802 = 9802

(ii) The mean number of packets containing one defective blade

= 10000 x me^–m = 196

(iii) The mean number of packets containing two defective blades

= 10000 x

= 2

Q.90 Show that at the point on the surface where x = y = z, we have . (7)

Ans:

It is given that x^xy^yz^z=c, taking log we get

x log x+ y log y +z log z=log c

Differentiating the above equation w.r.t. y respectively, we get

, now differentiating w.r.t. x we get

At the point x=y=z, we get

Q.91 Find the volume of greatest rectangular parallelopiped that can be inscribed inside the ellipsoid . (7)

Ans:

Let edges of parallelopipid be 2x, 2y, 2z parallel to the coordinate axes. The volume V is given by V = 8xyz.

Let

Using Lagrange’s multiplier method, for maxima and minima, let

F = V + λ, where λ is a constant. For stationary values,

, = 0

Q.92 Change the order of integration and hence evaluate . (7)

Ans:

On changing order of integration, the elementary strip has to be taken parallel to x

axis for which the region of integration has to be divided into two regions R₁and R₂. The region R₁: 0 ≤ x ≤ y, 0 ≤ y ≤ 1 and the region R₂: 0 ≤ x ≤ 2-y, 1 ≤ y ≤ 2.

= 2log2-1.

Q.93 Find the volume common to cylinders and . (7)

Ans:

The volume V is given as

Q.94 Solve the differential equations

(i) . (8)

(ii) . (6)

Ans:

(i)The auxiliary equation is m² - 4m + 4 = 0, which gives m = - 2, -2.

Thus C.F. is given as (C₁ + C₂ x) e^-2x.

, since 2 is a root of order 2.

Therefore, general solution is

.

(ii)Let Y = y + k, X = x + h,

where h, k are so chosen so that

-7h+3k+7=0

-3h+7k+3=0

Solving, we get h = 1, k = 0. Let Y = VX,

Integrating, we get

Q.95 Discuss the consistency of the following system of equations for various values of

(7)

and, if consistent solve it.

Ans:

The augmented matrix is given as

(A : B) =

Applying R₃ àR₃ – 2R₁,

If l ¹ 7, rank (A:B) = 3 ¹ rank A = 2. Hence system is inconsistent.

If l = 7, rank (A:B) = 2 = rank A . Hence system is consistent. The solution is obtained as follows :

Set x₃ = t₁, x₄ = t₂, t₁, t₂, arbitrary, then

x₂ = 4t₁- t₂+ 1, x₁ = 3t₁+ t₂+ 3.

Q.96 Find the characteristic equation of the matrix and hence, find the matrix represented by (7)

Ans:

The characteristic equation of A is |A-lI| = 0

ð l³ - 5l² + 7l – 3 = 0, is the characteristic equation.

By Cayley Hamilton theorem, A must satisfy this equation i.e.

A³ – 5A² + 7A – 3I = 0

Thus A⁸ – 5A⁷ +7A⁶ -3A⁵ +A⁴ – 5A³ + 8A² – 2A +I

= (A⁵ +A) (A³ –5 A² + 7A - 3I) + A² +A + I = A² + A + I

=

Q.97 Show that the vector field is irrotational as well as solenoidal. Find the scalar potential. (6)

Ans:

vector is solenoidal.

Therefore is irrotational. Thus , where is a scalar function.

Q.98 Evaluate where and S is the region of the plane 2x + 2y + z = 6 in the first octant. (8)

Ans:

Unit normal to the plane 2x + 2y + z = 6 is along the vector , is given as .

Thus projection of given plane z=6-2x-2y on z=0 is region bounded by x=0, y=0,

y=3-x.

Q.99 The odds that a Ph.D. thesis will be favourably reviewed by three independent examiners are 5 to 2, 4 to 3 and 3 to 4. What is the probability that a majority approve the thesis? (7)

Ans:

Let p₁ , p₂ , p₃ be the probabilities that thesis is approved by examiner A,B,C

p₁ = 5/7, p₂ = 4/7, p₃ = 3/7. A majority approves thesis if atleast two examiners are favorable.

P = p₁p₂q₃ + p₁p₃q₂ + p₂p₃q₁+ p₁p₂p₃

= =209/343.

Q.100 If the probabilities of committing an error of magnitude x is given by compute the probable error from the following data :

. (7)

Ans:

=0.0126,

We know that probable error is given as (2/3)=0.0084.

