forked from Sasserisop/MATH201
revised up to and including method of undetermined coefficients
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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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#start of lecture 11
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last lecture we did cauchy euler equations:
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$ax^2{\frac{d^2y}{dx^2}+bx{\frac{ dy }{ dx }}+cy=f(x)},\ x>0$
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where $a,\ b,\ c$ are still constants and $\in \mathbb{R}$
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1) $x=e^t$
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$a{\frac{d^2y}{dt^2}}+(b-a){\frac{dy}{dt}}+cy=f(e^t)$ <- lousy notation, the y here isnt quite the same as in the above definition.
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2) $y=x^r$
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$ar^2+(b-a)r+C=0$
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three cases:
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(i) $r_1\ne r_{2}$
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then: $y_{h}(x)=c_{1}x^{r_{1}}+c_{2}x^{r_{2}}$
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(ii) $r_{1}=r_{2}=r$
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then: $y_{h}(x)=c_{1}x^r+c_{2}x^r\ln(x)$
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(iii) $r_{1,2}=\alpha+i\beta$
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then: $y_{h}(x)=x^2(c_{1}\cos(\ln \beta x)+c_{2}\sin \ln(\beta x))$
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now find one particular solution for a non homogenous soultion, using variation of parameters, combine the y_h and y_p to get y(x).
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not all equations can fall into cauchy euler type.
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$y''+p(x)y'+q(x)y=f(x)$ (1) <- no general solution procudure always
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but, if $y_{1}(x)$ solves $y''+p(x)y'+q(x)y=0$
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then we can find the general solution to the non homogenous equation (1) by guessing it in the form $y(x)=v(x)y_{1}(x)$
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$y'=v'y_{1}+vy_{1}'$
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$y''=v''y_{1}+2v'y_{1}'+vy_{1}''$
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$(v''y_{1}+2v_{1}'y_{1}'+y_{1}''v)+p(x)(v'y_{1}+vy_{1}')+q(x)vy_{1}=f(x)$
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$v\cancelto{ 0 }{ (y_{1}''+p(x)y_{1}'+q(x)y_{1}) }+v''y_{1}+(2y_{1}'+p(x)y_{1})v'=f$
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$y_{1}v''+()$
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$v''+\left( \frac{2y_{1}'}{y_{1}}+p \right)v'=\frac{f}{y_{1}}$
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$v'=u$
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$u'+\left( \frac{2y_{1}'}{y_{1}}+p \right)u=\frac{f}{y_{1}}$<- this is a linear first order equation
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how to solve linear first order equation? we compute the integrating factor $\mu$
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$\mu=e^{\int(2y_{1}'/y_{1}+p)dx}=e^{\ln(y_{1})^2}e^{\int P(x) \, dx}=y_{1}^2e^{\int p(x) \, dx}$
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isnt this nice? some kind of magic. We made some guesses and we arrived somewhere.
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#ex find the general solution to the equation:
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$y''+4xy'+(4x^2+2)y=8e^{-x(x+2)}$
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if $y_{1}(x)=e^{-x^2}$ is one solution.
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therefore were finding the solution of the form: $y(x)=v(x)y_{1}=v(x)e^{-x^2}$
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$v'=u$
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$u'+\left( \frac{2y_{1}'}{y_{1}}+4x \right)u=\frac{8{e^{-x^2}e^{-2x}}}{e^{-x^2}}$ <-(p(x)=4x)
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$u'+\left( \frac{2{e^{-x^2}(-2x)}}{e^{-x^2}}+4x \right)u=8e^{-2x}$
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$u'=8e^{-2x}$
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$u=-4e^{-2x}+c_{1}$
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$v'=u=-4e^{-2x}+c_{1}$
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$v(x)=2e^{-2x}+c_{1}x+c_{2}$
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general solution:
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$$y(x)=v(x)y_{1}(x)=(2e^{-2x}+c_{1}x+c_{2})e^{-x^2}$$
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## Free vibrations
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$mr^2+br+k=0$ characteristic polynomail
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(i) $r_{1}\ne r_{2}$ $b^2-4mk>0$
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$y_{h}(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}$
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$r_{1,2}=-\frac{b}{2m}\pm \frac{\sqrt{ b^2-4mk }}{2m}<0$
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then the limit of the homogenous solution is 0 as t->$\infty$ (over damped case)
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(ii) $r_{1}=r_{2}=-\frac{b}{2m}$
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$r_{1}=r_{2}=-\frac{b}{2m}$
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$y_{h}(t)=e^-\frac{b}{2m}+c_{2}te^{-b/2m}t$ limit =0 as t approches inf critically damped
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#start of lec 8 (sept 22)
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last lecture we talked about $ay''+b'y+cy=f(t)$
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in the case when $f(t)=0$ :
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1) $ay''+b'y+cy=0$
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then $ar^2+br+c=0$ and solve with quadratic formula
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general solutions are:
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if $r_{1}\ne r_{2}\Rightarrow y_{h}(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}$ <- **overdamped**
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if $r_{1}=r_{2}\Rightarrow y_{h}(t)=c_{1}e^{rt}+c_{2}te^{rt}$ <- **critically damped**
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if $r_{1,2}\in \mathbb{C}\Rightarrow y_{h}(t)=e^{\alpha t}(c_{1}\cos(\beta t)+c_{2}\sin(\beta t))$ <- **underdamped**
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where h means homogenous, (when $f(t)=0$ the equation is homogenous.)
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in the case when $f(t)\ne 0$ :
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2) If $y_{p}(t)$ solves 1) then the general solution to $y(t)$ is $y(t)=y_{h}(t)+y_{p}(t)$
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theorem: if $p(t),\ q(t),\ f(t)$ are continuous on $I$ then the following IVP has a unique solution: $y''+p(t)y'+q(t)y=f(t)\quad \text{where}\quad y''(t_{o}),\ y'(t_{o}),\ y(t_{o})\in I$
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---
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# Method of undetermined coefficients:
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#ex #mouc Find the general solution for:
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$$y''-4y'+4y=3t+9$$
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The equation is certainly non-homogenous.
