MATH201/content/Variation of parameters (lec 9-10).md

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:
2. 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.
3. 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)