diff --git a/content/Cauchy-Euler equations (lec 10).md b/content/Cauchy-Euler equations (lec 10).md index fffff2f..4784e82 100644 --- a/content/Cauchy-Euler equations (lec 10).md +++ b/content/Cauchy-Euler equations (lec 10).md @@ -18,8 +18,8 @@ compute 2nd derivative of y wrt to x: $\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2} \frac{dt}{dx}\cdot \frac{dt}{dx}+\frac{dy}{dt}{\frac{d^2t}{dx^2}}=\frac{1}{x^2}{\frac{d^2y}{dt^2}}-\frac{\frac{1}{x^2}dy}{dt}$ $\underset{ \text{Important} }{ x^2{\frac{d^2y}{dx^2}}=\frac{d^2y}{dt^2}-\frac{dy}{dt} }$ plugging those derivatives in we get: -$$a\frac{d^2y}{dt^2}+(b-a){\frac{dy}{dt}}+cy=f(e^t)$$ -^ this is a constant coefficient equation now! We can solve it now using prior tools. +$$a\frac{d^2y}{dt^2}+(b-a){\frac{dy}{dt}}+cy(t)=f(e^t)$$ +^ this is a constant coefficient second order non-homogenous equation now! We can solve it now using prior tools. ## Example: #ex #second_order #second_order_nonhomogenous #cauchy-euler @@ -39,5 +39,24 @@ but we want solution in terms of x: $y(x)=c_{1}e^{-\ln(x)}+c_{2}\ln(x)e^{-\ln(x)}+\frac{1}{2}\ln(x)^2e^{-\ln(x)}$ <- This is rather lousy notation, the y here isn't the same as the y above. Conceptually though, it's all oke doke. $$y(x)=c_{1}x^{-1}+c_{2}\ln(x)x^{-1}+\frac{1}{2}{\ln(x)^2}x^{-1}$$ We are done. +#end of lecture 10 #start of lecture 11 -#end of lecture 10 \ No newline at end of file +Last lecture we did Cauchy Euler equations: +$$ax^2{\frac{d^2y}{dx^2}+bx{\frac{ dy }{ dx }}+cy=f(x)} \qquad x>0$$ +where $a,\ b,\ c$ are constants and $\in \mathbb{R}$ +substitute $x=e^t$ +$a{\frac{d^2y}{dt^2}}+(b-a){\frac{dy}{dt}}+cy=f(e^t)$ <- lousy notation, the y here isn't quite the same as in the above definition. +substitute: $y=x^r$ +after calculating derivatives, plugging in, and simplifying we obtain the polynomial equation: +$ar^2+(b-a)r+C=0$ +Three cases: +**(i)** $r_1\ne r_{2}$ then: +$y_{h}(t)=c_{1}e^{rt}+c_{2}e^{rt}$ +$y_{h}(x)=c_{1}x^{r_{1}}+c_{2}x^{r_{2}}$ (lousy notation, because the two $y_{h}$ do not equal each other) +**(ii)** $r_{1}=r_{2}=r$ then: +$y_{h}(t)=c_{1}e^{rt}+c_{2}te^{rt}$ +$y_{h}(x)=c_{1}x^r+c_{2}x^r\ln(x)$ (derived by reduction of order.) +**(iii)** $r_{1,2}=\alpha\pm i\beta$ then: +$y_{h}=e^\alpha(c_{1}\cos(\beta t)+c_{2}\sin(\beta t))$ +$y_{h}(x)=x^\alpha(c_{1}\cos(\beta\ln x)+c_{2}\sin(\beta \ln x))$ +Now compute your particular solution, $y_{p}$, and combine with $y_{h}$ to obtain your general solution. \ No newline at end of file diff --git a/content/Free vibrations (lec 11-12).md b/content/Free vibrations (lec 11-12).md index 32051bb..d802edb 100644 --- a/content/Free vibrations (lec 11-12).md +++ b/content/Free vibrations (lec 11-12).md @@ -1,60 +1,4 @@ -#start of lecture 11 -last lecture