Solving Differential equations with Laplace transform $\displaystyle y''+4y'+3y=e^{-t}$, given $\displaystyle y(0)=y'(0)=1$
My Attempt: 
Taking Laplace transforms on both sides
$\displaystyle $
$\displaystyle [s^2\bar y-sy(0)-y'(0)]+4[s\bar y-y(0)]+3\bar y=\frac{1}{s+1} $
$\displaystyle [s^2+4s+3]\bar y=\frac{1}{s+1}+s+5 $
$\displaystyle \bar y=\frac{s^2+6s+6}{(s+1)(s^2+4s+3)}$
Resolving into partial fractions, 
$\displaystyle \frac{A}{s+1}+\frac{B}{(s+1)^2}+\frac{C}{s+3}=\frac{s^2+6s+6}{(s+1)(s^2+4s+3)}$
I get A=7/2; B=1/2 and C=1
Taking inverse,
$\displaystyle y=\frac{7}{2}e^{-t}+\frac{1}{2}te^{-t}+e^{-3t}$
The given answer is $\displaystyle y=\frac{7}{4}e^{-t}-\frac{1}{2}te^{-t}-\frac{3}{4}e^{-3t}$
I can't find where I am going wrong. Please help.

The value of $C$ is wrong.
It should be $A=\frac{7}{4}, B=\frac{1}{2}, C=-\frac{3}{4}$
