62 lines
3.6 KiB
Markdown
62 lines
3.6 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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$$a\frac{d^2y}{dt^2}+(b-a){\frac{dy}{dt}}+cy(t)=f(e^t)$$
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^ this is a constant coefficient second order non-homogenous 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 #start of lecture 11
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Last lecture we did Cauchy Euler equations:
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$$ax^2{\frac{d^2y}{dx^2}+bx{\frac{ dy }{ dx }}+cy=f(x)} \qquad x>0$$
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where $a,\ b,\ c$ are constants and $\in \mathbb{R}$
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substitute $x=e^t$
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$a{\frac{d^2y}{dt^2}}+(b-a){\frac{dy}{dt}}+cy=f(e^t)$ <- lousy notation, the y here isn't quite the same as in the above definition.
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substitute: $y=x^r$
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after calculating derivatives, plugging in, and simplifying we obtain the polynomial equation:
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$ar^2+(b-a)r+C=0$
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Three cases:
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**(i)** $r_1\ne r_{2}$ then:
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$y_{h}(t)=c_{1}e^{rt}+c_{2}e^{rt}$
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$y_{h}(x)=c_{1}x^{r_{1}}+c_{2}x^{r_{2}}$ (lousy notation, because the two $y_{h}$ do not equal each other)
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**(ii)** $r_{1}=r_{2}=r$ then:
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$y_{h}(t)=c_{1}e^{rt}+c_{2}te^{rt}$
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$y_{h}(x)=c_{1}x^r+c_{2}x^r\ln(x)$ (derived by reduction of order.)
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**(iii)** $r_{1,2}=\alpha\pm i\beta$ then:
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$y_{h}=e^\alpha(c_{1}\cos(\beta t)+c_{2}\sin(\beta t))$
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$y_{h}(x)=x^\alpha(c_{1}\cos(\beta\ln x)+c_{2}\sin(\beta \ln x))$
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Now compute your particular solution, $y_{p}$, and combine with $y_{h}$ to obtain your general solution. |