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First we have to find general solution to the homogenous equation (ie: find $y_{h}(t))$:
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1) $y''-4y'+4y=0$
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characteristic eq: $r^2-4r+4=0$
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$r=2$ (repeated root)
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$y_{h}(t)=c_{1}e^{2t}+c_{2}te^{2t}$
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Good. Now we need $y_{p}(t):$
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Look at the equation again: $y''+{-4}y'+4y=3t+9$
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We are looking for a particular polynomial where the power is not greater than 1. Because if for example $y_{p}(t)=t^2$ then the LHS would be a degree 2 polynomial and yet the RHS is only a degree one polynomial.
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So we guess that the equation will be of the form:
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2) $y_{p}(t)=At+B$
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$y_{p}'=A,\ y_{p}''=0$
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$0-4A+4(At+B)=3t+9$
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$4A=3,\ -4A+4B=9$
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$A=\frac{3}{4},\ B=3$
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$y_{p}(t)=\frac{3}{4}t+3$ <- our guess worked!
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general solution: $$y(t)=c_{1}e^{2t}+c_{2}te^{2t}+\frac{3}{4}t+3$$
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So the big takeaway from this example is if the RHS of the eq is a polynomial of degree u, we try to find a solution as a polynomial of degree u
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#ex #second_order_nonhomogenous #mouc
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Find the general solution of the following:
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$$y''-4y'+4y=2e^{2t}$$
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1) $y_h(t)=c_{1}e^{2t}+c_{2}te^{2t}$ (computed earlier)
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2) $y_p(t)=\ ?$
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we observe the RHS is some exponential, we need the function + its derivative + its second derivative to equal that, we have no option but suspect that its $Ae^{2t}$
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but then the LHS becomes 0! -> $4Ae^{2t}-4\cdot 2Ae^{2t}+4Ae^{2t}=0$
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so $Ae^{2t}$ is a wrong guess.
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So what do we do? Let's try $Ate^{2t}$ take $c_{2}=A, c_{1}=0$, this does not work again. LHS becomes 0 again -> $A(t4e^{2t}+2e^{2t}+2e^{2t})-4A(t2e^{2t}+e^{2t})+4Ate^{2t}=0$
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so let's try $At^2e^{2t}$:
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$A(\cancel{ t^24e^{2t} }+2e^{2t}2t+e^{2t}2+2t2e^{2t})-4A(\cancel{ t^22e^{2t} }+e^{2t}2t)+\cancel{ 4At^2e^{2t} }$
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$=8Ate^{2t}+2Ae^{2t}-8Ate^{2t}$
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$=2Ae^{2t}=2e^{2t},\ A=1$ This one works!
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we know the homogenous solution and the particular solution. Sum them together to get the general solution:
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$$y(t)=c_{1}e^{2t}+c_{2}te^{2t}+t^2e^{2t}$$
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Moral of story? if RHS is constant times $e^{2t}$ we guess with an exponent with a constant, if its homogenous we multiply by t, if still not a valid solution then we multiply by t again.
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#ex #IVP #second_order_nonhomogenous #mouc
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$$y''+2y'+2y=2e^{-t}+5\cos t \qquad y(0)=3,\ y'(0)=1$$
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We wanna solve this IVP! We know from earlier that it must have a unique solution.
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1) set RHS to 0: $r^2+2r+2=0$
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$r_{1,2}=-1\pm i$
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>someone mentions sqrt(i). sqrt(i) is interesting, but not the topic for today.
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$y_{h}(t)=e^{-t}(c_{1}\cos(t)+c_{2}\sin(t))$
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2) $y_{p}(t)=\ ?$
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RHS is much more complicated, sum of 2 functions. Lets use principle of super position:
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$y_{p}(t)=y_{p_{1}}(t)+y_{p_{2}}(t)$
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where $y_{p_{1}}$ solves $y''+2y'+2y=2e^{-t}$
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$y_{p_{2}}$ solves $y''+2y'+2y=5\cos (t)$
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lets try $y_{p_{1}}=Ae^{-t}$ Does this work? look at it, A must be zero but if A is zero you still get problems.
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$y_{p_{1}}'=-Ae^{-t}$
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$y_{p_{1}}''=Ae^{-t}$ plug in these three and we find that A=2
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second equation, not so easy:
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solution of $\cos(t)$ doesn't quite work because the LHS will obtain a $\sin$ term. Lets try this instead:
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$y_{p_{2}}=A\cos(t)+B\sin(t)$
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$y_{p_{2}}'=-A\sin(t)+B\cos (t)$
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$y_{p_{2}}''=-A\cos t-B\sin t$
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$(A+2B)\cos(t)+(-2A+B)\sin(t)=5\cos(t)$
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$A+2B=5$
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$-2A+B=0$ -> solving the system of linear equations yields: A=1, B=2
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but $y_{p_{1}}\ne y_{p_{2}}$ because of the $e^{-t}$ term.
