71 lines
3.4 KiB
Markdown
71 lines
3.4 KiB
Markdown
#start of lec 9
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1) $ay''+by'+cy=P_m(t)e^{rt}$
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$y_{p}(t)=t^s(b_{m}t^m+b_{m-1}t^{m-1}+\dots+b_{0})e^{rt}$
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s=0, if r is not a root
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s=1 if r is a single root
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s=2 if r is a double root
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where P is a polynomial degree m.
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2) $ay''+by'+cy=P_{m}(t)e^{\alpha t}\cos(\beta t)+P_{m}(t)e^{\alpha t}\sin(\beta t)$
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3) $y_{p}(t)=t^s[(A_{k}t^k+A_{K-1}t^{k-1}+\dots+A_{0})e^{\alpha t}\cos(\beta t)+(B_{k}t^k+B_{k-1}t^{k-1}+\dots+B_{0})e^{\alpha t}\sin(\beta t)]$
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s=0 if $\alpha+i\beta$ is not a root
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s=1 if $\alpha+i\beta$ is a root
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variation of parameters:
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$ay''+by'+cy=f(t)$
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1) $y_{h}=c_{1}y_{1}(t)+c_{2}y_{2}t$ <- h is homogenous, ie: $f(t)=0$
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lagrange proposed: find a particular solution of y_p
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$y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$ <- btw $y_{1}$ and $y_{2}$ are often called a fundamental pair.
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we put y_p into the equation and make it equal to the RHS
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$y'_{p}=v_{1}y_{1}+v_{1}y_{1}'+v_{2}'y_{2}+v_{2}y_{2}'$
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to avoid second derivatives in the equation and problems with uniqueness lagrange imposed:
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1) $v_{1}y_{1}+v_{2}'y_{2}=0$ this simplifies our work down the road as well.
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so $y_{p}''=v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}''$
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$a(v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}'')+b(v_{1}y_{1}'+v_{2}y_{2}')+c(v_{1}y_{1}+v_{2}y_{2})=f(t)$
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$v_{1}(ay_{1}''+\cancelto{ 0 }{ by_{1}' }+cy_{1})+v_{2}(ay_{2}''+\cancelto{ 0 }{ by_{2}'C }+cy_{2})+a(v_{1}'y_{1}'+v_{2}'y_{2}')$
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2) $v_{1}'y_{1}'+v_{2}'y_{2}'=\frac{f(t)}{a}$
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$\det \begin{pmatrix}y_{1} & y_{2} \\y_{1}'& y_{2}'\end{pmatrix}$ = rronsky = $W[y_{1},y_{2}]\ne 0$ this can never be 0!
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by definition $y_1$ and $y_2$ are linearly independant solutions so the above can never be 0!
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$v_{1}'=\frac{{f(t)y_{2}t}}{aW[y_{1},y_{2}]}$; $v_{2}'=-\frac{{f(t)y_{1}(t)}}{aW[y_{1},y_{2}]}$ <- integrate both sizes to get v1,2. When integrating, you don't need to add a generic constant.
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#ex #second_order #IVP
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$y''+4y=2\tan(2t)-e^t \qquad y(0)=0 \qquad y'(0)=\frac{4}{5}$
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can we use undetermined coefficients? yes and no
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find general solution to homogenous countepart
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1) $y''+4y=0$ -> $r^2+4=0$ -> $r_{1,2}=\pm 2i$
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$y_{h}(t)=c_{1}\cos(2t)+c_{2}\sin(2t)$
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2 $y''+4y=-e^t$ <- use method of undetermined coefficients
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$y_{p}'(t)=Ae^{t}$
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$5Ae^t=-e^t$
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$A=-\frac{1}{5}$
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$y_{p}'(t)=-\frac{1}{5}e^t$
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(ii) $y''+4y=2\tan(2t)$ <- cant use method of undetermined coefficients
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$y_{p}^2(t)=v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)$
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plugging in:
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we get a system of eq:
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$\cos(2t)v_{1}'+\sin(2t)v_{2}'=0$
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$-2\sin(2t)v_{1}'+2\cos(2t)v_{2}'=2\tan(2t)$
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> we know these two will give a unique solution.
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>to solve system of eq multiply each by:
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>$2\cos(2t)$
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>$\sin(2t)$
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$2(\sin^2(2t)+\cos^2(2t))v_{2}'=2\tan(2t)\cos(2t)$
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$v_{2}'=\sin(2t)$
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$v_{2}(t)=-\frac{1}{2}\cos(2t)$ no constant of integration, we want one solution only
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$v_{1}'=-{\frac{\sin^2(2t)}{\cos(2t)}}$
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$v_{1}=-\int \frac{\sin^2(2t)}{\cos(2t)} \, dx$
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$v_{1}=-\int \frac{{1-\cos^2(2t)}}{\cos(2t)} \, dt$
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$v_1=-\int sec(2t) \, dx+\int \cos(2t) \, dt$
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$v_{1}(t)=-\frac{1}{2}\ln\mid sec(2t)+\tan(2t)\mid+\frac{1}{2}\sin(2t)$
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$y_{p}^2(t)=v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)$
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$y(t)=y_{h}(t)+y_{p}'(t)+y_{p}^2(t)$
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=$c_{1}\cos(2t)+c_{2}\sin(2t)+v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)-\frac{1}{5}e^t$
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is our general answer.
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IVP solution:
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$y(0)=0=c_{1}-y_{p}(0)=c_{1}-\frac{1}{5}\Rightarrow c_{1}=\frac{1}{5}$
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skipping some differentiation: $y'(0)=2c_{2}+y_{p}'(0)=2c_{2}+v_{1}'(0)+2v_{2}(0)-\frac{1}{5}=\frac{4}{5}\Rightarrow c_{2}=1$
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$y(t)=\frac{1}{5}\cos(2t)+\sin(2t)-\frac{1}{5}e^t+v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)$
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#end of lecture 9 |