Question

Use the Laplace transform to solve the given initial-value problem.

y' + 3y = e^5t,  y(0) = 2

y(t) =

Answer 1

y' + 3y = e^5t,  y(0) = 2

Taking Laplace transform on both sides

$L(y')+3L(y) = L(e^{5t})$

$sY(s)-y(0)+3Y(s) = \dfrac{1}{(s-5)}$

$sY(s)-2+3Y(s) = \dfrac{1}{(s-5)}$

$Y(s)(s+3) - 2 = \dfrac{1}{(s-5)}$

$Y(s) = \dfrac{2}{(s+3)} + \dfrac{1}{(s+3)(s-5)}$

$Y(s) = \dfrac{(2s-9)}{(s+3)(s-5)}$

Resolving into partial fraction we get,

$\dfrac{(2s-9)}{(s+3)(s-5)} = \dfrac{A}{(s+3)} + \dfrac{B}{(s-5)}$

$(2s-9) = A(s-5) + B(s+3)$

We get, $A = \dfrac{15}{8}$

$B = \dfrac{1}{8}$

Hence, we get

$Y(s) = \dfrac{15}{8(s+3)} + \dfrac{1}{8(s-5)}$

Now, taking the inverse Laplace Transform

We get,

$y(t) = \dfrac{15(e^{-3t})}{8} + \dfrac{e^{5t}}{8}$