Solve $y''-3y'-4y=3\mathit{e}^{2t}$ using variation of parameters formula.
I just want to know whether I am on the right track with my solution.
I started by solving the differential equation $y''-3y'-4y=0$ by finding its companion matrix and the eigenvalues of that matrix. I got the solution set $\left \{ \varphi_{1},\varphi_{2} \right \}=\left \{ \mathit{e}^{-t},\mathit{e}^{4t} \right \}$.
The variation of parameters theorem is the following:
So I am guessing that $\phi_{h}(t)=c_{1}\mathit{e}^{-t}+c_{2}\mathit{e}^{4t}$ for some constants $c_{1},c_{2}$. I then calculated the sum of integrals:
$W(\phi_{1},\phi_{2})(t)=5\mathit{e}^{3t}$
$W_{1}(\phi_{1},\phi_{2})(t)=-\mathit{e}^{4t}$
$W_{2}(\phi_{1},\phi_{2})(t)=\mathit{e}^{-t}$
Then $\phi(t)=c_{1}\mathit{e}^{-t}+c_{2}\mathit{e}^{4t}-\frac{3}{5}\int_{\tau}^{t}\mathit{e}^{2s}ds+\frac{3}{5}\int_{\tau}^{t}\mathit{e}^{2s}ds=\phi_{h}(t)$.
I am not sure if this is how to find the solution or not. Is my $\phi_{h}$ correct, and is it really true that the solution is just $\phi_{h}$?
Any help and guidance appreciated!
Solutions of the homogeneous part: $$y''-3y'-4y=0$$ are $y_1=e^{-t}, y_2=e^{4t}$ The soloution of the in-homogeneous ODE: $$Y''-3Y-4Y= 3 e^{2t} =f(t)$$ by Vvariation of parameters $(C_1,C_2)$ is given by $$Y(t)=C_(t) e^{-t}+ C_2(t) e^{4t},$$wjere $$C_1=\int\frac{-y_2(t) f(t)}{W(t)} dt+D_1, C_2= \int \frac{y_1(t) f(t)}{W(t(} dt +D_2$$ $$W(t)=[y_1y'_2-y_1'y_2]=3e^{3t}.~~~(1)$$ $$C_1=-\int \frac{e^{4t}~ 3 e^{2t}}{5e^{3t}} dt + D_1=-\frac{1}{5}e^{3t}+D_1$$ $$C_2=\int \frac{e^{-t}~3e^{2t}}{5e^{3t}}+ dt+D_2 =-\frac{3}{10} e^{-2t}+D_2$$ Finally we have from Eq. (1) $$Y(t)=[-\frac{1}{5}e^{3t}+D_1]e^{-t}+[-\frac{3}{10}e^{-2t}+D_2]e^{4t}$$ $$\implies Y(t) =D_1 e^{-t}+D_2e^{4t}]-\frac{1}{2}e^{2t}.$$