Solve the differential equation $y''-10y'+21y=15e^{4t}$ using Laplace transform

Question

I need to solve the following with Laplace transform: $$y''-10y'+21y=15e^{4t}$$ $$y(0)=3,y'(0)=0$$

After Laplace transform I got: $$L(y)=\frac{15}{(S-4)(S^2-13S+21)}+\frac{30}{(S^2-13S+21)}$$ and now, what do I need to do?

Answer 1

Your LT is incorrect: The LT of $y''$ is $s^2 \hat{y}-s y(0)-y'(0) = s^2 \hat{y}-3 s$, so that the LT is

$$\hat{y}(s) = \frac{15}{(s-4) (s^2-10 s+21)} + 3 \frac{s-10}{s^2-10 s+21}$$

Note that $s^2-10 s+21 = (s-7) (s-3)$. The poles are simple, so the inverse LT may be computed simply via the residue theorem, or a lookup, e.g.,

$$y(t) = (-5+6) e^{4 t} + \left (\frac{15}{4}+\frac{21}{4} \right ) e^{3 t} + \left (\frac{5}{4} - \frac{9}{4} \right ) e^{7 t} = 9 e^{3 t}+e^{4 t}-e^{7 t}$$