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added lec 12
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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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@ -0,0 +1,120 @@
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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 polynomial
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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 approaches inf (critically damped)
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#end of lec 11 #start of lec 12 (oct 2 2023)
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![[Drawing 2023-10-02 13.02.06.excalidraw]]
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let $\omega =\frac{\sqrt{ 4mk-b^2 }}{2m}$ (angular frequency)
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then the underdamped case is:
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$y(t)=(c_{1}\cos \omega t+c_{2}\sin \omega t)e^{\frac{-b}{2m}t}$
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we know the trig identity:
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$\sin(\alpha+\beta)=\sin \alpha\cos \beta+\cos \alpha \sin \beta$
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cant make c_1 c_2 sin or cos so what we do?
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do a power transform to convert cartesian into cylindrical coordinates
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$c_{1}=A\sin \phi$
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$c_{2}=A\cos \phi$
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then:
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$Ae^{-bt/2m}(\sin \phi \cos \omega t+\cos \phi \sin \omega t)$
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$=Ae^{-bt/2m}\sin(\omega t+\phi)$ where $\phi$ is the phase shift.
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and $\frac{\omega}{2\pi}$ is the natural frequency
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$\frac{2\pi}{\omega}$ is the period
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but this is all classical mechanics, but beatifully the world of electronic circuits of R L C also has these equations. Biology too. Nature is beautiful and harmonic.
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btw we know $A=\sqrt{ c_{1}^2+c_{2}^2 }$
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and $\tan \phi=\frac{c_{1}}{c_{2}}$
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so we can get A and phi from c_1 and c_2.
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this under damped case also reaches 0 as t->$\infty$
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this system in the drawing is in free vibrattion (RHS=0 means no external force=free vibration.)
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#ex
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$y''+by'+25y=0 \qquad y(0)=1\quad y'(0)=0$
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1) b=0 -> no friction in the system (undamped)
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$b^2-4mk$
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$y(t)=c_{1}\cos 5t+c_{2}\sin 5t$
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$y(0)=c_1=1$
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$y'(0)=0=c_{2}$
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then $\sin 5t\Rightarrow y(t)=\cos(5t)=\sin\left( 5t+\frac{\pi}{2} \right)$ (by trig identity)
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important take away from undamped case: amplitude is constant 1, oscillates forever.
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2) b=6
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compute $b^2-4mk=36-4*25=-64$
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$r_{1,2}=-\frac{6}{2}\pm4i$
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$y(t)=e^{-3t}(c_{1}\cos4t+c_{2}\sin4t)$
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still under damped situation.
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$y(0)=1=c_{1}$
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$y'(0)=0=-3c_{1}+4c_{2}\Rightarrow c_{2}=\frac{3}{4}$
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$A=\frac{5}{4}$
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$\tan \phi=\frac{4}{3}$
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$\phi \approx 0.9273\dots$
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$$y(t)=\frac{5}{4}e^{-3t}\sin(4t+\phi)$$
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"I know engineers loves calculators, I know mathematicians hates calculators, and that's probably the only difference between mathematicians and engineers." -Peter (referring to calculating arctan(4/3) on an exam)
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3) b=10
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$r_{1,2}=-5$
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$y(t)=(c_{1}+c_{2}t)e^{-5t}$
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$y(0)=1=c_{1}$
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$y'(0)=c_{2}-5c_{1}=0$
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$c_{2}=5$
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$y(t)=(1+5t)e^{-5t}\rightarrow0_{as\ t\to\infty}$
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$y(t)=(1+5t)e^{-5t}>0$
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4) b=12
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$r_{1,2}=-6\pm \sqrt{ 11 }$
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$y(t)=c_{1}e^{(-6\pm \sqrt{ 11 })t}+c_{2}e^{(-6-\sqrt{11 })t}$
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$y(0)=c_{1}+c_{2}=1$
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$y'(0)=(-6+\sqrt{ 11 })c_{1}+(-6-\sqrt{ 11 })c_{2}=0$
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$c_{1}=\frac{11+6\sqrt{ 11 }}{22}$
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$c_{2}=\frac{{11-6\sqrt{ 11 }}}{22}$
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this is an over damped case.
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lets look at the graphs: (graphs featuring the three cases shown on projector.)
