$3y''+4y'+y=sin(t)e^{-t}, y(0)=1 y'(0)=0$ I can get that the two solutions are $e^{\frac{-t}{3}}$ and $e^{-t}$ but am getting integrals that don't work out when I calculate $u_1$ and $u_2$.  
For $W[y_1,y_2](t)$ I am getting 
$-e^{\frac{4t}{3}}+\frac{e^{-2t/3}}{3}$, which I don't think is right since this shouldn't be a complex problem.
Any help please would be appreciated.
 A: For the homogeneous case, $3u''(t)+4u'(t)+u(t)=0$, the general solution is ($C_1$ and $C_2$ arbitrary constants):
$$ u(t)=C_1e^{-t/3}+C_2e^{-t}$$
Now for the inhomogeneous case $3v''(t)+4v'(t)+v(t)=\sin(t)e^{-t}$, guess a solution of the form $e^{-t}(A\cos(t)+B\sin(t))$, with $A, B$ constants, and work out $A$ and $B$ by differentiating and using the differential equation. You will find $A=\frac{2}{13}$ and $B=-\frac{3}{13}$ and hence the particular solution:
$$v(t)=\frac{1}{13}e^{-t}(2\cos(t)-3\sin(t))$$
It follows that the general solution to the differential equation is:
$$y(t)=C_1e^{-t/3}+C_2e^{-t}+\frac{1}{13}e^{-t}(2\cos(t)-3\sin(t))$$
Now fill in the initial values to determine the constants $C_1$ and $C_2$ to be $\frac{24}{13}$ and $-1$ respectively. So the final answer is:
$$y(t)=\frac{24}{13}e^{-t/3}-e^{-t}+\frac{1}{13}e^{-t}(2\cos(t)-3\sin(t))$$
A: I am assuming that by $W[y_1,y_2]$ you mean the Wronskian.  It should be
$$\det\pmatrix{e^{-t/3}&e^{-t}\cr -\frac13e^{-t/3}&-e^{-t}\cr}
  =-\frac23e^{-4t/3}\ .$$