Q.101 Solve by method of separation of variables . (5)

Ans:

Let Z=X(x)Y(y), then given differential equation becomes

X’’Y-2X’Y+XY’=0, where X’’,Y’, X’,Y’’ are first and second order derivatives.

Q.102 Solve for conduction of heat along a rod without radiation subject to

(i) u is not infinite for (ii) for x = 0, x =

(iii) for t = 0 between x = 0 and x = . (9)

Ans:

Let U=X(x)T(t), then given differential equation becomes

. Condition (i) is satisfied.

If k² =0, X=ax+b, T=c, thus by condition (ii), we get a=0

Thus U = bc.

Now U= (A cos kx + B sin kx) C

By (ii) condition, we get B=0, kl=nπ. Thus

C is solution for all n.

Since x-x² =

Q.103 Obtain the series solution of equation . (8)

Ans:

Since x = 0 is a regular singular point

Let

Now the given differential equation becomes

The terms with lowest power of x is x^r-1. Its coefficient equated to zero gives C₀r(r-2) = 0. Because C₀ ¹ 0

=> r = 0, r = 2

The coefficient of x^m+r is equated to zero gives

,………

Q.104 Express in terms of and . (6)

Ans:

We know

For n = 1, 2 , 3

Q.105 Express as = where . (7)

Ans:

If f(x) =∑c_nP_n(x), then

where we have used the fact that

Q.106 Find analytic function whose real part is . (7)

Ans:

Let , and f(z) = u + iv

since u is an analytic function, thus it must satisfies

C-R equations, thus

Using Milne’s Thomson method, Let x = z , y = 0

\ where c is an arbitrary constant.

Q.107 Show that under the transformation , real axis in the z-plane is mapped into the circle Which portion of the z plane corresponds to the interior of the circle? (7)

Ans:

For the real axis in the z plane y=0, i.e.=1, Also <1 implies y>0. hence the result.

Q.108 Let where C is ellipse . Find value of F(3.5) and . (7)

Ans:

, since 3.5 is a point which lies outside C, thus F(3.5) = 0 by Cauchy theorem.

Also -1 lies within C, by Cauchy Integral Formula

.

Q.109 Compute and for the function

(6)

Ans:

Q.110 Let v be a function of (x, y) and x, y are functions of defined by

where Show that . (8)

Ans:

Because x + y = 2e^qcosj, x – y = 2ie^qsinj

Adding & Subtracting, we get

2x = 2e^q (cosj + isinj) = 2e^{q + ij}

=> x = e^{q + ij}

Similarly y = e^{q - ij}

Let v = f (x, y), x = g (q, j), y = h (q, j)

Q.111 Expand near (1, 1) upto 3^rd degree terms by Taylor’s series. (7)

Ans:

f(x, y) = x^y , f(1, 1) = 1

f_x (x, y) = yx^y-1 , f_x(1, 1) = 1

f_y (x, y) = x^ylog x , f_y(1, 1) = 0

(x, y) = y (y-1) x^y-2 , (1, 1) = 0

(x, y) = x^y (log x)² , (1, 1) = 0

f_xy (x, y) = x^y-1 + yx^y-1 log x , f_xy (1, 1) = 1

(x, y) = y (y-1) (y-2) x^y-3 , (1, 1) = 0

(x, y) = (2y-1)x^y-2 + y (y-1) x^y-2 log x , (1, 1) = 1

(x, y) = yx^y-1 (log x)² + 2 log x x^y-1 , (1, 1) = 0

By Taylor’s Theorem

Q.112 Find the extreme value of subject to the conditions and . (7)

Ans:

Let f = x² + y² + z² + xy + xz + zy

g = x + y + z – 1 = 0

h = x + 2y + 3z – 3 = 0

Let l₁, l₂ be two constants. Using Lagrange’s multiplier method, we get

F = f + l₁g + l₂h OR

F = x² + y² + z² + xy + xz + zy + l₁(x + y + z – 1) + l₂(x + 2y + 3z – 3)

For extreme values,

, x + y + z = 1, x + 2y + 3z = 3.

