Two linearly independent solutions of the differential equationy" - 2y' + y = 0 are y1 = ex and y2 = xex.Then a particular solution of y" - 2y' + y = ex sin x isa)y1cos x + y2(sin x - x cos x)b)y1sin x + y2(x cos x - sin x)c)y1(x cos x - sin x) - y2 cos xd)y1(x sin x - cos x) + y2cos xCorrect answer is option 'C'. Can you explain this answer?

Question

Answer

Let's consider the differential equation:

y'' + p(x)y' + q(x)y = 0

To find two linearly independent solutions of this differential equation, we can use the method of variation of parameters.

Step 1: Find the complementary function (CF)
To find the CF, we set p(x) = 0 and q(x) = 0 in the differential equation.

y'' = 0

Integrating twice, we get:

y = Ax + B

where A and B are constants.

Step 2: Find the particular integral (PI)
To find the PI, we assume a solution of the form:

y = u1(x)(Ax + B) + u2(x)(Ax + B)

where u1(x) and u2(x) are functions to be determined.

Differentiating the assumed solution twice, we get:

y' = u1'(Ax + B) + u1(A) + u2'(Ax + B) + u2(A)

y'' = u1''(Ax + B) + 2u1'(A) + u2''(Ax + B) + 2u2'(A)

Substituting these expressions into the differential equation, we get:

(u1'' + 2u1' + u2'' + 2u2')*(Ax + B) + p(x)*(u1'(Ax + B) + u1(A) + u2'(Ax + B) + u2(A)) + q(x)*(u1(Ax + B) + u2(Ax + B)) = 0

Simplifying, we get:

(Ax + B)*(u1'' + 2u1' + u2'' + 2u2') + p(x)*(u1'(Ax + B) + u2'(Ax + B)) + q(x)*(u1(Ax + B) + u2(Ax + B)) = 0

Matching the coefficients of (Ax + B) on both sides, we get:

u1'' + 2u1' + u2'' + 2u2' = 0

Matching the coefficients of 1 on both sides, we get:

u1'(A) + u2'(A) = 0

Matching the constant terms on both sides, we get:

u1(A) + u2(A) = 0

These three equations can be solved to find the functions u1(x) and u2(x).

Step 3: The general solution
The general solution of the differential equation is given by:

y = CF + PI

Substituting the expressions for CF and PI, we get:

y = Ax + B + u1(x)(Ax + B) + u2(x)(Ax + B)

Simplifying, we get:

y = (A + u1(x)A + u2(x)A)x + (B + u1(x)B + u2(x)B)

This is the general solution of the differential equation. The functions u1(x) and u2(x) can be determined by solving the system of equations obtained in Step 2.

Similar Mathematics Doubts

Let y1(x) and y2(x) be linearly independent solution of the differential equation y" + P(x) y + Q(x)y = 0, where P(x) and Q(x) are continuous functions on an interval I. Then y3(x) = ay1(x) + by2(x) and y4(x) = cy1(x) dy2(x) are linearly independent solutions of the given differential equations if

If y1 and y2 are two solutions o f y" + p{x)y + q(x)y - 0, then for general solution of this given equation, y1 and y2 are

Let y1(x) and y2(x) are two linearly independent solutions of the differential equation (sin x - x cos x) y” - x sin xy’ + y sin x = 0 for x∈ confider the wrenskian ω(x) = y1.ifis equal t o .... ([.] denotesgreatest integer function) Correct answer is '3'. Can you explain this answer?

Ifare three solutions of a non - homogeneous linear differential equationwhere P(x), Q(x) and R(x) are continuous function on [a, b] with a > 0, then itsparticular solution w.r. to the conditions y(0) = 0 y.O) = 1

Which of the following statements is incorrect regarding the singular solutions of the differential equations?

Top Courses for Mathematics

Suggested Free Tests

Related Content

Schools

Competitive Examinations

Quick Access Links

Technical Exams