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#end of lec 12
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@ -11,7 +11,7 @@ if $r_{1}=r_{2}\Rightarrow y_{h}(t)=c_{1}e^{rt}+c_{2}te^{rt}$ <- **critically da
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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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But what about 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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@ -106,4 +106,4 @@ case ii) $ay''+by'+cy=P_{m}(t)e^{\alpha t}\cos(\beta t)+P_{m}(t)e^{\alpha t}\sin
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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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s=1 if $\alpha+i\beta$ is a root.
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@ -3,7 +3,7 @@
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{"id":"767241b95828457e","type":"text","text":"Integrate both sides","x":-57,"y":-360,"width":352,"height":80},
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{"id":"3572d3ccf3a666dc","type":"text","text":"# Solution","x":-43,"y":-520,"width":325,"height":90},
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{"id":"ac4eb08e6ceeccbd","type":"text","text":"# How to solve (almost) any differential equation:","x":-561,"y":-760,"width":1296,"height":136},
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{"id":"76786ab85409e204","type":"text","text":"updated Sept 20","x":778,"y":-729,"width":301,"height":74},
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{"id":"76786ab85409e204","type":"text","text":"updated Oct 1","x":778,"y":-729,"width":301,"height":74},
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{"id":"c6a536ee57248cfd","type":"file","file":"Math 201/Lectures/Separable equations (lec 1).md","x":-102,"y":-170,"width":441,"height":140},
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{"id":"df03f8f1000d374a","type":"text","text":"If $y'=G(ax+by)$ \n\nsubstitute u=ax+by\n$\\frac{du}{dx}=a+b\\frac{ dy }{ dx}$","x":-251,"y":121,"width":341,"height":165},
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{"id":"1044e4c2c0610a1e","type":"text","text":"if $y'=G\\left( \\frac{y}{x} \\right)$ \n\nsubstitute $u=\\frac{y}{x}$\n$\\frac{dy}{dx}=u+x{\\frac{du}{dx}}$","x":150,"y":121,"width":315,"height":165},
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@ -20,11 +20,6 @@
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{"id":"ece672db8e16ac5a","type":"text","text":"substitute $x=u+k$\n$y=v+l$","x":9,"y":769,"width":242,"height":156},
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{"id":"120b3008bd08d69a","type":"file","file":"Math 201/Lectures/Linear coefficient equations (lec 4).md","x":-81,"y":1035,"width":422,"height":143},
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{"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},
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{"id":"3f081acda4f30a27","type":"text","text":"solve for $r_{1}$ & $r_{2}$ using quadratic formula","x":1315,"y":183,"width":274,"height":125},
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{"id":"4ffaa5c9a7e8d22b","type":"text","text":"use principle of super position","x":1329,"y":29,"width":245,"height":96},
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{"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},
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{"id":"cd31ca74652b6936","type":"text","text":"substitute $y(t)=e^{rt}$ and its derivatives in the equation","x":1293,"y":556,"width":317,"height":122},
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{"id":"e063ab92aef817e4","type":"text","text":"divide both sides of equation by $e^{rt}$","x":1322,"y":382,"width":260,"height":110},
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{"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},
