Use the laplace transform to solve the given initial-value problem. y' + 3y = e^5t, y(0) = 2

Solution:

Given, y' + 3y = e^5t

Laplace transform can be given by

$F(s)=\int_{0}^{\infty }f(t).e^{-st}.dt$

Taking Laplace transform on both sides,

L(y’) + 3L(y) = L(e^5t)

sY(s) - y(0) + 3Y(s) = 1 / (s - 5)

sY(s) - 2 + 3Y(s) = 1 / (s - 5)

Taking out common term,

Y(s)(s + 3) - 2 = 1 / (s - 5)

Y(s)(s + 3) = [1 / (s - 5)] + 2

Y(s) = [1 / (s - 5)(s + 3)] + (2/(s + 3))

Y(s) = [1 + 2(s - 5)] / [(s + 3)(s - 5)]

Y(s) = [1 +2s - 10] / [(s + 3)(s - 5)]

Y(s) = (2s - 9) / [(s + 3)(s - 5)]

Resolving into partial fraction we get,

(2s - 9) / [(s + 3)(s - 5)] = (A / (s + 3)) + (B / (s - 5)) ------------------ (1)

(2s - 9) = A(s - 5) + B(s + 3) 

Now, (s - 5) = 0 

s = 5

(2(5) - 9) = A(0) + B(5 + 3)

10 - 9 = 8B

B = 1/8

Now, s + 3 = 0

s = -3

(2(-3) - 9) = A(-3 - 5) + B(0)

-6 - 9 = A(-8)

8A = 15

A = 15 / 8

Substitute the values of A and B in (1)

Y(s) = [(15 / 8) / (s + 3)] + [(1 / 8) / (s - 5)]

Y(s) = [15 / 8(s + 3)] + [1 / 8(s - 5)]

Taking inverse Laplace Transform,

y(t) = [15(e^-3t) / 8] + [e^5t / 8]

Therefore, y(t) = [15(e^-3t) / 8] + [e^5t / 8]

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Use the laplace transform to solve the given initial-value problem. y' + 3y = e^5t, y(0) = 2

Summary:

Using laplace transform to solve the given initial-value problem y' + 3y = e^5t, y(0) = 2. The value of y(t) = [15(e^-3t) / 8] + [e^5t / 8]