MATH201/content/Undetermined coefficients (lec 9).md

#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:
2. 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}')
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} = 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