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{"id":"ba1ef733f104894d","type":"text","text":"$M(x,y)=\\frac{\\partial}{\\partial x} \\int N \\, dy+g(x)$\n","x":875,"y":354,"width":340,"height":77},
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{"id":"3cc6d2966364d19c","type":"text","text":"$F(x,y)=\\int M \\, dx+g(y)$","x":520,"y":497,"width":315,"height":94},
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@ -35,9 +30,28 @@
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{"id":"d172c6ad7421b458","type":"file","file":"Math 201/Lectures/Exact equations (lec 4-5).md","x":674,"y":903,"width":393,"height":133},
|
||||
{"id":"d3ffeeccd9e88489","type":"text","text":"$M(x,y)dx+N(x,y)dy=0$","x":674,"y":1036,"width":342,"height":83},
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{"id":"937768cc91c15daa","type":"text","text":"$(a_{1}x+b_{1}y+c_{1})dx+(a_{2}x+b_{2}y+c_{2})dy=0$","x":-81,"y":1178,"width":516,"height":63},
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||||
{"id":"59f62d39b48e7b57","x":1663,"y":-306,"width":250,"height":60,"type":"text","text":"under-damped"},
|
||||
{"id":"7089887c01722e83","x":1880,"y":-173,"width":250,"height":60,"type":"text","text":"over-damped"},
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||||
{"id":"b3e73030feee12da","x":1843,"y":33,"width":287,"height":50,"type":"text","text":"critically-damped"}
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{"id":"b3e73030feee12da","type":"text","text":"case 2)\ncritically-damped\n$r_{1}=r_{2}=r$","x":1420,"y":177,"width":300,"height":140},
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{"id":"59f62d39b48e7b57","type":"text","text":"case 3)\nunder-damped\n$r_{1,2}=\\alpha\\pm \\beta i$","x":1760,"y":177,"width":280,"height":140},
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{"id":"7089887c01722e83","type":"text","text":"case 1)\nover-damped $r_{1}\\ne r_{2}$","x":1140,"y":177,"width":240,"height":140},
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||||
{"id":"fb1b8ea1fa7e997b","type":"text","text":"general solution:\n$y_{h}(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}$","x":1079,"y":-83,"width":281,"height":103},
|
||||
{"id":"21dddcaf48c1f19e","type":"text","text":"general solution:\n$y_{h}(t)=c_{1}e^{rt}+c_{2}te^{rt}$","x":1410,"y":-83,"width":320,"height":90},
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||||
{"id":"e85838ac18ea4a1e","type":"text","text":"general solution:\n$y_{h}(t)=e^{\\alpha t}(c_{1}\\cos\\beta t+c_{2}\\sin\\beta t)$","x":1760,"y":-83,"width":380,"height":103},
|
||||
{"id":"cd7490f8cce0b6e0","type":"file","file":"Math 201/Lectures/Second order homogenous linear equations (lec 5-7).md","x":1336,"y":808,"width":465,"height":190},
|
||||
{"id":"e063ab92aef817e4","type":"text","text":"obtain characteristic equation","x":1420,"y":619,"width":298,"height":78},
|
||||
{"id":"d9167b6da862a39f","type":"file","file":"Math 201/Lectures/Free vibrations (lec 11-12).md","x":1467,"y":-875,"width":433,"height":146},
|
||||
{"id":"9d6ed2c739b0615e","type":"file","file":"Math 201/Lectures/Variation of parameters (lec 9-10).md","x":1467,"y":-655,"width":530,"height":153},
|
||||
{"id":"2b3a2c1a51539afd","type":"text","text":"idk how to solve these yet","x":1139,"y":-802,"width":241,"height":102},
|
||||
{"id":"3f081acda4f30a27","type":"text","text":"solve for $r_{1}$ & $r_{2}$ using quadratic formula $\\frac{{-b\\pm \\sqrt{ b^2-4ac }}}{2a}$","x":1365,"y":419,"width":408,"height":125},
|
||||
{"id":"50996e52b63ee5e0","type":"text","text":"$ay''+by'+cy=0$","x":1336,"y":998,"width":250,"height":60},
|
||||
{"id":"174f52023f715bf2","type":"file","file":"Math 201/Lectures/Method of undetermined coefficients (lec 8-9).md","x":2240,"y":1143,"width":498,"height":196},
|
||||
{"id":"0daa567ae142c34f","type":"text","text":"split into the homogenous solution and particular solution","x":2313,"y":927,"width":352,"height":131},
