revised reduction of order and cauchy
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@ -18,8 +18,8 @@ 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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$$a\frac{d^2y}{dt^2}+(b-a){\frac{dy}{dt}}+cy(t)=f(e^t)$$
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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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@ -39,5 +39,24 @@ 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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#end of lecture 10
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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.
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@ -1,60 +1,4 @@
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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 isn't quite the same as in the above definition.
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2) $y=x^r$
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$y=e^{rt}$
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$y'=te^{rt}$
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$y''=t^2e^{rt}$
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plug in derivative terms into equation:
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$at^2e^{rt}+(b-a)te^{rt}+ce^{rt}=0$
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divide both sides by $e^{rt}$
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$ar^2+(b-a)r+C=0$
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^ We have a polynomial! Solve for r using quadratic formula.
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Notice, y is a function of x which is a function of t.
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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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using reduction of order, you can derive: $y_{h}(x)=c_{1}x^r+c_{2}x^r\ln(x)$
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(iii) $r_{1,2}=\alpha+i\beta$
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then: $y_{h}(x)=x^\alpha(c_{1}\cos(\ln \beta x)+c_{2}\sin \ln(\beta x))$
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now find one particular solution for a non homogenous solution, using variation of parameters, combine the y_h and y_p to get y(x).
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# Reduction of order
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$y''+p(x)y'+q(x)y=f(x)$ (1) <- no general solution procedure 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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Isn't this nice? some kind of magic. We made some guesses and we arrived somewhere.
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#ex #reduction_of_order 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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# Free vibrations
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Free vibrations are when there are no externally applied forces acting upon an oscillatory system. RHS=0.
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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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@ -81,7 +25,7 @@ $Ae^{-bt/2m}(\sin \phi \cos \omega t+\cos \phi \sin \omega t)$
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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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but this is all classical mechanics, but beautifully 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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@ -0,0 +1,41 @@
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# Reduction of order
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#reduction_of_order
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Consider the equation:
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$$y''+p(x)y'+q(x)y=f(x)$$this equation (1) does not have a general solution procedure always.
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But, if $y_{1}(x)$ solves the homogenous counterpart: $y''+p(x)y'+q(x)y=0$
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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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let's calculate the derivatives wrt. 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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plugging in:
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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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$v\cancelto{ 0 }{ (y_{1}''+p(x)y_{1}'+q(x)y_{1}) }+v''y_{1}+(2y_{1}'+p(x)y_{1})v'=f(x)$
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$y_{1}v''+(2y_{1}'+p(x)y_{1})=f(x)$
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$v''+\left( \frac{2y_{1}'}{y_{1}}+p(x) \right)v'=\frac{f(x)}{y_{1}}$
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substitute $v'=u$
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$u'+\left( \frac{2y_{1}'}{y_{1}}+p(x) \right)u=\frac{f(x)}{y_{1}}$<- This is now a linear first order equation #de_L_type2
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This can be solved with prior tools now, We compute the integrating factor $\mu$
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$\mu=e^{\int(2y_{1}'/y_{1}+p)dx}=e^{\ln(y_{1}^2)}e^{\int P(x) \, dx}=y_{1}^2\cdot e^{\int p(x) \, dx}$
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From there, continue on as you would with any linear first order equation.
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Isn't this nice? some kind of magic. We made some guesses and we arrived somewhere.
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#ex #second_order_nonhomogenous #reduction_of_order
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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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> Ouch look at those x terms. And the exponent on the RHS. This isn't even in Cauchy Euler form!
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we guess the general solution will be in the form: $y(x)=v(x)y_{1}=v(x)e^{-x^2}$
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substitute: $v'=u$
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plug into formula derived above:
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$u'+\left( \frac{2y_{1}'}{y_{1}}+4x \right)u=\frac{8{e^{-x^2}e^{-2x}}}{e^{-x^2}}$ <-(note: $p(x)=4x$)
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$u'+\underbrace{ \left( \frac{2{e^{-x^2}(-2x)}}{e^{-x^2}}+4x \right) }_{ =0 }u=8e^{-2x}$
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$u'=8e^{-2x}$
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> Lucky us! This is just a separable equation. No need to treat it like a linear equation.
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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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We are done.
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@ -10,7 +10,8 @@ I have written these notes for myself, I thought it would be cool to share them.
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[Second order homogenous linear equations (lec 5-7)](second-order-homogenous-linear-equations-lec-5-7.html)
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[Method of undetermined coefficients (lec 8-9)](method-of-undetermined-coefficients-lec-8-9.html)
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[Variation of parameters (lec 9-10)](variation-of-parameters-lec-9-10.html)
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[Cauchy-Euler equations (lec 10)](cauchy-euler-equations-lec-10.html)
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[Cauchy-Euler equations (lec 10-11)](cauchy-euler-equations-lec-10-11.html)
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[Reduction of order (lec 11)](reduction-of-order-lec-11.html)
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[Free vibrations (lec 11-12)](free-vibrations-lec-11-12.html) (raw notes, not reviewed or revised yet.)
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[Resonance & AM (lec 13-14)](resonance-am-lec-13-14.html)
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[Laplace transform (lec 14-16)](laplace-transform-lec-14-16.html) (raw notes, not reviewed or revised yet.)
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