How to solve this by Laplace transform

Question

How would we solve this differential equation by Laplace? $$(D^2+4D+3)y(t)=(2D+1)x(t)$$ $$y(0^-)=1, y'(0^-)=2$$ if $x(t) = u(t)$ ie the unit step function $$$$ after doing the transform I am at this point $$Y(S)= \frac{S^2+8S+1}{S(S+1)(S+3)}$$ but doing it this way is not giving me the correct answer which should be $$y(t)=\color{red}{\frac{1}{3}}(1+9e^{-t}-7e^{-3t})$$ I am attaching a copy of my solution of why I think I am wrong

Answer 1

You are almost there, just need to expand your expression for $Y(S)$ in partial fractions,

$$ Y(S) = \frac{A}{s} + \frac{B}{s+1} + \frac{C}{s+3} $$

and you will get $A = 1/3$, $B = 3$, $C = -7/3$

$$ Y(s) = \frac{1}{3}\left[\frac{1}{s} + \frac{9}{s+1} - \frac{7}{s+3} \right] $$

Then you just need to remember that $L\{e^{at}\}(s) = 1/(s-a)$, so that

$$ y(t) = \color{red}{\frac{1}{3}}(1 +9e^{-t}-7e^{-3t}) $$

you can actually check that this follows the initial conditions $y(0)=1$ and $y'(0)=2$.