Two linearly independent solutions of the differential equation y" - 2y' + y = 0 are y₁ = e^x and y₂ = xe^x. Then a particular solution of y
"
- 2y' + y + e^x sin x is

a)
y₁ cos x + y₂ (sin x - x cos x)

b)
y₁ sin x + y₂ (x cos x - sin x)

c)
y₁ (x cos x - sin x) - y₂ cos x

d)
y₁ (x sin x - cos x) + y₂ cos x

Correct answer is option 'C'. Can you explain this answer?

Question

Two linearly independent solutions of the differential equation y" - 2y' + y = 0 are y₁ = e^x and y₂ = xe^x. Then a particular solution of y

"

- 2y' + y + e^x sin x is

a)
y₁ cos x + y₂ (sin x - x cos x)
b)
y₁ sin x + y₂ (x cos x - sin x)
c)
y₁ (x cos x - sin x) - y₂ cos x
d)
y₁ (x sin x - cos x) + y₂ cos x

Correct answer is option 'C'. Can you explain this answer?

Answer

To find two linearly independent solutions of the differential equation y'' + 4y' + 4y = 0, we can assume a solution of the form y = e^(rx), where r is a constant.

Plugging this into the differential equation, we get:

r^2e^(rx) + 4re^(rx) + 4e^(rx) = 0

Dividing through by e^(rx), we have:

r^2 + 4r + 4 = 0

This is a quadratic equation in r. Solving for r, we get:

(r + 2)^2 = 0

r + 2 = 0

r = -2

Therefore, one solution of the differential equation is y1 = e^(-2x).

To find a second linearly independent solution, we can use the method of reduction of order. Let's assume a second solution of the form y2 = v(x)e^(-2x), where v(x) is another function to be determined.

Taking the first derivative of y2, we have:

y2' = v'(x)e^(-2x) - 2v(x)e^(-2x)

Taking the second derivative of y2, we have:

y2'' = v''(x)e^(-2x) - 4v'(x)e^(-2x) + 4v(x)e^(-2x)

Plugging y2 and its derivatives into the differential equation, we get:

(v''(x)e^(-2x) - 4v'(x)e^(-2x) + 4v(x)e^(-2x)) + 4(v'(x)e^(-2x) - 2v(x)e^(-2x)) + 4(v(x)e^(-2x)) = 0

Simplifying, we have:

v''(x)e^(-2x) = 0

v''(x) = 0

Integrating twice, we have:

v'(x) = c1

v(x) = c1x + c2

Therefore, the second solution of the differential equation is y2 = (c1x + c2)e^(-2x).

The two linearly independent solutions of the differential equation are y1 = e^(-2x) and y2 = (c1x + c2)e^(-2x), where c1 and c2 are arbitrary constants.

Similar Mathematics Doubts

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 is

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?

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