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The general solution is:
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$y(t)=c_{1}e^{-t}\cos(t)+c_{2}e^{-t}\sin t+2e^{-t}+\cos t+2\sin t$
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now we solve the IVP:
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$y(0)=3=c_{1}+3=3\implies c_{1}=0$
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$y'(t)=c_{2}e^{-t}\cos(t)+\sin(t)(-1)e^{-t}-2e^t-\sin(t)+2\cos(t)$
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$y'(0)=c_{2}+0-2+0+2$
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$y'(0)=1=c_{2}$
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final solution to IVP:
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$$y(t)=e^{-t}(\sin t+2)+\cos t+2\sin t$$
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#end of lec 8 #start of lec 9
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*remember in a previous example when we had to guess that $y_{p}=At^2e^{2t}$? Here is a generalized algorithm that can find $y_{p}$ when the RHS falls under the following form. Reducing the guess work to zero:*
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# Generalized guesses for undetermined coefficients:
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case i) $ay''+by'+cy=P_m(t)e^{rt}$
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where $P_{m}(t)=a_{m}t^m+a_{m-1}t^{m-1}+ \dots +a_{0}$ *i.e. P is a polynomial degree m.*
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Then we guess the particular solution is of the form: $y_{p}(t)=t^s(b_{m}t^m+b_{m-1}t^{m-1}+\dots+b_{0})e^{rt}$
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where:
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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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case ii) $ay''+by'+cy=P_{m}(t)e^{\alpha t}\cos(\beta t)+P_{m}(t)e^{\alpha t}\sin(\beta t)$
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Then we guess the particular solution is of the form: $y_{p}(t)=t^s[(A_{k}t^k+A_{K-1}t^{k-1}+\dots+A_{0})e^{\alpha t}\cos(\beta t)+(B_{k}t^k+B_{k-1}t^{k-1}+\dots+B_{0})e^{\alpha t}\sin(\beta t)]$
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where:
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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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#start of lec 8 (sept 22)
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last lecture we talked about $ay''+b'y+cy=f(t)$
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in the case when $f(t)=0$ :
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1) $ay''+b'y+cy=0$
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then $ar^2+br+c=0$ and solve with quadratic formula
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general solution is: $y_{h}(t)=c_{1}e^{r_{1}(t)}+c_{2}e^{r_{2}t}$ where h means homogenous, ( because when =0 its homogenous)
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if $r_{1}=r_{2}$ then $y_{h}(t)=c_{1}e^{r(t)}+c_{2}e^{rt}$
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if imaginary roots:
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$y_{h}(t)=e^{\alpha t}(c_{1}\cos(\beta t)+c_{2}\sin(\beta t))$
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2) If $y_{p}(t)$ solves 1) then its general solution is $y(t)=y_{h}(t)+y_{p}(t)$
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theorem: if $p(t),\ g(t),\ f(t)$ are continuous on $I$ then the IVP $y''+p(t)y'+q(t)y=f(t), y(t_{o}),\ y'(t_{o})=y_{1} t_{o}\in I$ has a unique solution
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method of undetermined coeffecients:
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#ex
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$y''\pm_{4}y'+4y=3t+9$ lets find general solution, its centainly non homogenous.
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first we have to find general solution to the homogenous equation:
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1) $y''-4y'+4y=0$
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characteristic eq: $r^2-4r+4=0$ what are the roots?
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$r=2$ (repeated solution)
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$y_{h}(t)=c_{1}e^{2t}+c_{2}te^{2t}$
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we are looking for a particular polynomial where the power is not greater than 1 (?)
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2) $y_{p}(t)=At+B$
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$y_{p}'=A,\ y_{p}''=0$
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$-4A+4(At+B)=3t+9$
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$4A=3,\ -4A+4B=9$
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$A=\frac{3}{4},\ B=3$
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$y(t)=\frac{3}{4}t+3$
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general solution: $$y(t)=c_{1}e^{2t}+c_{2}te^{2t}+\frac{3}{4}t+3$$
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so big takeaway is if the RHS of eq is a polynomial of degree u, we try to find a solution as a polynomial of degree u
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#ex
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$$y''-4y'+4y=e^{2t}$$
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find general solution.
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1) $y_h(t)=c_{1}e^{2t}+c_{2}te^{2t}$ (computed earlier)
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2) $y_p(t)$
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we observe the RHS is some exponential, we need the derivative + its second derivative to equal that, we have no option but suspect that its $Ae^{2t}$
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but then the LHS becomes 0!
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so $Ae^{2t}$ is a wrong guess.
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so what do we do? try $Ate^{2t}$ take $c_{2}=A, c_{1}=0$, this does not work again. LHS becomes 0 again
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so try $At^2e^{2t}$
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$2Ae^{2t}=2e^{2t},\ A=1$ This one works!
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we know the homogeenous solution.
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$y(t)=c_{1}e^{2t}+c_{2}te^{2t}+t^2e^{2t}$ is the general solution
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moral of sotry? if RHS is constant times $e^2t$ we guess with an exponent with a constant, if its homogenous we multiply by t, if still not a valid solution then we multiply by t again.
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Ex:
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$y''+2y'+2y=2e^{-t}+\cos t,\ y(0)=3,\ y'(0)=1$ I wanna solve this IVP! it must have a unique solution.
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1) set RHS to 0: $r^2+2r+2=0$
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$r_{1,2}=-1\pm i$ sqrt(i) is interesting, but not the topic for today.
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$y_{h}(t)=e^{-t}(c_{1}\cos(t)+c_{2}\sin(t))$
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2) $y_{p}(t)=$
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RHS is much more complicated, sum of 2 functions. Lets use principle of super position
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$y_{p}(t)=y_{p_{1}}+y_{p_{2}}$
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where $y_{p_{1}}$ solves $y''+2y'+2y=2e^{-t}$
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$y_{p_{2}}$ solves $y''+2y'+2y=5\cos (t)$
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lets try $y_{p_{1}}=Ae^{-t}$ does this work? look at it, A must be zero but if A is zero you still get problems.
