A-01/C-01/T-01

Code: AE-01/AC-01/AT-01 Subject: MATHEMATICS-I

PART - I, VOL – I

TYPICAL QUESTIONS & ANSWERS

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

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

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

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

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

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

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

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

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(4dx + 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

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

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

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

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

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

PART – II, VOL – I

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(Y) is differentiable.

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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

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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)

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

Ans:

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

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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.

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,

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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.

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

\ 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 + c₂ze^z

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

Substituting in the equation(1), we get

Hence the general solution

y = c₃ Sin2z + c₄Cos2z

y = c₃ Sin2(logx) + c₄Cos2(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.

= .

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

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.

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

Setting this coefficient to zero, we get

(1) may be written as

\ For m ³ 0, we get

when r = 0

when r = ½

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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.

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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,

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

Ans:

Now we know that

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= .

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 .

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

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)^ù

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

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

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

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

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

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

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

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

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

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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

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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

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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

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

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

Ans:

A.E. is

m = 1, 2, 3.

y(0) = 0

y¹ (0) = 1

y¹¹ (0) = -1

=> .

Thus

Solving .

Thus solution of initial value problem is .

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

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:

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

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)

Ans:

M = ,

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

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

Thus equation is

Now M₁ = .

(ii)

Ans:

Integrating, we get

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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

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

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

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

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

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

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).

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

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.

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

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.

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

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

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

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

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

(i)

Ans:

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.

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(ii)

Ans:

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.

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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.

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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.

;

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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.

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

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

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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.

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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 R₁and 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)

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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

Code: AE-01/AC-01/AT-01 MATHEMATICS-I

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