Solve the initial value problem $y′′+36y=e^{−t}$ 
Solve the initial value problem $y′′+36y=e^{−t}$
$y(0)=y_0$,
$y'(0)=y_0'$
Suppose we know that $y(t)\to0$ as $t\to\infty$. Determine the solution and the initial conditions.

So far I have come up with
$m^2+36=0$
$m=6i$
$m=-6i$
Guess solution is $y(t)=c_1 \sin(6t)+c_2\cos(6t)+e^{-t}$
 A: When you do $m^2+36=0$ and you get $m=\pm6i$, you're actually looking for the solutions of the homogeneous associated differential equation; in this case, $y''+36y=0$. So what you got are two independent solutions that generate any solution of the homogeneous equation: $c_1\sin(6t)+c_2\cos(6t)$. Now you have to look for a particular solution $\phi_p(t)$ of the differential equation you're given, since all solutions are of the form $\phi_p(t)+c_1\sin(6t)+c_2\cos(6t)$.
To obtain $\phi_p(t)$ suppose it is of the form $\phi_p(t)=a(t)\sin(6t)+b(t)\cos(6t)$, where $a(t),b(t)$ are some suitable functions.
If we impose two conditions $\begin{cases}a'(t)\sin(6t)+b'(t)\cos(6t)=0\\a'(t)(\sin(6t))'+b'(t)(\cos(6t))'=e^{-t}\end{cases}$
we'll get that $\phi_p(t)$ is indeed a solution (this is a general method; you can check it works).
Then we need $a(t),b(t)$ such that $\begin{cases}a'(t)\sin(6t)+b'(t)\cos(6t)=0\\6a'(t)\cos(6t)-6b'(t)\sin(6t)=e^{-t}\end{cases}$
Note that this can be written as $\pmatrix{\sin(6t)&\cos(6t)\\6\cos(6t)&-6\sin(6t)}\pmatrix{a'(t)\\b'(t)}=\pmatrix{0\\e^{-t}}$, so you can solve for $a'(t),b'(t)$ with the inverse of the matrix.
$\pmatrix{\sin(6t)&\cos(6t)\\6\cos(6t)&-6\sin(6t)}^{-1}=A^{-1}=\frac{1}{\det A}\text{adj}(A^\mathsf t)$
We have

$\det A=-6\sin^2(6t)-6\cos^2(6t)=-6\\A^\mathsf t=\pmatrix{\sin(6t)&6\cos(6t)\\\cos(6t)&-6\sin(6t)}\\\text{adj}(A^\mathsf t)=\pmatrix{-6\sin(6t)&-\cos(6t)\\-6\cos(6t)&\sin(6t)}$

So we get

$\pmatrix{\sin(6t)&\cos(6t)\\6\cos(6t)&-6\sin(6t)}^{-1}=\frac{-1}{6}\pmatrix{-6\sin(6t)&-\cos(6t)\\-6\cos(6t)&\sin(6t)}=\frac{1}{6}\pmatrix{6\sin(6t)&\cos(6t)\\6\cos(6t)&-\sin(6t)}$

Then $\pmatrix{a'(t)\\b'(t)}=A^{-1}\pmatrix{0\\e^{-t}}$, so

$\pmatrix{a'(t)\\b'(t)}=\frac{1}{6}\pmatrix{6\sin(6t)&\cos(6t)\\6\cos(6t)&-\sin(6t)}\pmatrix{0\\e^{-t}}=\frac{1}{6}\pmatrix{e^{-t}\cos(6t)\\-e^{-t}\sin(6t)}\Rightarrow\begin{cases}a'(t)=\frac{e^{-t}\cos(6t)}{6}\\b'(t)=\frac{-e^{-t}\sin(6t)}{6}\end{cases}$

So $a(t)=\int a'(t)\,\text{d}t$ and $b(t)=\int b'(t)\,\text{d}t$ (I think this is the most tedious part). Once you have obtained them, you get the particular solution $\phi_p(t)$ (to be honest, at this point I used Wolfram Alpha, but when I first learned this I had to do quite a few integrals). We get $\phi_p(t)=\dfrac{e^{-t}}{37}$.
So a general solution $\varphi(t)$ is of the form $\varphi(t)=\dfrac{e^{-t}}{37}+c_1\cos(6t)+c_2\sin(6t)$.
If we know that $\phi(t)\xrightarrow{t\to+\infty}0$ then we impose $\varphi(t)=\dfrac{e^{-t}}{37}+c_1\cos(6t)+c_2\sin(6t)\xrightarrow{t\to+\infty}0$. But $\dfrac{e^{-t}}{37}\xrightarrow{t\to+\infty}0$, and $\cos(6t)$ and $\sin(6t)$ doesn't stop oscillating; so we need to get rid of them in order to get to $0$ when we do $t\to+\infty$. The only possibility is $c_1=c_2=0$.
Therefore the solution is $\varphi(t)=\dfrac{e^{-t}}{37}$, so the initial conditions are $\varphi(0)=\frac{1}{37},\varphi'(0)=-\frac{1}{37}$.
A: Your guess solution is incorrect because $e^{-t}$ does not satisfy the ODE.$$y''+36y=e^{-t}+36e^{-t}=37e^{-t}\ne e^{-t}$$which suggests the particular solution is $\frac{e^{-t}}{37}$, giving our guess solution as $y(t)=c_1\sin 6t+c_2\cos 6t+\frac{e^{-t}}{37}$.
As $t\to\infty$, we want that $y$ should vanish. The third term $e^{-t}/37$ vanishes but we can't say $c_1\sin 6t+c_2\cos 6t$ vanishes unless coefficients of both $\sin6t,\cos 6t$ are $0$. Thus the solution we desire is $e^{-t}/37$. Can you find $y_0,y'_0$ now?