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$y_{p_{1}}'=-Ae^{-t}$
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$y_{p_{1}}''=Ae^-t$ plug in these three and we find that A=2
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second equation, not so easy:
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solution of cos t doenst quite work
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$y_{p_{2}}=A\cos(t)+B\sin(t)$
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$y_{p_{2}}'=-A\sin(t)+b\cos (t)$
|
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$y_{p_{2}}''=-A\cos t-B\sin t$
|
||||
$(A+2B)\cos(t)+(-2A+B)\sin(t)=5\cos(t)$
|
||||
$A+2B=0$
|
||||
$-2A+B=0$ -> A=1, B=2
|
||||
but $y_{p_{1}}\ne y_{p_{2}}$ because of the $e^{-t}$ term
|
||||
$y(t)=c_{1}e^{-t}\cos(t)+c_{2}e^{-t}\sin t+2e^{-t}+\cos t+2\sin t$
|
||||
$y(0)=3=c_{1}+3=3\implies c_{1}=0$
|
||||
$y'(0)=1=c_{2}$
|
||||
final solution $y(t)=e^{-t}(\sin t+2)+\cos t+2\sin t$
|
||||
|
||||
If we have an equation of the from:
|
||||
1) $ay''+by'+cy=P_{m}(t)e^{rt}$
|
||||
where $p_{m}(t)=a_{m}t^m+a_{m-1}t^{m-1}+ \dots +a_{0}$
|
||||
then the guess is: $y_{p}(t)=t^s(b_{mt}t^m+b_{m-1}t^{m-1}+\dots+b_{0})e^{rt}$
|
||||
(i) s=0 if r is not a characteristic polynomial
|
||||
(ii)) s=1 if r is a single root
|
||||
(iii) s=2 if r is a double root
|
||||
we will talk about this more in the coming lecture.
|
||||
#end of lec 8
|
||||
|
|
@ -16,9 +16,9 @@ The equation $my''+by'+ky=0$ is a homogenous second order equation. (in this cas
|
|||
It's called second order because we have second derivative in the equation.
|
||||
|
||||
|
||||
#ex #second_order
|
||||
#ex #second_order_homogenous
|
||||
### $$y''-4y'+3y=0$$
|
||||
(This is obviously homogenous as stated by prof, although I don't understand why that is.)
|
||||
(This equation is homogenous as the RHS is equal to 0)
|
||||
Imagine there's no y' (meaning no friction) you kind want the derivates to equal itself, an exponential!
|
||||
We guess the solution is of the form $y(t)=e^{rt}$
|
||||
$y(t)=e^{rt}$
|
||||
|
|
@ -38,10 +38,10 @@ and we're done.
|
|||
---
|
||||
#end of lec 5 #start of lec 6
|
||||
|
||||
#ex #IVP #second_order Same equation from last lecture, but now an IVP:
|
||||
#ex #IVP #second_order_homogenous Same equation from last lecture, but now an IVP:
|
||||
|
||||
$$y(t)=c_{1}e^{t}+c_{2}e^{3t} \quad c_{1},c_{2}\in\mathbb{R} \quad\text{let } y(0)=0,\ y'(0)=4\quad \text{ What is } c_{1}, c_{2}?$$
|
||||
> Lemma: $y'(t)=c_{1}y_{1}+c_{2}y_{2}$
|
||||
> Lemma: if $y(t)=c_{1}e^{t}+c_{2}e^{3t}$ then $y'(t)=c_{1}y_{1}+c_{2}y_{2}$
|
||||
> proof: let $y_{1}=e^{r_{1}t}\qquad y_{2}=e^{r_{2}t}$
|
||||
> $y(t)=c_{1}e^t+c_{2}e^{3t}$
|
||||
> $y'(t)=c_{1}r_{1}e^{r_{1}t}+r_{2}c_{2}e^{r_{2}t}$
|
||||
|
|
@ -60,11 +60,12 @@ Solving the linear system of equations gives: $c_{1}=-2,\ c_{2}=2$ which gives t
|
|||
$$y'(t)=-2e^t+2e^{3t}$$
|
||||
|
||||
---
|
||||
# Principle of super position
|
||||
Remember from the example above where I said the "general solution is the sum of the two possibilities"? Let's explore and see why that is:
|
||||
Recap: suppose we have an equation of the form $ay''+by'+cy=0$
|
||||
$y(t)=e^{rt}$
|
||||
then $ar^2+br+c=0$
|
||||
case i) $r_{1},r_{2}=\frac{{-b\pm \sqrt{ b^2-4ac }}}{2a}, {r_{1}}\ne r_{2}$
|
||||
case i) $r_{1},r_{2}=\frac{{-b\pm \sqrt{ b^2-4ac }}}{2a}, {r_{1}}\ne r_{2}$ (over damped case, there are 3 possible cases, see below.)
|
||||
$y_{1}(t)=e^{r_{1}t}\qquad y_{2}(t)=e^{r_{2}t}$
|
||||
$y(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}$ is also a solution. But why? Principle of super position.
|
||||
## Principle of super position:
|
||||
|
|
@ -84,9 +85,9 @@ $y_2(t)=e^{r_{2}t}$ solves $ay''+by'+cy=0$
|
|||
$f_{1}(t)=f_{2}(t)=0$
|
||||
This following can be concluded:
|
||||
$y(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}$ must solve $ay''+by'+cy=0$ by principle of super position.
|
||||
> Yay! Note this is only true when $f_{1}(t)=f_{2}(t)=0$ aka your RHS in the second order equation must be 0.
|
||||
> Yay! Note this is only true when $f_{1}(t)=f_{2}(t)=0$ aka your RHS in the second order equation must be 0. aka the equation is homogenous.
|
||||
|
||||
case ii) $r_{1}=r_{2}=\frac{-b}{2a}$ if $b^2-4ac=0$
|
||||
case ii) $r_{1}=r_{2}=\frac{-b}{2a}$ if $b^2-4ac=0$ (critically damped)
|
||||
if we assume $y_{1}=e^{r_{1}t}, y_{2}=e^{r_{1}t}$ like before then we get:
|
||||
$y(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{1}t}=ce^{r_{1}t}$ <- this doesn't seem like it works! We need two integration constants for a second order equation.
