#start of lec 9
1) $ay''+by'+cy=P_m(t)e^{rt}$
$y_{p}(t)=t^s(b_{m}t^m+b_{m-1}t^{m-1}+\dots+b_{0})e^{rt}$
s=0, if r is not a root
s=1 if r is a single root
s=2 if r is a double root
where P is a polynomial degree m.

2) $ay''+by'+cy=P_{m}(t)e^{\alpha t}\cos(\beta t)+P_{m}(t)e^{\alpha t}\sin(\beta t)$
3) $y_{p}(t)=t^s[(A_{k}t^k+A_{K-1}t^{k-1}+\dots+A_{0})e^{\alpha t}\cos(\beta t)+(B_{k}t^k+B_{k-1}t^{k-1}+\dots+B_{0})e^{\alpha t}\sin(\beta t)]$
s=0 if $\alpha+i\beta$ is not a root
s=1 if $\alpha+i\beta$ is a root

variation of parameters:
$ay''+by'+cy=f(t)$
1) $y_{h}=c_{1}y_{1}(t)+c_{2}y_{2}t$ <- h is homogenous, ie: $f(t)=0$
lagrange proposed: find a particular solution of y_p
$y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$ <- btw $y_{1}$ and $y_{2}$ are often called a fundamental pair.
we put y_p into the equation and make it equal to the RHS
$y'_{p}=v_{1}y_{1}+v_{1}y_{1}'+v_{2}'y_{2}+v_{2}y_{2}'$
to avoid second derivatives in the equation and problems with uniqueness lagrange imposed:
1) $v_{1}y_{1}+v_{2}'y_{2}=0$ this simplifies our work down the road as well.
so $y_{p}''=v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}''$
$a(v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}'')+b(v_{1}y_{1}'+v_{2}y_{2}')+c(v_{1}y_{1}+v_{2}y_{2})=f(t)$
$v_{1}(ay_{1}''+\cancelto{ 0 }{ by_{1}' }+cy_{1})+v_{2}(ay_{2}''+\cancelto{ 0 }{ by_{2}'C }+cy_{2})+a(v_{1}'y_{1}'+v_{2}'y_{2}')$
2) $v_{1}'y_{1}'+v_{2}'y_{2}'=\frac{f(t)}{a}$
$\det \begin{pmatrix}y_{1} & y_{2} \\y_{1}'& y_{2}'\end{pmatrix}$ = rronsky = $W[y_{1},y_{2}]\ne 0$ this can never be 0!
by definition $y_1$ and $y_2$ are linearly independant solutions so the above can never be 0!
$v_{1}'=\frac{{f(t)y_{2}t}}{aW[y_{1},y_{2}]}$; $v_{2}'=-\frac{{f(t)y_{1}(t)}}{aW[y_{1},y_{2}]}$ <- integrate both sizes to get v1,2. When integrating, you don't need to add a generic constant.

#ex #second_order #IVP 
$y''+4y=2\tan(2t)-e^t \qquad y(0)=0 \qquad y'(0)=\frac{4}{5}$
can we use undetermined coefficients? yes and no
find general solution to homogenous countepart
1) $y''+4y=0$ -> $r^2+4=0$ -> $r_{1,2}=\pm 2i$
$y_{h}(t)=c_{1}\cos(2t)+c_{2}\sin(2t)$
2 $y''+4y=-e^t$ <- use method of undetermined coefficients
$y_{p}'(t)=Ae^{t}$
$5Ae^t=-e^t$
$A=-\frac{1}{5}$
$y_{p}'(t)=-\frac{1}{5}e^t$
(ii) $y''+4y=2\tan(2t)$ <- cant use method of undetermined coefficients
$y_{p}^2(t)=v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)$
plugging in:
we get a system of eq:
	$\cos(2t)v_{1}'+\sin(2t)v_{2}'=0$
	$-2\sin(2t)v_{1}'+2\cos(2t)v_{2}'=2\tan(2t)$
> we know these two will give a unique solution.
>to solve system of eq multiply each by:
>$2\cos(2t)$
>$\sin(2t)$

$2(\sin^2(2t)+\cos^2(2t))v_{2}'=2\tan(2t)\cos(2t)$
$v_{2}'=\sin(2t)$
$v_{2}(t)=-\frac{1}{2}\cos(2t)$ no constant of integration, we want one solution only
$v_{1}'=-{\frac{\sin^2(2t)}{\cos(2t)}}$
$v_{1}=-\int \frac{\sin^2(2t)}{\cos(2t)} \, dx$
$v_{1}=-\int \frac{{1-\cos^2(2t)}}{\cos(2t)} \, dt$
$v_1=-\int sec(2t) \, dx+\int \cos(2t) \, dt$
$v_{1}(t)=-\frac{1}{2}\ln\mid sec(2t)+\tan(2t)\mid+\frac{1}{2}\sin(2t)$
$y_{p}^2(t)=v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)$
$y(t)=y_{h}(t)+y_{p}'(t)+y_{p}^2(t)$
=$c_{1}\cos(2t)+c_{2}\sin(2t)+v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)-\frac{1}{5}e^t$
is our general answer.

IVP solution:
$y(0)=0=c_{1}-y_{p}(0)=c_{1}-\frac{1}{5}\Rightarrow c_{1}=\frac{1}{5}$
skipping some differentiation: $y'(0)=2c_{2}+y_{p}'(0)=2c_{2}+v_{1}'(0)+2v_{2}(0)-\frac{1}{5}=\frac{4}{5}\Rightarrow c_{2}=1$
$y(t)=\frac{1}{5}\cos(2t)+\sin(2t)-\frac{1}{5}e^t+v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)$

#end of lecture 9