# 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. To do so, find the derivatives first:
$y'_{p}=v_{1}'y_{1}+v_{1}y_{1}'+v_{2}'y_{2}+v_{2}y_{2}'$
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.
$y'_{p}=\cancel{ v_{1}'y_{1} }+v_{1}y_{1}'+\cancel{ v_{2}'y_{2} }+v_{2}y_{2}'$
so $y_{p}''=v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}''$
now we plug into the second order equation:
$a(v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}'')+b(v_{1}y_{1}'+v_{2}y_{2}')+c(v_{1}y_{1}+v_{2}y_{2})=f(t)$
$v_{1}(\cancel{ ay_{1}''+ by_{1}' +cy_{1} })+v_{2}(\cancel{ ay_{2}''+ by_{2}'+cy_{2} })+a(v_{1}'y_{1}'+v_{2}'y_{2}')$
Both $y_{1}$ and $y_{2}$ are solutions to the homogenous counterpart. So the first two terms above equal to zero.
2) $v_{1}'y_{1}'+v_{2}'y_{2}'=\frac{f(t)}{a}$
$\det \begin{pmatrix}y_{1} & y_{2} \\y_{1}'& y_{2}'\end{pmatrix}$ = Wronskian = $W[y_{1},y_{2}]\ne 0$ 
by definition $y_1$ and $y_2$ are linearly independent 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 $v_{1}$ and $v_{2}$. 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 counterpart
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
$y_{p}^1(t)=Ae^{t}$
$5Ae^t=-e^t$
$A=-\frac{1}{5}$
$y_{p}^1(t)=-\frac{1}{5}e^t$
(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
$v_{1}'=-{\frac{\sin^2(2t)}{\cos(2t)}}$
$v_{1}=-\int \frac{\sin^2(2t)}{\cos(2t)} \, dt$
$v_{1}=-\int \frac{{1-\cos^2(2t)}}{\cos(2t)} \, dt$
$v_1=-\int sec(2t) \, dt+\int \cos(2t) \, dt$
$v_{1}(t)=-\frac{1}{2}\ln\mid sec(2t)+\tan(2t)\mid+\frac{1}{2}\sin(2t)$
$y_{p}^2(t)=v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)$
$y(t)=y_{h}(t)+y_{p}^1(t)+y_{p}^2(t)$
=$c_{1}\cos(2t)+c_{2}\sin(2t)+v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)-\frac{1}{5}e^t$
is the 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
#start of lecture 10
# Variation of parameters
last lec we did some variation of parameters
$ay''+by'+cy=f(t)$
1) $y_{h}(t)=c_{1}y_{1}(t)+c_{2}y_{2}(t)$
2) $y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$
	$y_{1}v_{1}'+y_{2}v_{2}'=0$
	$y_{1}v_{1}'+y_{2}'v_{2}'=\frac{b}{a}$ or f/a?
is the system of equations we will need to solve. You can also memorize a formula but peter likes remembering this system of equations and moving on from there.
#ex #variation_of_parameters
$$y''-2y'+y=e^t\ln(t)+2\cos(t)$$
i) $y_{h}(t)=?$
$r^2-2r+1=0$
$r_{1,2}=1$
$y_{h}(t)=c_{1}e^t+c_{2}te^t$
2) $y_{p}(t)=?$
$y''-2y'+y=2\cos (t)$
$y_{p}''=A\cos(t)+B\sin(t)$ is our first guess. but it does not solve the homogenous eq.
$y_{p}'=-\sin(t)$ (obtained by using method of undetermined coefficients, computation not shown.)
$y''-2y'+y=e^t\ln(t)$ cant use undetermined coefficients, use variation of parameters
$y''_{p}(t)=v_{1}y_{1}+v_{2}y_{2}$
$=v_{1}e^t+v_{2}te^t$
compute v1 and v2, using the linear system:
eq1) $e^t+v_{1}'+te^tv_{2}'=0$
eq2) $e^tv_{1}'+(te^t+e^t){v_{2}'}=e^t{\ln t}$
subtract eq1 from eq2 $v_{2}'=\ln(t)$
$v_{2}(t)=\int \ln(t) \, dt$
integrate by parts
$=t\ln(t)-\int t\frac{1}{t} \, dt$
$=t\ln(t)-t$ no constant of integration.
compute $v_{1}$ now:
$v_{1}'=-tv_{2}'$
$=-t\ln t$
integrate to get v_1:
$v_{1}=-\int t\ln t \, dt$
integrate by parts (btw integration by parts will be the most important integration technique in this course):
$v_{1}=-\frac{1}{2}(t^2\ln t)-\int t^2\frac{1}{t} \, dt$
$=-\frac{1}{2}\left( t^2\ln t-\frac{t^2}{2} \right)=-\frac{1}{2}t^2\ln t+\frac{1}{4}t^2$
$y_{p}''(t)=(\frac{1}{2}t^2\ln t+\frac{1}{4}t^2)e^t+(t\ln t-t)te^t$
$y_{p}(t)=-\sin(t)+\frac{1}{2}t^2\ln(t)e^t-\frac{3}{4}t^2e^t$
general solution is produced by adding the homogenous eq with $y_{p}(t)$
general: 
$$y(t)=c_{1}e^t+c_{2}te^t+y_{p}(t)$$