forked from Sasserisop/MATH201
43 lines
2.5 KiB
Markdown
43 lines
2.5 KiB
Markdown
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 very important!*
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#cauchy-euler equations are equations in the form:
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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: x=0 is not interesting as the derivative terms disappear.
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How to solve? There are two approaches:
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Textbook only use 2nd method, prof doesn't like this. You can find both methods in the profs notes. Btw, do 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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$\frac{dy}{dx}=\frac{dy}{dt}{\frac{dt}{dx}}=\frac{ dy }{ dt }{\frac{1}{x}}\Rightarrow \underset{ \text{Important} }{ x\frac{dy}{dx}=\frac{dy}{dt} }$
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compute 2nd derivative of y wrt to x:
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$\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2} \frac{dt}{dx}\cdot \frac{dt}{dx}+\frac{dy}{dt}{\frac{d^2t}{dx^2}}=\frac{1}{x^2}{\frac{d^2y}{dt^2}}-\frac{\frac{1}{x^2}dy}{dt}$
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$\underset{ \text{Important} }{ x^2{\frac{d^2y}{dx^2}}=\frac{d^2y}{dt^2}-\frac{dy}{dt} }$
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plugging those derivatives in we get:
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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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## Example:
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#ex #second_order #second_order_nonhomogenous #cauchy-euler
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Find the general solution for:
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$$x^2{\frac{d^2y}{dx^2}}+3x{\frac{dy}{dx}}+y=x^{-1},\ x>0$$
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substitute: $x=e^t$
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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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$y(t)=c_{1}e^{-t}+c_{2}te^{-t}+\frac{1}{2}t^2e^{-t}$
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but we want solution in terms of x:
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$y(x)=c_{1}e^{-\ln(x)}+c_{2}\ln(x)e^{-\ln(x)}+\frac{1}{2}\ln(x)^2e^{-\ln(x)}$ <- This is rather lousy notation, the y here isn't the same as the y above. Conceptually though, it's all oke doke.
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$$y(x)=c_{1}x^{-1}+c_{2}\ln(x)x^{-1}+\frac{1}{2}{\ln(x)^2}x^{-1}$$
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We are done.
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#end of lecture 10 |