|
||||
$y_{1}(t)=e^{-bt/2a}, y_{2}(t)=te^{-bt/2a}$ for time being we take this as true, we can prove it later.
|
||||
|
|
@ -96,83 +97,94 @@ we can check later at home, but also, how was the idea for this found? He will t
|
|||
### linear algebra 101: linear independence makes unit vectors, which forms a basis.
|
||||
definition: if $y_{1}, y_{2}$ are solutions to $a(t)y''+b(t)y+c(t)=0$ on some interval $I_{1}$
|
||||
then they are called linearly independent if none of them is a constant multiple of the other.
|
||||
Theorem: If $y_{1}(t), y_{2}(t)$ are linearly independent solutions to $ay''+by'+cy=0$ then any other solution can be written as $y(t)=c_{1}y_{1}(t)+c_{2}y_{2}(t)$
|
||||
Theorem: If $y_{1}(t), y_{2}(t)$ are linearly independent solutions to $ay''+by'+cy=0$ then any other solution can be written as $y(t)=c_{1}y_{1}(t)+c_{2}y_{2}(t)$ (due to super position)
|
||||
|
||||
how do we know the two solutions are linearly independent? Test for linear independence:
|
||||
$y_{1}, y_{2}$ are solutions to $a(t)y''+b(t)y+c(t)y=0$ on some interval $I_{1}$
|
||||
then they are called linearly independent iff
|
||||
then they are called linearly independent iff:
|
||||
$W(y_{1},y_{2})(t)=\det \begin{pmatrix}y_{1} & y_{2} \\y_{1}' & y_{2}'\end{pmatrix}\ne 0$
|
||||
(i) $b^2-4ac>0 \Rightarrow r_{1}\ne r_{2}$
|
||||
case i) $b^2-4ac>0 \Rightarrow r_{1}\ne r_{2}$
|
||||
$y_{1}=e^{r_{1}t}, y_{2}=e^{r_{2}}t$
|
||||
$W(y_{1},y_{2})=\det\begin{pmatrix}e^{r_{1}t} & e^{r_{2}t} \\r_{1}e^{r_{1}t} & r_{2}e^{r_{2}t}\end{pmatrix}$
|
||||
$=e^{t(r_{1}+r_{2})}(r_{2}-r_{1})\ne 0$
|
||||
(ii) $b^2-4ac=0\Rightarrow$
|
||||
$=e^{t(r_{1}+r_{2})}(r_{2}-r_{1})\ne 0$ (because this can only be 0 when $r_{1}= r_{2}$ but we know $r_{1}\ne r_{2}$)
|
||||
case ii) $b^2-4ac=0\Rightarrow$
|
||||
$r_{1}=r_{2}=-\frac{b}{2a}=r$
|
||||
$y_{1}(t)=e^{rt}, y_{2}(t)=te^{rt}$
|
||||
$W(y_{1},y_{2})=\det\begin{pmatrix}e^{rt} & te^{rt} \\re^{rt} & e^{rt}(1+rt)\end{pmatrix}$
|
||||
$=e^{2rt}(1+rt)-rte^{rt}e^{2rt}$
|
||||
$=e^{2rt}(1+rt)-rte^{2rt}$
|
||||
$=e^{2rt}\ne 0$
|
||||
> But what about case iii) ? it wasn't covered here. But now we proven that for case i and ii their two respective solutions are linearly independent and therefore we know we can safely apply the principle of super position on them and obtain their general solutions.
|
||||
|
||||
#ex #IVP #second_order_homogenous
|
||||
$$y''-2y'+y=0 \qquad y(0)=1,\ y'(0)=0$$
|
||||
|
||||
#ex #IVP #second_order
|
||||
$$y''-2y'+y=0, y(0)=1, y'(0)=0$$
|
||||
|
||||
$e^{rt}(\underset{ = }{ r^2-2r+1 })=0$
|
||||
$(r-1)^2=0\Rightarrow r_{1}=r_{2}=1$
|
||||
$e^{rt}({ r^2-2r+1 })=0$
|
||||
$(r-1)^2=0\Rightarrow r_{1}=r_{2}=1$ (this is case ii, critically damped)
|
||||
$y(t)=c_{1}e^t+c_{2}te^t$
|
||||
$y(0)=c_{1}=1$
|
||||
$y(t)=e^t+c_{2}te^t$
|
||||
$y'(0)=0=1+c_{2}\Rightarrow c_{2}=1$
|
||||
|
||||
|
||||
$y'(0)=0$
|
||||
$=e^0+c_{2}(0e^0+1\cdot e^0)$
|
||||
$=1+c_{2}$
|
||||
$\Rightarrow c_{2}=-1$
|
||||
$$y(t)=e^t-te^t$$
|
||||
Wow, that was a lot today, my notes look like a mess. I'll have to clean this up and understand what's going on later.
|
||||
#end of lecture 6
|
||||
#end of lecture 6 #start of lecture 7 (sept 20)
|
||||
|
||||
|
||||
#start of lecture 7 (sept 20)
|
||||
from last class: if $r_{1}=r_{2}=-\frac{b}{2a}=r$
|
||||
$y_{1}=e^{rt}, \quad y_{2}=te^{rt}$
|
||||
general solution:
|
||||
$$y(t)=c_{1}y_{1}(t)+c_{2}y_{2}(t)$$
|
||||
Last lecture we covered case i) and case ii).