=> 2x + y + z + l₁ + l₂ = 0 => x + l₁ + l₂ + 1 = 0

2y + x + z + l₁ + 2l₂ = 0 y + l₁ + 2l₂ + 1 = 0 (A)

2z + x + y + l₁ + 3l₂ = 0 z + l₁ + 3l₂ + 1 = 0

Adding (A) and using x + y + z = 1, we get

3l₁ + 6l₂ + 4 = 0

Multiplying equation (ii) of ‘A’ by 2 and (iii) by 3 and adding all and using

x +2 y +3 z = 1, we get 6l₁ + 14l₂ + 9 = 0

Solving, 3l₁ + 6l₂ + 4 = 0

6l₁ + 14l₂ + 9 = 0, we get

l₁ = –1 /3, l₂ = –1 /2

From (A), we get

x = –1/6, y = 1/3, z = 5/6

Therefore, (–1/6, 1/3, 5/6) is a point of extremum, with extreme value

F(–1/6, 1/3, 5/6)= (-1/6)² + (1/3)² + (5/6)² – 1/6*1/3 – 1/6 * 5/6 + 1/3 * 5/6 =11/12

Q.113 Find the rank of the matrix

(6)

Ans:

Applying R₁ ↔ R₃, R₂ ↔R₄

R₃ R₄

R₃ àR₃ – 3R₁, R₄ à R₄ – 9R₁

R₄ àR₄ – R₂, R₃ à R₂ – 2R₃

R₄ à 9R₃ + 5R₄

Thus |A| 0 Hence, rank of A = 4.

Q.114 Let

be a linear transformation from to and

be a linear transformation from to .

Find the linear transformation from to by inverting appropriate matrix and matrix multiplication. (8)

Ans:

Let

y₁ = 1.25 z₁ + 3 z₂ – 2.3 z₃

y₂ = 0.75 z₁ + 2 z₂ – 2.3 z₃

y₃ = 0.5 z₁ - z₂ + z₃

Q.115 Prove that the eigenvalues of a real matrix are real or complex conjugates in pairs and further if the matrix is orthogonal, then eigenvalues have absolute value 1. (6)

Ans:

Let A be a square matrix of order n.

Then |A – lI | = (-1)ⁿlⁿ + k₁l^n-2 + ----- + k_n = 0

where k’s are expressible in terms of elements a_ij of matrix A. The roots of this equation are eigen values of matrix A. Since this is n^th polynomial in l which has n distinct roots which are either real or complex conjugates.

Hence, eigen values of matrix are either real or complex conjugates.

If l is an eigen value of orthogonal matrix then 1/l is an eigen value of A^-1. Because A is an orthogonal matrix. Therefore A^-1 is same as A’.

Therefore 1/l is eigen value of A’. But A and A’ have same eigen values.

Hence, 1/l is also an eigen value of A. The product of eigen value of orthogonal matrix = 1 and hence if the order of A is odd it must have 1 as eigen value. Since product of eigen value of matrix A is equal to its determinant. Therefore |A| = ±1.

Q.116 Find eigenvalues and eigenvectors of the matrix . (8)

Ans:

|A – lI | = 0

=> l³ + l² – 21l - 45 = 0

=> l = 5, -3, - 3

Eigen values are 5, -3, -3

For l = 5, eigen vectors are obtained from

=> -7x + 2y - 3z = 0

2x – 4y – 6z = 0

-x –2y – 5z = 0

Solving we get,

i.e. (1, 2, -1)’ is an eigen vector

For l = -3, eigen vectors are obtained from

i.e. x +2y – 3z = 0

There are two linearly independent eigen vectors for l = -3. These are obtained by putting x = 0 and y = 0 respectively in the equation.

Let x = 0 then 2y – 3z = 0

i.e. is an eigen vector

Let y = 0, then x – 3z = 0

is an eigen vector.

Eigen vectors corresponding to 5, -3 , -3 are

[1, 2, -1]’ , [0, 3, 2]’ and [3, 0, 1]’.

Q.117 Find a matrix X such that is a diagonal matrix, where . Hence compute . (8)

Ans:

A =

| A - lI | = 0

=> l² - 7l + 6 = 0

=> l = 1, 6

For l = 1,

Therefore, x + y = 0

Eigen vector is (1, -1)’

For l = 6,

Therefore, x - 4y = 0

Eigen vector is (4, 1)’

Therefore, modal matrix is X = and

X^-1 =

D = X^-1AX = which is diagonal matrix

Also A = XDX^-1

A⁵⁰ = XD⁵⁰X^-1

X^-1

Q.118 Prove that a general solution of the system

can be written as

++ where are arbitrary. (6)

