#start of lec 8 (sept 22)
last lecture we talked about $ay''+b'y+cy=f(t)$

in the case when $f(t)=0$ :
1) $ay''+b'y+cy=0$
then $ar^2+br+c=0$ and solve with quadratic formula
general solution is: $y_{h}(t)=c_{1}e^{r_{1}(t)}+c_{2}e^{r_{2}t}$ where h means homogenous, ( because when =0 its homogenous)

if $r_{1}=r_{2}$ then $y_{h}(t)=c_{1}e^{r(t)}+c_{2}e^{rt}$
if imaginary roots:
$y_{h}(t)=e^{\alpha t}(c_{1}\cos(\beta t)+c_{2}\sin(\beta t))$
2) If $y_{p}(t)$ solves 1) then its general solution is $y(t)=y_{h}(t)+y_{p}(t)$
theorem: if $p(t),\ g(t),\ f(t)$ are continuous on $I$ then the IVP $y''+p(t)y'+q(t)y=f(t), y(t_{o}),\ y'(t_{o})=y_{1} t_{o}\in I$ has a unique solution
method of undetermined coeffecients:
#ex
$y''\pm_{4}y'+4y=3t+9$ lets find general solution, its centainly non homogenous.
first we have to find general solution to the homogenous equation:
1) $y''-4y'+4y=0$
characteristic eq: $r^2-4r+4=0$ what are the roots?
$r=2$ (repeated solution)
$y_{h}(t)=c_{1}e^{2t}+c_{2}te^{2t}$
we are looking for a particular polynomial where the power is not greater than 1 (?)
2) $y_{p}(t)=At+B$
$y_{p}'=A,\ y_{p}''=0$
$-4A+4(At+B)=3t+9$
$4A=3,\ -4A+4B=9$
$A=\frac{3}{4},\ B=3$
$y(t)=\frac{3}{4}t+3$
general solution: $$y(t)=c_{1}e^{2t}+c_{2}te^{2t}+\frac{3}{4}t+3$$
so big takeaway is if the RHS of eq is a polynomial of  degree u, we try to find a solution as a polynomial of degree u

#ex 
$$y''-4y'+4y=e^{2t}$$
find general solution.
1) $y_h(t)=c_{1}e^{2t}+c_{2}te^{2t}$ (computed earlier)
2) $y_p(t)$ 
we observe the RHS is some exponential, we need the derivative + its second derivative to equal that, we have no option but suspect that its $Ae^{2t}$
but then the LHS becomes 0!
so $Ae^{2t}$ is a wrong guess.
so what do we do? try $Ate^{2t}$ take $c_{2}=A, c_{1}=0$, this does not work again. LHS becomes 0 again
so try $At^2e^{2t}$
$2Ae^{2t}=2e^{2t},\ A=1$ This one works!
we know the homogeenous solution.
$y(t)=c_{1}e^{2t}+c_{2}te^{2t}+t^2e^{2t}$ is the general solution
moral of sotry? if RHS is constant times $e^2t$ we guess with an exponent with a constant, if its homogenous we multiply by t, if still not a valid solution then we multiply by t again.
Ex:
$y''+2y'+2y=2e^{-t}+\cos t,\ y(0)=3,\ y'(0)=1$ I wanna solve this IVP! it must have a unique solution.
1) set RHS to 0: $r^2+2r+2=0$
 $r_{1,2}=-1\pm i$ sqrt(i) is interesting, but not the topic for today.
 $y_{h}(t)=e^{-t}(c_{1}\cos(t)+c_{2}\sin(t))$
 2) $y_{p}(t)=$
 RHS is much more complicated, sum of 2 functions. Lets use principle of super position
 $y_{p}(t)=y_{p_{1}}+y_{p_{2}}$
 where $y_{p_{1}}$ solves $y''+2y'+2y=2e^{-t}$
 $y_{p_{2}}$ solves $y''+2y'+2y=5\cos (t)$
 lets try $y_{p_{1}}=Ae^{-t}$ does this work? look at it, A must be zero but if A is zero you still get problems.
 $y_{p_{1}}'=-Ae^{-t}$
 $y_{p_{1}}''=Ae^-t$ plug in these three and we find that A=2
 
second equation, not so easy:
solution of cos t doenst quite work
$y_{p_{2}}=A\cos(t)+B\sin(t)$
$y_{p_{2}}'=-A\sin(t)+b\cos (t)$
$y_{p_{2}}''=-A\cos t-B\sin t$
$(A+2B)\cos(t)+(-2A+B)\sin(t)=5\cos(t)$
$A+2B=0$
$-2A+B=0$ -> A=1, B=2
but $y_{p_{1}}\ne y_{p_{2}}$ because of the $e^{-t}$ term
$y(t)=c_{1}e^{-t}\cos(t)+c_{2}e^{-t}\sin t+2e^{-t}+\cos t+2\sin t$
$y(0)=3=c_{1}+3=3\implies c_{1}=0$
$y'(0)=1=c_{2}$
final solution $y(t)=e^{-t}(\sin t+2)+\cos t+2\sin t$

If we have an equation of the from:
1) $ay''+by'+cy=P_{m}(t)e^{rt}$
where $p_{m}(t)=a_{m}t^m+a_{m-1}t^{m-1}+ \dots +a_{0}$
then the guess is: $y_{p}(t)=t^s(b_{mt}t^m+b_{m-1}t^{m-1}+\dots+b_{0})e^{rt}$
(i) s=0 if r is not a characteristic polynomial
(ii)) s=1 if r is a single root
(iii) s=2 if r is a double root
we will talk about this more in the coming lecture.
#end of lec 8