|
||||
however there's a third option:
|
||||
$b^2-4ac<0$
|
||||
then we have complex roots
|
||||
$r_{1,2}=-\frac{b}{2a}\pm\frac{i\sqrt{ 4ac-b^2 }}{2a}=\alpha+i\beta$ <- Complex conjugates. And due to fundamental theorem of algebra, there are only 2 roots.
|
||||
Then we have complex roots:
|
||||
$r_{1,2}=-\frac{b}{2a}\pm\frac{i\sqrt{ 4ac-b^2 }}{2a}=\alpha\pm i\beta$ <- Complex conjugates. And due to fundamental theorem of algebra, there are only 2 roots in this degree 2 polynomial equation.
|
||||
|
||||
$e^{r_{1}t}=e^{(\alpha+i\beta)t}=e^{\alpha t}+e^{i\beta t}$
|
||||
side note: there are no numbers that are more than two components that are "useful", even quaternions
|
||||
$e^{i\beta t}=e^{i\theta}$
|
||||
let $e^{i\beta t}=e^{i\theta}$
|
||||
expand into power series:
|
||||
$=1+\frac{i\theta}{1!}+\frac{{(i\theta)^2}}{2!}=\frac{{(i\theta)^3}}{3!}\dots$
|
||||
$=1+\frac{i\theta}{1!}-\frac{\theta^2}{2!}-\frac{i\theta^3}{3!}+\frac{\theta^4}{4!}+\frac{i\theta^5}{5!}+\dots$
|
||||
$=\left( 1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\dots \right)+i\left( \frac{\theta}{1!}-\frac{\theta^3}{2!}+\frac{\theta^5}{3!}\dots\right)$
|
||||
$e^{i\theta}=\cos(\theta)+i\sin(\theta) \quad \Box$ We have proven the Euler formula
|
||||
>Recall from math 101: $e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$
|
||||
|
||||
$y(t)=e^{rt}=e^{\alpha t}(\cos \beta t+i\sin \beta t)$
|
||||
$=1+\frac{i\theta}{1!}+\frac{{(i\theta)^2}}{2!}+\frac{{(i\theta)^3}}{3!}+\dots$
|
||||
$=1+\frac{i\theta}{1!}-\frac{\theta^2}{2!}-\frac{i\theta^3}{3!}+\frac{\theta^4}{4!}+\frac{i\theta^5}{5!}-\dots$
|
||||
$=\left( 1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\dots \right)+i\left( \frac{\theta}{1!}-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\dots\right)$
|
||||
>Recall from math 101: $\cos(x)=\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}\quad\text{and}\quad \sin(x)=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}$
|
||||
|
||||
$e^{i\theta}=\cos(\theta)+i\sin(\theta) \quad \Box$ We have proven the Euler formula.
|
||||
|
||||
we guess the solution is of the form: $y(t)=e^{rt}=e^{(\alpha+i\beta)t}=e^{\alpha t}(\cos \beta t+i\sin \beta t)$
|
||||
Lemma: If $u(t)+iv(t)$ solves $ay''+by'+cy=0$ then $u(t),\ v(t)$ are also solutions.
|
||||
Proof:
|
||||
$a(u+iv)''+b(u+iv)'+c(u+iv)=0$
|
||||
$\underbrace{ { (au''+bu'+cu) } }_{ =0 }+i\underbrace{ (av''+bv'+cv) }_{ =0 }=0$
|
||||
$\underbrace{ { (au''+bu'+cu) } }_{ =0 }+i\underbrace{ (av''+bv'+cv) }_{ =0 }=0$ <- since the RHS is zero, both the real and imaginary parts must also equal zero.
|
||||
$y_{1}(t)=e^{\alpha t}\cos(\beta t),\ y_{2}(t)=e^{\alpha t}\sin(\beta t)$
|
||||
$\alpha=-\frac{b}{2a},\quad \beta={\frac{\sqrt{ 4ac-b^2 }}{2a}}$
|
||||
then by principle of super position:
|
||||
$y(t)=c_{1}y_{1}+c_{2}y_{2}$
|
||||
now we have to test the two solutions are linearly independent
|
||||
$W[y_{1},y_{2}]=\det\begin{pmatrix}y_{1} & y_{2} \\ y_{1}' & y_{2}'\end{pmatrix}\ne0$ <- remember to do/show this at home
|
||||
$W[y_{1},y_{2}]=\det\begin{pmatrix}y_{1} & y_{2} \\ y_{1}' & y_{2}'\end{pmatrix}\ne0$
|
||||
$=\det \begin{pmatrix}e^{\alpha t}\cos(\beta t)& e^{\alpha t}\sin(\beta t)\\ -\beta\sin(\beta t)e^{\alpha t}+\cos(\beta t)\alpha e^{\alpha t} & \beta\cos(\beta t)e^{\alpha t}+\sin(\beta t)\alpha e^{\alpha t}\end{pmatrix}$
|
||||
$=e^{2\alpha t}[\cos\beta t(\beta \cos(\beta t)+\alpha\sin(\beta t))-\sin \beta t(-\beta \sin(\beta t)+\alpha\cos(\beta t))]$
|
||||
$=e^{2\alpha t}(\beta \cos^2(\beta t)+\beta \sin^2(\beta t))$
|
||||
$=e^{2\alpha t}\beta\ne 0$
|
||||
> Yay! We have shown the two solutions are indeed always linearly independent.