Ans:

The system of equation can be written as AX = B

Rank(A) = Rank(A,B) = 3. Thus system is consistent. Homogeneous system AX=0, has 5 – 3 = 2 linearly independent solutions. Clearly (, -1, 3, 1, 0), (, 1, -2, 0, 1) are linearly independent and satisfy the homogeneous system AX = 0. Also (, 2, 0, 0, 0) is a particular solution of non-homogeneous system AX = B. Thus general solution of non homogeneous system is (x₁, x₂, x₃, x₄, x₅) = (, 2, 0, 0, 0) + a (, -1, 3, 1, 0) + b (, 1, -2, 0, 1), where a, b are arbitrary.

Q.119 Let Recognise the region R of integration on the r.h.s. and then evaluate the integral on the right in the order indicated. (7)

Ans:

For I₁, region of integration is bonded by the lines x = 1, x = 2, y=0 y = 1 i.e. region R₁ in figure. For I₂, region of integration is bonded by the lines x = y, x = 2, y = 1, y = 2 i.e. region R₂ in figure.

Now the region R of integration i.e. union of R₁ and R₂ is bonded by the lines y = 0, y = x, x = 1, x = 2

.

Q.120 Compute the volume of the solid bounded by the surfaces and . (7)

Ans:

Let V be the volume of solid. The two surfaces intersect at z = 1. Therefore

Let . Then dydx = rdrdθ, r varies from 0 to and θ varies from 0 to 2π. Then

Q.121 Let be an integrating factor for differential equation Mdx+Ndy=0 and is a solution of this equation, then show that is also an integrating factor of this equation, G being a non-zero differentiable function of . (6)

Ans:

Since m be an integrating factor for differential equation Mdx + Ndy = 0. Thus m(Mdx + Ndy) = 0 is an exact differential equation.

Also dj = m (Mdx + Ndy) (given)

Because, j = constant, is a solution.

Let G(j) be any function of j

Therefore G(j)dj = m G(j) (Mdx + Ndy).

Let

Since terms on left is an exact differential, the terms on right must be an exact differential. Therefore, m G(j) is an integrating factor of differential equation.

Q.122 Solve the initial value problem . (8)

Ans:

is the differential equation.

I.F. = e^x . Hence solution is

=> e^x (y ln x + 1) + cy = 0

Q.123 Find general solution of differential equation . (7)

Ans:

y” + y’ = sec x

can be written as (D² + D) y = sec x

i.e. D (D + 1) y = sec x

Therefore, auxiliary equation is m² + m = 0

m = 0 , -1

C.F. = C₁ + C₂e^-x

Therefore,

Q.124 Solve the boundary value problem

. (7)

Ans:

The given differential equation is x²y” – xy’ + y =

i.e. (q(q-1) - q + 1) y = ,

C.F. = x (C₁ + C₂ log x)

Therefore, y = x (C₁ + C₂ log x) +

Q.125 Solve the differential equation . (5)

Ans:

y^iv + 32y” + 256y = 0

i.e. (D⁴ + 32D² +256) y = 0

A.E. is m⁴ + 32m² + 256 = 0

i.e. (m² + 16)² = 0

=> m = ± 4i, ± 4i

Therefore, y = (C₁ + C₂ x) (C₃ cos 4x + C₄ sin 4x)

Q.126 Using power series method find a fifth degree polynomial approximation to the solution of initial value problem

. (9)

Ans:

Let x = 0 be an ordinary point

Let be solution about x = 0

Then given differential equation becomes

equating coefficient of xⁿ to zero, we get

Also y(0) = 2 => a₀ = 2

y'(0) = -1 => a₁ = -1

Therefore, a₀ = 2, a₁ = -1, a₂ = 1, a₃ = 0, a₄ = ¼ , a₅ = 3/20 , _______

+…..

Q.127 Let denote the Bessel’s function of first kind. Find the generating function of the sequence . Hence prove that

(7)

Ans:

J_n(x) is the coefficient of zⁿ in expansion of.

