Use the laplace transform to solve the given initial-value problem. y' + 5y = e^4t, y(0) = 2

Solution:

Given, y' + 5y = e^4t

Laplace transform can be given by

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

Taking Laplace transform on both sides,

L(y’) + 5L(y) = L(e^4t)

sY(s) - y(0) + 5Y(s) = 1 / (s - 4)

sY(s) - 2 + 5Y(s) = 1 / (s - 4)

Taking out common term,

Y(s)(s + 5) - 2 = 1 / (s - 4)

Y(s)(s + 5) = [1 / (s - 4)] + 2

Y(s) = [1 / (s - 4)(s + 5)] + (2 / (s + 5))

Y(s) = [1 + 2(s - 4)] / [(s + 5)(s - 4)]

Y(s) = [1 + 2s- 8] / [(s + 5)(s - 4)]

Y(s) = (2s - 7) / [(s + 5)(s - 4)]

Resolving into partial fraction we get,

(2s - 7)/[(s + 5)(s - 4)] = (A / (s + 5)) + (B / (s - 4)) ------------------ (1)

(2s - 7) = A(s - 4) + B(s + 5)

Now, (s - 4) = 0 

s = 4

(2(4) - 7) = A(0) + B(4 + 5)

8 - 7 = 9B

B = 1 / 9

Now, s + 5 = 0

s = -5

(2(-5) - 7) = A(-5 - 4) + B(0)

-10 - 7 = A(-9)

9A = 17

A = 17 / 9

Substitute the values of A and B in (1)

Y(s) = [(17 / 9) / (s + 5)] + [(1 / 9) / (s - 4)]

Y(s) = [17 / 9(s + 5)] + [1 / 9(s - 4)]

Taking inverse Laplace Transform,

y(t) = [17(e^-4t) / 9] + [e^4t / 9]

Therefore, y(t) = [17(e^-4t) / 9] + [e^4t / 9].

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Use the laplace transform to solve the given initial-value problem. y' + 5y = e^4t, y(0) = 2

Summary:

Using laplace transform to solve the given initial-value problem y' + 5y = e^4t, y(0) = 2. The value of y(t) = [17(e^-4t) / 9] + [e^4t / 9].