|
||||
|
||||
#ex #IVP
|
||||
#ex #IVP #second_order_homogenous
|
||||
$$y''-2y'+5y=0 \quad y(0)=0 \quad y'(0)=2$$
|
||||
$r^2-2r+5=0$<-characteristic equation
|
||||
$r_{1,2}=\frac{{-b\pm \sqrt{ b^2-4ac }}}{2a}$
|
||||
$r_{1,2}={1\pm \frac{\sqrt{ -4b }}{2}}=1\pm2i$
|
||||
$y_{1}=e^t\cos(2t)$ $y_{2}=e^t\sin(2t)$
|
||||
general solution: $y(t)=e^{\alpha t}(c_{1}\cancel{ \cos 2 t }+c_{2}\sin 2 t)$
|
||||
$y(0)=0=c_{1}$
|
||||
$y'(0)=c_{2}(e^t)$
|
||||
*I missed stuff here that he erased*
|
||||
general solution is:
|
||||
general solution: $y(t)=e^t(c_{1} \cos 2 t +c_{2}\sin 2 t)$
|
||||
$y(0)=0=c_{1}$ <- this helps us calculate y' easier as we can cross out the $\cos(2t)$ term before taking the derivative.
|
||||
$y'(t)=\frac{d}{dt}(e^tc_{2}\sin(2t))=e^tc_{2}2\cos(2t)+c_{2}\sin(2t)e^t$
|
||||
$y'(0)=2=c_{2}2$
|
||||
therefore: $c_{1}=0,\ c_{2}=1$ and the solution to the IVP is:
|
||||
$$y(t)=e^t\sin(2t)$$
|
||||
it has a nice graph, where if it was a circuit it would blow up
|
||||
or if it was a bridge it would collapse
|
||||
The solution has a nice graph, where if it was a circuit it would blow up,
|
||||
or if it was a bridge it would oscillate and eventually collapse.
|
||||
|
||||
## Something more difficult now:
|
||||
## In the next lecture:
|
||||
something more difficult now:
|
||||
$ay''+by'+cy=f(t)$ Again, a mass-spring system without any external force.
|
||||
if f(t)=0 we can find the solution easily and use superposition to get the general solution
|
||||
$ay''+by'+cy=0$
|
||||
|
|
@ -185,4 +197,4 @@ $y(t)=c_{1}y_{1}(t)+c_{2}y_{2}(t)+y_{p}(t)$ must solve $ay''+by'+cy=f(t)$
|
|||
|
||||
Theorem: If $a(t),\ b(t),\ c(t)$ are continuous on $I$ , then IVP: $a(t)y''+b(t)y'+c(t)y=f(t)$ ; $y(t_{o})=y_{o}$ \ , $y'(t_{o})=y_{1}$ has a unique solution.
|
||||
we will do the proofs next class.
|
||||
#end of lecture 7
|
||||
#end of lecture 7
|
||||
|
|
@ -44,4 +44,8 @@ $C=-1$
|
|||
So, the answer is:
|
||||
$$y=\arctan(x^2+1)$$
|
||||
|
||||
#end of Lecture 1
|
||||
#end of Lecture 1
|
||||
|
||||
$\int \frac{1}{x^5} \, dx$
|
||||
$a\in\mathbb{C}$
|
||||
$\Rightarrow$
|
||||
|
|
|
|||
|
|
@ -22,7 +22,7 @@
|
|||
{"id":"bebd67e847df16e1","type":"text","text":"Shortcut:\n$$I(x)=e^{\\int (1-n)P(x) \\, dx }$$\n$$y^{1-n}=\\frac{1}{I(x)}\\left( \\int (1-n)I(x)Q(x) \\, dx +C\\right)$$","x":-1333,"y":-430,"width":505,"height":260},
|
||||
{"id":"3f081acda4f30a27","type":"text","text":"solve for $r_{1}$ & $r_{2}$ using quadratic formula","x":1315,"y":183,"width":274,"height":125},
|
||||
{"id":"4ffaa5c9a7e8d22b","type":"text","text":"use principle of super position","x":1329,"y":29,"width":245,"height":96},
|
||||
{"id":"cd7490f8cce0b6e0","type":"file","file":"Math 201/Lectures/Second order linear equations (lec 5-7).md","x":1215,"y":780,"width":474,"height":145},
|
||||
{"id":"cd7490f8cce0b6e0","type":"file","file":"Math 201/Lectures/Second order homogenous linear equations (lec 5-7).md","x":1215,"y":780,"width":474,"height":145},
|
||||
{"id":"cd31ca74652b6936","type":"text","text":"substitute $y(t)=e^{rt}$ and its derivatives in the equation","x":1293,"y":556,"width":317,"height":122},
|
||||
{"id":"e063ab92aef817e4","type":"text","text":"divide both sides of equation by $e^{rt}$","x":1322,"y":382,"width":260,"height":110},
|
||||
{"id":"d98da52cb7139c25","type":"text","text":"$N(x,y)=\\frac{\\partial}{\\partial y} \\int M \\, dx+g(y)$\n","x":514,"y":354,"width":328,"height":77},
|
||||
|
|
|
|||
|
|
@ -1,20 +1,7 @@
|
|||
#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:
|
||||
# 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
|
||||
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
|
||||
$y'_{p}=v_{1}y_{1}+v_{1}y_{1}'+v_{2}'y_{2}+v_{2}y_{2}'$
|
||||
|
|
@ -24,14 +11,14 @@ so $y_{p}''=v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}''$
|
|||
$a(v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}'')+b(v_{1}y_{1}'+v_{2}y_{2}')+c(v_{1}y_{1}+v_{2}y_{2})=f(t)$
|
||||
$v_{1}(ay_{1}''+\cancelto{ 0 }{ by_{1}' }+cy_{1})+v_{2}(ay_{2}''+\cancelto{ 0 }{ by_{2}'C }+cy_{2})+a(v_{1}'y_{1}'+v_{2}'y_{2}')$
|
||||
2) $v_{1}'y_{1}'+v_{2}'y_{2}'=\frac{f(t)}{a}$
|
||||
$\det \begin{pmatrix}y_{1} & y_{2} \\y_{1}'& y_{2}'\end{pmatrix}$ = 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!