Coefficient of zⁿ , n ≥ 0

Similarly we can get the result for n < 0. Set z = i . Then

Comparing real and imaginary parts and by using we get

cos (x) = J₀ (x) - 2 J₂ (x) + 2 J₄ (x) +……….

sin (x) = 2[ J₁ (x) + J₃ (x)+ J₅ (x)+ ---------]

Q.128 Show that for Legendre polynomials

(7)

Ans:

We know that (2n-1) x P_n-1 = n P_n + (n-1) P_n-2

Multiplying by P_n(x) both sides and integrating, we get

Q.129 For the function show that

. (8)

Ans:

For obtainingandwe need and . For obtaining these derivatives we use the definition of and

,

Thus

Q.130 Find the absolute maximum and minimum values of the function

over the rectangle in the first quadrant bounded by the lines x = 2, y = 3 and the coordinate axes.

(8)

Ans

The function f can attain maximum/ minimum values at the critical points or on the boundary of the rectangle OABC, such that O (0,0), A(2,0), B(2,3), C (0,3). We have . The critical point is (x,y)=(1,2/3). Now, since and r >0. The point (1,2/3) is a point of relative minimum. The minimum value is f(1,2/3)=-4.On the boundary line OA, we have y = 0 and f(x,y) = f(x,0) = g(x) = , which is a function of one variable. Setting we get x = 1. Now, . Therefore, at x =1, the function has a minimum. The minimum value is g(1)=0. Also, at the corners (0,0), (2,0) we have f(0,0)=g(0)= 4, f(2,0)=4. On the boundary line AB, we have x = 2 and f(x,y) = h(y) = , which is a function of one variable. Setting we get y =2/3. Now, . Therefore, at y=2/3, the function has a minimum. The minimum value is f(2,2/3)=0. Also, at the corners (2,3) we have f(2,3)=49. On the boundary line BC, we have y = 3 and f(x,y) = g(x) = , which is a function of one variable. Setting we get x =1. Now, . Therefore, at x=1, the function has a minimum. The minimum value is f(1,3)=45. Also, at the corners (0,3) we have f(0,3)=49.On the boundary line CO, we have x = 0 and f(x,y) = h(y) = , which is the same case as for x=2.Therefore, the absolute minimum value is -4 at (1,2/3) and the absolute maximum value is 49 at (2,3) and (0,3).

Q.131 If , find an approximate value of f(1.1,0.8) using the Taylor’s series quadratic approximation. (8)

Ans:

Using the Taylor series quadratic approximation, one can write

------------------- (1)

Here h=0.1, k= -0.2 Thus

----- (2)

Now

, . Thus

= -(0.05); ;

. On using the values of ,

in Eqn (2), we get

.

Q.132 Evaluate the integral by changing to polar coordinates, where R is the region in the x-y plane bounded by the circles and =9. (8)

Ans:

Using x = r cosθ, y = r sinθ, we get dx dy = r dr dθ, and

Q.133 Find the solution of the differential equation (y-x+1)dy – (y+x+2) dx = 0. (6)

Ans:

When written in the form ; the differential equation (d.e.) belongs to the category of reducible homogeneous d.e. of first order and can be integrated by reducing to the homogeneous form. Before we indulge into this let us first examine the given differential equation by writing it in the form M(x, y)dx + N(x. y)dy = 0

Here M = -(y + x +2); N = y – x + 1

; , Since

Therefore the given equation is exact, consequently, we write it as follows

ydy – xdy + dy – ydx – xdx – 2dx = 0

or ydy + dy – (xdy + ydx) – xdx – 2dx = 0

Integrating

In fact, on observation for its exactness should have been made before classifying it to any other category. If one fails to make this observation then it can be reduced to homogenous form y making the transformation x = X + h, y = Y + k which yields

Choose h, k such that h + k + 2 = 0, k – h + 1 = 0. Thus, we get h = , k = and the d.e. reduces to YdY – XdY – YdX – XdX = 0

Integrating ; .

Q.134 Solve the differential equation . (6)

Ans:

On dividing throughout by cot3x, the given differential equation can be written as

-------- (1)

Eqn(2) is a linear differential equation of the form

; where ;

I.F. = .

Multiplying (1) throughout by cos3x and integrating, we get

.

Q.135 Show that the functions 1, sinx, cosx are linearly independent. (4)

Ans:

For functions 1, sin x, cos x to be linearly independent the Wronskian of the functions given by

has to be non-zero. Here it is (-1) 0. Hence the result.