|
||||
{"id":"e5369edab4d8a607","type":"text","text":"find $y_{h}(t)$\n(the easy part)","x":1911,"y":895,"width":229,"height":98},
|
||||
{"id":"208cafba1dc71798","type":"text","text":"Use intuition and guess and check. (Practice improves intuition)","x":2068,"y":247,"width":332,"height":132},
|
||||
{"id":"eded7f0b26c3ebc5","type":"text","text":"use generalized algorithm:\ncase 1) $ay''+by'+cy=P_{m}(t)e^{rt}$\nthen: $y_{p}(t)=t^s(b_{m}t^m+b_{m-1}t^{m-1}+\\dots+b_{0})e^{rt}$\nwhere:\ns=0, if r is not a root,\ns=1 if r is a single root,\ns=2 if r is a double root\n\ncase 2)\n$ay''+by'+cy=P_{m}(t)e^{\\alpha t}\\cos(\\beta t)+P_{m}(t)e^{\\alpha t}\\sin(\\beta t)$\nthen: $y_{p}(t)=t^s[(A_{k}t^k+A_{K-1}t^{k-1}+\\dots+A_{0})e^{\\alpha t}\\cos(\\beta t)+(B_{k}t^k+B_{k-1}t^{k-1}+\\dots+B_{0})e^{\\alpha t}\\sin(\\beta t)]$\nwhere:\ns=0 if $\\alpha+i\\beta$ is not a root,\ns=1 if $\\alpha+i\\beta$ is a root.","x":2489,"y":-156,"width":1054,"height":575},
|
||||
{"id":"d327b2b2e59fc4cc","type":"text","text":"general solution will be $y(t)=y_{h}(t)+y_{p}(t)$","x":2108,"y":-261,"width":336,"height":105},
|
||||
{"id":"b9455e68aa57c635","type":"text","text":"$ay''+by'+cy=f(t)$","x":2239,"y":1339,"width":283,"height":59},
|
||||
{"id":"5ef326cd80c96f9a","type":"text","text":"if $f(t)$ is a sum of two functions, split it up and find $y_{p_{1}}$, $y_{p_{2}}, \\dots$ separately, then $y_{p}$ will be the sum of the all the $y_{p_{n}}$'s you find.","x":2276,"y":658,"width":426,"height":188},
|
||||
{"id":"ac0242af88518588","type":"text","text":"find $y_{p}(t)$ (hard part)","x":2389,"y":520,"width":200,"height":99}
|
||||
],
|
||||
"edges":[
|
||||
{"id":"7f7a9f03d9ef74ef","fromNode":"c6a536ee57248cfd","fromSide":"top","toNode":"767241b95828457e","toSide":"bottom"},
|
||||
|
|
@ -55,13 +69,10 @@
|
|||
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{"id":"c71b296bcb5e22a4","fromNode":"ece672db8e16ac5a","fromSide":"top","toNode":"109d20a863d93116","toSide":"bottom"},
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|
@ -72,7 +83,28 @@
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]
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}
|
||||
|
|
@ -3,17 +3,20 @@ $ay''+by'+cy=f(t)$
|
|||
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 $y_{p}$
|
||||
$y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$ <- 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
|
||||
$y'_{p}=v_{1}y_{1}+v_{1}y_{1}'+v_{2}'y_{2}+v_{2}y_{2}'$
|
||||
to avoid second derivatives in the equation and problems with uniqueness lagrange imposed:
|
||||
1) $v_{1}y_{1}+v_{2}'y_{2}=0$ this simplifies our work down the road as well.
|
||||
we put $y_p$ into the equation and make it equal to the RHS. To do so, find the derivatives first:
|
||||
$y'_{p}=v_{1}'y_{1}+v_{1}y_{1}'+v_{2}'y_{2}+v_{2}y_{2}'$
|
||||
to avoid second derivatives in the equation and problems with uniqueness, Lagrange imposed:
|
||||
1) $v_{1}'y_{1}+v_{2}'y_{2}=0$ this simplifies our work down the road as well.
|
||||
$y'_{p}=\cancel{ v_{1}'y_{1} }+v_{1}y_{1}'+\cancel{ v_{2}'y_{2} }+v_{2}y_{2}'$
|
||||
so $y_{p}''=v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}''$
|
||||
now we plug into the second order equation:
|
||||
$a(v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}'')+b(v_{1}y_{1}'+v_{2}y_{2}')+c(v_{1}y_{1}+v_{2}y_{2})=f(t)$
|
||||
$v_{1}(ay_{1}''+\cancelto{ 0 }{ by_{1}' }+cy_{1})+v_{2}(ay_{2}''+\cancelto{ 0 }{ by_{2}'C }+cy_{2})+a(v_{1}'y_{1}'+v_{2}'y_{2}')$
|
||||
$v_{1}(\cancel{ ay_{1}''+ by_{1}' +cy_{1} })+v_{2}(\cancel{ ay_{2}''+ by_{2}'+cy_{2} })+a(v_{1}'y_{1}'+v_{2}'y_{2}')$
|
||||
Both $y_{1}$ and $y_{2}$ are solutions to the homogenous counterpart. So the first two terms above equal to zero.
|
||||
2) $v_{1}'y_{1}'+v_{2}'y_{2}'=\frac{f(t)}{a}$
|
||||
$\det \begin{pmatrix}y_{1} & y_{2} \\y_{1}'& y_{2}'\end{pmatrix}$ = 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.