|
||||
$\det \begin{pmatrix}y_{1} & y_{2} \\y_{1}'& y_{2}'\end{pmatrix}$ = Wronskian = $W[y_{1},y_{2}]\ne 0$
|
||||
by definition $y_1$ and $y_2$ are linearly independent 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
|
||||
find general solution to homogenous counterpart
|
||||
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
|
||||
|
|
@ -68,4 +55,47 @@ $y(0)=0=c_{1}-y_{p}(0)=c_{1}-\frac{1}{5}\Rightarrow c_{1}=\frac{1}{5}$
|
|||
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
|
||||
#end of lecture 9
|
||||
#start of lecture 10
|
||||
# Variation of parameters
|
||||
last lec we did some variation of parameters
|
||||
$ay''+by'+cy=f(t)$
|
||||
1) $y_{h}(t)=c_{1}y_{1}(t)+c_{2}y_{2}(t)$
|
||||
2) $y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$
|
||||
$y_{1}v_{1}'+y_{2}v_{2}'=0$
|
||||
$y_{1}v_{1}'+y_{2}'v_{2}'=\frac{b}{a}$ or f/a?
|
||||
is the system of equations we will need to solve. You can also memorize a formula but peter likes remembering this system of equations and moving on from there.
|
||||
#ex #variation_of_parameters
|
||||
$$y''-2y'+y=e^t\ln(t)+2\cos(t)$$
|
||||
i) $y_{h}(t)=?$
|
||||
$r^2-2r+1=0$
|
||||
$r_{1,2}=1$
|
||||
$y_{h}(t)=c_{1}e^t+c_{2}te^t$
|
||||
2) $y_{p}(t)=?$
|
||||
$y''-2y'+y=2\cos (t)$
|
||||
$y_{p}''=A\cos(t)+B\sin(t)$ is our first guess. but it does not solve the homogenous eq.
|
||||
$y_{p}'=-\sin(t)$ (obtained by using method of undetermined coefficients, computation not shown.)
|
||||
$y''-2y'+y=e^t\ln(t)$ cant use undetermined coefficients, use variation of parameters
|
||||
$y''_{p}(t)=v_{1}y_{1}+v_{2}y_{2}$
|
||||
$=v_{1}e^t+v_{2}te^t$
|
||||
compute v1 and v2, using the linear system:
|
||||
eq1) $e^t+v_{1}'+te^tv_{2}'=0$
|
||||
eq2) $e^tv_{1}'+(te^t+e^t){v_{2}'}=e^t{\ln t}$
|
||||
subtract eq1 from eq2 $v_{2}'=\ln(t)$
|
||||
$v_{2}(t)=\int \ln(t) \, dt$
|
||||
integrate by parts
|
||||
$=t\ln(t)-\int t\frac{1}{t} \, dt$
|
||||
$=t\ln(t)-t$ no constant of integration.
|
||||
compute $v_{1}$ now:
|
||||
$v_{1}'=-tv_{2}'$
|
||||
$=-t\ln t$
|
||||
integrate to get v_1:
|
||||
$v_{1}=-\int t\ln t \, dt$
|
||||
integrate by parts (btw integration by parts will be the most important integration technique in this course):
|
||||
$v_{1}=-\frac{1}{2}(t^2\ln t)-\int t^2\frac{1}{t} \, dt$
|
||||
$=-\frac{1}{2}\left( t^2\ln t-\frac{t^2}{2} \right)=-\frac{1}{2}t^2\ln t+\frac{1}{4}t^2$
|
||||
$y_{p}''(t)=(\frac{1}{2}t^2\ln t+\frac{1}{4}t^2)e^t+(t\ln t-t)te^t$
|
||||
$y_{p}(t)=-\sin(t)+\frac{1}{2}t^2\ln(t)e^t-\frac{3}{4}t^2e^t$
|
||||
general solution is produced by adding the homogenous eq with $y_{p}(t)$
|
||||
general:
|
||||
$$y(t)=c_{1}e^t+c_{2}te^t+y_{p}(t)$$
|
||||
|
|
@ -7,9 +7,12 @@ I have written these notes for myself, I thought it would be cool to share them.
|
|||
[Bernoulli equations (lec 3)](bernoulli-equations-lec-3.html)
|
||||
[Linear coefficient equations (lec 4)](linear-coefficient-equations-lec-4.html)
|
||||
[Exact equations (lec 4-5)](exact-equations-lec-4-5.html)
|
||||
[Second order linear equations (lec 5-7)](second-order-linear-equations-lec-5-7.html) (raw notes, not reviewed or revised yet.)
|
||||
[More second order stuff (lec 8)](more-second-order-stuff-lec-8.html) (raw notes, not reviewed or revised yet.)
|
||||
[Undetermined coefficients (lec 9)](undetermined-coefficients-lec-9.html) (raw notes, not reviewed or revised yet.)
|
||||
[Second order homogenous linear equations (lec 5-7)](second-order-homogenous-linear-equations-lec-5-7.html)
|
||||
[Method of undetermined coefficients (lec 8-9)](method-of-undetermined-coefficients-lec-8-9.html)
|
||||
[Variation of parameters (lec 9-10)](variation-of-parameters-lec-9-10.html) (raw notes, not reviewed or revised yet.)
|
||||
[Free vibrations (lec 11)](free-vibrations-lec-11) (raw notes, not reviewed or revised yet.)
|
||||
</br>
|
||||
[How to solve any DE, a flow chart](Solve-any-DE.png)
|
||||
</br>
|
||||
I'd like to add a search by tag feature. I'm also thinking of hosting the source code for all this on a git server. That way, people can contribute and fix my notes for me :P
|
||||
It would also allow people to contribute or host their own notes, which would be pretty cool. (Side note: I am against sharing instructor materials without their approved consent.)
|
||||
Loading…
Reference in New Issue