Q.136 Using method of undetermined coefficients, find the general solution of the equation . (8)

Ans:

For obtaining the general solution of

-------------------- (1)

We have to determine , the complementary function that is the solution of (1) without the RHS and the P.I. Here for determining the P.I. we have to use the method of undetermined coefficients. For we have to write the auxiliary/characteristic equation which is . The roots of the equation are m = 2+3i , 2-3i . The complementary function is . We note that appears both in the complementary function and the right hand side r(x). Therefore, we choose . Consequently we have

,

Substituting in the given equations, we get

Comparing, both sides we get . Therefore, the particular integral is. The general solution is .

Q.137 Solve . (8)

Ans:

The given differential equation has to be transformed to the differential equation with constant coefficients by changing the independent variable x to t using the transformation

. Thus,

The given d.e. assumes the following form:

--------------- (1)

Characteristic equation of (1) is --------------- (2)

Roots of (2) are .

Thus, C.F. =

P.I. =

Q.138 In an L-C-R circuit, the charge q on a plate of a condenser is given by

. The circuit is tuned to resonance so that . If initially the current I and the charge q be zero, show that, for small values of R/L, the current in the circuit at time t is given by (Et/2L)sinpt. (8)

Ans:

Given differential equation is. It’s A.E. is which gives . As R/L is small, ≈ 0 therefore, to the first order in R/L, , where .Thus, , rejecting terms in etc.

Thus P.I. = .

Thus the complete solution is .

Initially when t=0, q=0, i=o, we get Thus, substituting these values of constants, we get

Q.139 Find a linear transformation T from _R3 into _R3 such that (8)

Ans:

Let the matrix . Thus

Solving, .

Q.140 Examine, whether the matrix A is diagonalizable. . If, so, obtain the matrix P such that is a diagonal matrix. (8)

Ans:

The characteristic equation of the matrix A is given by

.

Eigenvector corresponding to λ = 5 is the solution of the system

. The solution of this system is .

Eigenvector corresponding to λ =-3 is the solution of the system

. The rank of the matrix is 1. Therefore, the system has two linearly independent solutions. Taking z = 0,y = 1, we get eigenvector as , and taking y = 0, z = 1, we get eigenvector as . Thus the matrix P is given by

Q.141 Investigate the values of µ and λ so that the equations

, has (i) no solutions (ii) a unique solution and (iii) an infinite number of solutions. (8)

Ans.

We have . The system admits of unique solution if and only if, the coefficient matrix is of rank 3. Thus Thus for a unique solution

λ ≠ 5 and μ may have any value. If λ = 5, the system will have no solution for those values of μ for which the matrices and are not of the same rank. But A is of rank 2 and K is not of rank 2 unless μ=9. Thus if λ = 5 and μ ≠ 9, the system will have no solution. If λ = 5 and μ=9, the system will have an infinite number of solutions.

Q.142 Find the power series solution about the point of the equation .

(11)

Ans:

The power series can be written as . Substituting in the given equation, we get

. Setting the coefficients of successive powers of x to zero, we get

, where are arbitrary constants. We obtain The power series solution is

Q.143 Express f(x)= in terms of Legendre Polynomial. (5)

Ans:

As

Q.144 Express in terms of and . (8)

Ans:

We know

Putting n=1, 2, 3, 4 successively, we get

Substituting the values, we get

Q.145 Ifshow that. (8)

Ans:

As. Then is given by

,

--

Q.146 Find the extreme value of the function f(x,y,z) = 2x + 3y + z such that x²+y²=5 and x + z =1

(8)

Ans.

We have the auxiliary function as

, ; ----- (1,2,3)

Using (3) in (1) we get -- (4).

Using (2) we get ----- (5).

Using Equations(4) and (5) in we get . For , we arrive at the following point.

For the extremum, substituting in =0, we get For , we get the point For we get the point

Q.147 Show that the function is continuous at (0,0) but its partial derivatives of first order does not exist at (0,0). (8)

Ans.

As

If we choose then Therefore Hence the given function is continuous at (0,0). Now at (0,0) does not exist. Therefore f_x does not exist at (0,0). Similarly does not exist. Therefore f_y does not exist at (0,0).

Q.148 Evaluate the integral where T is region bounded by the cone

and the planes z=0 to z=h in the first octant. (8)

Ans.

The required region can be written as thus

Q.149 Show that the approximate change in the angle A of a triangle ABC due to small changes in the sides a, b, c respectively, is given by where ∆ is the area of the triangle. Verify that (8)

Ans.

For any triangle ABC, we have under usual notations

, ,

Differentiating the first of the above, we get

=

=

=

=

Or

QED (Ist part)

]

Adding the above three expressions for we get

Q.150 If Show that (8)

Ans.