|
||||
$v_{1}'=\frac{{f(t)y_{2}t}}{aW[y_{1},y_{2}]}$; $v_{2}'=-\frac{{f(t)y_{1}(t)}}{aW[y_{1},y_{2}]}$ <- integrate both sizes to get $v_{1}$ and $v_{2}$. When integrating, you don't need to add a generic constant.
|
||||
|
||||
#ex #second_order #IVP
|
||||
$y''+4y=2\tan(2t)-e^t \qquad y(0)=0 \qquad y'(0)=\frac{4}{5}$
|
||||
|
|
@ -21,11 +24,11 @@ can we use undetermined coefficients? yes and no
|
|||
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
|
||||
$y_{p}'(t)=Ae^{t}$
|
||||
2) $y''+4y=-e^t$ <- use method of undetermined coefficients
|
||||
$y_{p}^1(t)=Ae^{t}$
|
||||
$5Ae^t=-e^t$
|
||||
$A=-\frac{1}{5}$
|
||||
$y_{p}'(t)=-\frac{1}{5}e^t$
|
||||
$y_{p}^1(t)=-\frac{1}{5}e^t$
|
||||
(ii) $y''+4y=2\tan(2t)$ <- cant use method of undetermined coefficients
|
||||
$y_{p}^2(t)=v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)$
|
||||
plugging in:
|
||||
|
|
@ -41,19 +44,19 @@ $2(\sin^2(2t)+\cos^2(2t))v_{2}'=2\tan(2t)\cos(2t)$
|
|||
$v_{2}'=\sin(2t)$
|
||||
$v_{2}(t)=-\frac{1}{2}\cos(2t)$ no constant of integration, we want one solution only
|
||||
$v_{1}'=-{\frac{\sin^2(2t)}{\cos(2t)}}$
|
||||
$v_{1}=-\int \frac{\sin^2(2t)}{\cos(2t)} \, dx$
|
||||
$v_{1}=-\int \frac{\sin^2(2t)}{\cos(2t)} \, dt$
|
||||
$v_{1}=-\int \frac{{1-\cos^2(2t)}}{\cos(2t)} \, dt$
|
||||
$v_1=-\int sec(2t) \, dx+\int \cos(2t) \, dt$
|
||||
$v_1=-\int sec(2t) \, dt+\int \cos(2t) \, dt$
|
||||
$v_{1}(t)=-\frac{1}{2}\ln\mid sec(2t)+\tan(2t)\mid+\frac{1}{2}\sin(2t)$
|
||||
$y_{p}^2(t)=v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)$
|
||||
$y(t)=y_{h}(t)+y_{p}'(t)+y_{p}^2(t)$
|
||||
$y(t)=y_{h}(t)+y_{p}^1(t)+y_{p}^2(t)$
|
||||
=$c_{1}\cos(2t)+c_{2}\sin(2t)+v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)-\frac{1}{5}e^t$
|
||||
is our general answer.
|
||||
is the general answer.
|
||||
|
||||
IVP solution:
|
||||
$y(0)=0=c_{1}-y_{p}(0)=c_{1}-\frac{1}{5}\Rightarrow c_{1}=\frac{1}{5}$
|
||||
skipping some differentiation: $y'(0)=2c_{2}+y_{p}'(0)=2c_{2}+v_{1}'(0)+2v_{2}(0)-\frac{1}{5}=\frac{4}{5}\Rightarrow c_{2}=1$
|
||||
$y(t)=\frac{1}{5}\cos(2t)+\sin(2t)-\frac{1}{5}e^t+v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)$
|
||||
$$y(t)=\frac{1}{5}\cos(2t)+\sin(2t)-\frac{1}{5}e^t+v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)$$
|
||||
|
||||
#end of lecture 9
|
||||
#start of lecture 10
|
||||
|
|
|
|||
|
|
@ -10,7 +10,7 @@ I have written these notes for myself, I thought it would be cool to share them.
|
|||
[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.html) (raw notes, not reviewed or revised yet.)
|
||||
[Free vibrations (lec 11-12)](free-vibrations-lec-11-12) (raw notes, not reviewed or revised yet.)
|
||||
</br>
|
||||
[How to solve any DE, a flow chart](Solve-any-DE.png)
|
||||
</br>
|
||||
Loading…
Reference in New Issue