Laplace transformations in differential equations

Question

Using Laplace transformations solve the differential equation $$y''+3y'-17y=e^{-3x}$$ $$y(0)=4 , y'(0)=7$$

I'm having some trouble understanding the basic concept and seeing an example done I'm sure will clear some confusion.

Answer 1

We have:

$$\mathcal{L}\{y''+3y'-17y=e^{-3x}\} = (s^2 y(s) - sy(0)-y'(0)) + 3 (sy(s) - y(0)) -17 y(s) = \dfrac{1}{s+3}$$

with: $y(0)=4, y'(0)=7$

Substitute for the IC's, solve for $y(s)$ and then find the inverse Laplace Transform.

We get:

$$y(s) = \dfrac{4 s^2+ 31 s + 58}{(s+3) (s^2+3s-17)}$$

Now we can write this using partial fractions as:

$$y(s) = -\dfrac{1}{17 (s+3)}+\dfrac{69s+323}{17 (s^2+3s-17)}$$

Now, you just need to take $\mathcal{L^{-1}}\{y(s)\} = y(t)$.

Can you take it from here?