It is given that

Adding , subtracting

=

=

Alternative operate operator on

Q.151 Using the method of variation of parameter method, find the general solution of differential equation (8)

Ans.

The characteristic equation of the corresponding homogeneous equation is m²+16=0. The complementary function is given by y=Acos4x + Bsin4x where y₁= cos4x and

y₂= sin4x are two linearly independent solutions of the homogeneous equation. The Wronskian is given by

(P.I = A(x)cos4x + B(x)sin4x; where A(x) and B(x) are given by A(x) as above.

Also sin4x=2sinxcosx and cos4x= 2cos²2x -1)

Q.152 Find the general solution of the equation . (8)

Ans.

The characteristic equation of the homogeneous equation is The roots of the equation are m = 2+3i , 2-3i . The complementary function is

= (a case of failure)

= .

Thus y(x) = C.F. + P.I.

.

Q.153 Find the general solution of the equation . (8)

Ans.

Put . Thus the given equation becomes which is a linear equation with constant coefficients. It’s A.E. is

Thus C.F. = , and

P.I. =

Q.154 Solve (8)

Ans.

The equation can be written as which is linear equation

Q.155 The set of vectors {x₁, x₂}, where x₁ = (1,3)^T, x₂ = (4,6)^T is a basis in R². Find a linear transformations T such that Tx₁ = (-2,2,-7)^T and Tx₂ = (-2,-4,-10)^T (8)

Ans.

Let the matrix .

;

;

;

Solving the above system of equations we get

Thus

A=

Q.156 Show that the matrix A is diagonalizable. . If, so, obtain the matrix P such that is a diagonal matrix. (8)

Ans.

The characteristic equation of the matrix A is given by .

Eigenvector corresponding to λ = 1 is the solution of the system

. The solution of this system is .

Eigenvector corresponding to λ =2, 3 are and Thus the matrix P is given by

Q.157 Investigate the values of λ for which the equations

are consistent, hence find the ratios of x:y:z when λ has the smallest of these value. (8)

Ans.

The system will be consistent if Thus if λ=0, x=y=z;

For , system reduces to a single equation 2x + 10y + 6z = 0.

Q.158 Find the first five non-vanishing terms in the power series solution of the initial value problem

(11)

Ans.

The power series can be written as . Substituting in the given equation,

.

Setting the coefficients of successive powers of x to zero where are arbitrary constants. We obtain The power series solution is

Q.159 Show that (5)

Ans.

Integrating by parts, we get

Q.160 Show that (8)

Ans.

We know Putting n=1/2, 3/2, successively, we get ,  Substituting the values, we get

Q.161 Show that (8)

Ans.

We have to use that ---------------- (1)

We know that Bessel functions , n ≠ 0, 1, 2 --- of various orders can be derived as coefficients of various powers of t in the expansion of the function ; that is

= ----------------- (2)

Put

Thus,

------------------ (3)

Using the value of in (2), we get

Equating real and imaginary parts in the generating function of Bessel’s equation, .

Squaring and integrating w.r.t. from 0 to and noting that ,. Thus

Adding, we get

Q.162 Compute for the function

Also discuss the continuity of at (0,0). (8)

Ans.

For obtaining and we need the partial derivatives and .

For obtaining these derivatives we use the definition of and :

; .

Thus are not continuous at (0,0).

Q.163 Find the minimum values of subject to the condition (8)

Ans.

Let . The necessary conditions for extremum is thus we get .At each of these points, the value of the given function is . Arithmetic Mean of ,, is AM =

the Geometric Mean of ,, is GM = Since we get . Hence, all the above points are the points of constrained minimum and the minimum value of is .

Q.164 The function is approximated by a first degree Taylor’s polynomial about the point (2,3). Find a square with centre at (2,3) such that the error of approximation is less than or equal to 0.1 in magnitude for all points within the square. (8)

Ans.

We have

The maximum absolute error in the first degree approximation is given by . Also it is given that , therefore we want to determine the value of such that

Q.165 Find the Volume of the ellipsoid (8)

Ans.

Volume = 8(volume in the first octant). The projection of the surface in the x-y plane is the region in the first quadrant of the ellipse .

Thus

Using the transformation x = aX, y = bY, z = cZ, the desired volume can be expressed as

where which is a sphere of radius 1. Using spherical polar coordinates , , .

