I'm having trouble with the following exercise
$$ y'' +4y - (4/e^x) = 0 $$
with the initial values:
$$ y(0) = 1 y'(0)=5$$
I used the formula $$ y'' = s^2Y(s) − s*f(0) − f'(0)$$
and got to:
$$ Y(s) = \left(\frac{4 + 5 e^x+se^x}{s^2e^x+e^x}\right) $$
But now I'm stuck. Can't find anything remotely similar in Laplace tables.
Thank you for your help
The Laplace transform of the term $-\frac{4}{e^x}=-4e^{-x}$ is $-\frac{4}{s+1}$ by the following rule: $$L\{e^{ax}\}=\frac{1}{s-a}$$
Hence $$Y(s)=\frac{4+(s+5)(s+1)}{(s^2+4)(s+1)}$$
You will then need to use partial fractions to separate the fraction.