forked from Sasserisop/MATH201
39 lines
2.2 KiB
Markdown
39 lines
2.2 KiB
Markdown
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we know how to solve second order equations where a,b,c are constants. Even if they're not constant some can be expressed as a linear equation. But not always will they be solvable. However, there is one class of second order equations with non constant coefficients that are always solvable.
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# Cauchy-Euler equations
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*if it has a name in it, its very important, if it has 2 names its very important!*
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$ax^2{\frac{d^2y}{dx^2}+bx{\frac{ dy }{ dx }}+cy=f(x)},\ x>0$
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where $a,\ b,\ c$ are still constants and $\in \mathbb{R}$
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note if x=0 is not interesting as the derivative terms disappear.
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how to solve? two approaches:
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textbook only use 2nd method. prof doesn't like this.
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you can find both methods in the profs notes.
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you know Stewart? multimillionaire, he's living in a mansion in Ontario.
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introduce change of variables:
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$x=e^t\Rightarrow t=\ln x$ (x is always +)
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(do $x=-e^t$ if you need it to be negative.)
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find derivatives with respect to t now. y is a function of t which is a function of x.
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$\frac{dy}{dx}=\frac{dy}{dt}{\frac{dt}{dx}}=\frac{ dy }{ dt }{\frac{1}{x}}\Rightarrow \underset{ Impor\tan t }{ x\frac{dy}{dx}=\frac{dy}{dt} }$
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compute 2nd derivative of y wrt to x:
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$\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2}\left( \frac{dt}{dx} \right)^2+\frac{dy}{dt}{\frac{d^2t}{dx^2}}=\frac{1}{x^2}{\frac{d^2y}{dt^2}}-\frac{\frac{1}{x^2}dy}{dt}$
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$\underset{ \mathrm{Im}por\tan t }{ x^22{\frac{d^2y}{dx^2}}=\frac{d^2y}{dt^2}-\frac{dy}{dt} }$
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$$a\frac{d^2y}{dt^2}+(b-a){\frac{dy}{dt}}+cy=f(e^t)$$
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^ this is a constant coefficient equation now! We can solve it now using prior tools.
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#ex
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solve:
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$$x^2{\frac{d^2y}{dx^2}}+3x{\frac{dy}{dx}}+y=x^{-1},\ x>0$$
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$x=e^t$
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transform using the technique we showed just earlier:
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$\frac{d^2y}{dt^2}+2{\frac{dy}{dt}}+y=e^{-t}$
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1) $r^2+2r+1=0$
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$r_{1,2}=-1$
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$y_{h}(t)=c_{1}e^{-t}+c_{2}te^{-t}$
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2) $y_{p}(t)=At^2e^{-t}$ <- using method of undetermined coefficients
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$A=\frac{1}{2}$
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general solution in terms of t:
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$y_{1}(t)=c_{1}e^t+c_{2}te^{-t}+\frac{1}{2}t^2e^{-t}$
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bottom line: solution in terms of t, but we want solution wrt to x:
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$y_{1}(x)=c_{1}e^{-\ln(x)}+c_{2}\ln(x)e^{-\ln(x)}+\frac{1}{2}\ln(x)^2e^{-\ln(x)}$
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$=c_{1}x^-1+c_{2}\ln(x)x^-1+\frac{1}{2}{\ln(x)^2}x^-1$
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#end of lecture 10
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