.

Q.166 Solve the differential equation (8)

Ans.

Here ;

For the given equation to be exact . Consequently, we determine

.

As , equation is not exact. Accordingly, we determine the 1.F. by examining

.

On multiplying throughout by and integrating (using the rule of exact) d.e. we get

, where k is constant integration.

Q.167 Using the method of variation of parameters, solve the differential equation . (8)

Ans.

Auxiliary equation is

. Its P.I. is given by

Using the method of variation of parameters A(x) and B(x) are given by

,

Where r(x) denotes the RHS of the given differential equation i.e. and W(x) is the Wronskian

.

Thus,

and the general solution is C.F. + P.I.

.

Q.168 Find the general solution of the equation . (8)

Ans.

The characteristic equation of the homogeneous equation is The roots of the equation are m = -1, -3. The complementary function is .

P.I. =

Thus

Here

Thus the first term of the P.I. =

Similarly,

; = A + B

.

Q.169 The eigenvectors of a 3 x 3 matrix A corresponding to the eigen values 1, 1, 3 are .Find the matrix A. (8)

Ans.

From the eigen values 1,1,3 we write the Diagonal matrix D as

; From the eigen vectors we write the Modal matrix

; For obtaining the matrix , we proceed as follows

Thus

Q.170 Test for consistency and solve the system of equations (8)

Ans.

The given system of equations can be expressed as

or AX = B

Using row transformations A can be expressed as

Which is of rank 2. The augumented matrix

is used of rank. Consequently, system is consistent. On solving we get

where z is a parameter. Thus

Q.171 Given that show that is a unitary matrix. (8)

Ans.

;

.

=

------ (1)

For proving that is a unitary matrix we need the transpose of the above matrix. Consequently

------------------------------------------- (2)

The product of (1) and (2) is a unitary matrix. I = .

Q.172 Show that the transformation

is non-singular. Find the inverse transformation. (8)

Ans.

Writing the given transformation in matrix form Y = AX.

. Therefore the given matrix A is non-singular and hence the given transformation is also regular. Thus, X = A^-1Y .

Hence we arrive at the following expressions for the inverse transformation

Q.173 If u = f(x,y), x = rcosθ, y = rsinθ, then show that

(8)

Ans.

---------------- (1)

Similarly,

---------------- (2)

Squaring (1)and (2) and adding, we get

=

Q.174 Find the power series solution about the origin of the equation

. (11)

Ans.

The point x = 0 is a regular singular point. The power series can be written as . Substituting in the given equation, we get

The indicial roots are r = -2, -3.

Setting the coefficients of x^r+1 to zero, we get

, For r = -2, =0 and for r = -3, is arbitrary constants. Thus the remaining terms are We obtain .   Thus . For r = -3, , --------

The power series solution for r = -3 is

For r= -2, we get =0, The power series solution for r = -2

Q.175 Find the value of . (5)

Ans.

There are two ways of obtaining the value of

(i) Through recurrence relation

(ii) Using Rodrigue’s formula

Through (i) we make use of the following recurrence relation

------------------ (1)

With ------------------ (2)

Putting n = 1; we get (Using equation 1)

------------------ (3)

For n = 2, equation (1) yields

---- (4)

Thus

= ; ~ 20.005

Method II :Using Rodrigue’s formula

Thus = 20.0025

Q.176 Prove the Orthogonal property of Legendre Polynomials. (8)

Ans.

The orthogonality property of the Legendre’s functions is defined by the relation

---------------- (1)

We first prove (1) form the case m ≠ n.

Let and . Thus, u and v satisfy respectively the following differential equations:

--------------- (2)

--------------- (3)

Multiplying equation (2) by v and equation (3) by u and subtracting, we get

Or --------------- (4)

Integrating equation (4) w r to x between the limits (-1) to (1) we get

--------------- (5)

Thus, for m ≠ n

Case III; m = n;

Above result can be proved either through Rodrigue’s formula

---------------- (6)

Or using generating function of the Legendres polynomial, that is

---------------- (7)

However, we use equation (7) to prove (1) for m = n.

Squaring (7) we get

---------------- (8)

Interpreting (8) w.r.to x between (-1) to (+1), we get

---------------- (9)

Using the orthogonality property for case m ≠ n, we get

Or = R.H.S.

Or = R.H.S.

Or = R.H.S.

Or = R.H.S.

Equating the coefficients of , we get