we did cauchy euler equations: -$ax^2{\frac{d^2y}{dx^2}+bx{\frac{ dy }{ dx }}+cy=f(x)},\ x>0$ -where $a,\ b,\ c$ are still constants and $\in \mathbb{R}$ -1) $x=e^t$ -$a{\frac{d^2y}{dt^2}}+(b-a){\frac{dy}{dt}}+cy=f(e^t)$ <- lousy notation, the y here isn't quite the same as in the above definition. -2) $y=x^r$ -$y=e^{rt}$ -$y'=te^{rt}$ -$y''=t^2e^{rt}$ -plug in derivative terms into equation: -$at^2e^{rt}+(b-a)te^{rt}+ce^{rt}=0$ -divide both sides by $e^{rt}$ -$ar^2+(b-a)r+C=0$ -^ We have a polynomial! Solve for r using quadratic formula. -Notice, y is a function of x which is a function of t. -Three cases: -(i) $r_1\ne r_{2}$ -then: $y_{h}(x)=c_{1}x^{r_{1}}+c_{2}x^{r_{2}}$ - -(ii) $r_{1}=r_{2}=r$ -using reduction of order, you can derive: $y_{h}(x)=c_{1}x^r+c_{2}x^r\ln(x)$ -(iii) $r_{1,2}=\alpha+i\beta$ -then: $y_{h}(x)=x^\alpha(c_{1}\cos(\ln \beta x)+c_{2}\sin \ln(\beta x))$ -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). - -# Reduction of order -$y''+p(x)y'+q(x)y=f(x)$ (1) <- no general solution procedure always -but, if $y_{1}(x)$ solves $y''+p(x)y'+q(x)y=0$ -then we can find the general solution to the non homogenous equation (1) by guessing it in the form $y(x)=v(x)y_{1}(x)$ -$y'=v'y_{1}+vy_{1}'$ -$y''=v''y_{1}+2v'y_{1}'+vy_{1}''$ -$(v''y_{1}+2v_{1}'y_{1}'+y_{1}''v)+p(x)(v'y_{1}+vy_{1}')+q(x)vy_{1}=f(x)$ -$v\cancelto{ 0 }{ (y_{1}''+p(x)y_{1}'+q(x)y_{1}) }+v''y_{1}+(2y_{1}'+p(x)y_{1})v'=f$ -$y_{1}v''+()$ -$v''+\left( \frac{2y_{1}'}{y_{1}}+p \right)v'=\frac{f}{y_{1}}$ -$v'=u$ -$u'+\left( \frac{2y_{1}'}{y_{1}}+p \right)u=\frac{f}{y_{1}}$<- this is a linear first order equation -how to solve linear first order equation? we compute the integrating factor $\mu$ -$\mu=e^{\int(2y_{1}'/y_{1}+p)dx}=e^{\ln(y_{1})^2}e^{\int P(x) \, dx}=y_{1}^2e^{\int p(x) \, dx}$ -Isn't this nice? some kind of magic. We made some guesses and we arrived somewhere. - -#ex #reduction_of_order find the general solution to the equation: -$$y''+4xy'+(4x^2+2)y=8e^{-x(x+2)}$$ -if $y_{1}(x)=e^{-x^2}$ is one solution. -therefore were finding the solution of the form: $y(x)=v(x)y_{1}=v(x)e^{-x^2}$ -$v'=u$ -$u'+\left( \frac{2y_{1}'}{y_{1}}+4x \right)u=\frac{8{e^{-x^2}e^{-2x}}}{e^{-x^2}}$ <-(p(x)=4x) -$u'+\left( \frac{2{e^{-x^2}(-2x)}}{e^{-x^2}}+4x \right)u=8e^{-2x}$ -$u'=8e^{-2x}$ -$u=-4e^{-2x}+c_{1}$ -$v'=u=-4e^{-2x}+c_{1}$ -$v(x)=2e^{-2x}+c_{1}x+c_{2}$ -general solution: -$$y(x)=v(x)y_{1}(x)=(2e^{-2x}+c_{1}x+c_{2})e^{-x^2}$$ - -## Free vibrations +# Free vibrations Free vibrations are when there are no externally applied forces acting upon an oscillatory system. RHS=0. $mr^2+br+k=0$ characteristic polynomial (i) $r_{1}\ne r_{2}$ $b^2-4mk>0$ @@ -81,7 +25,7 @@ $Ae^{-bt/2m}(\sin \phi \cos \omega t+\cos \phi \sin \omega t)$ $=Ae^{-bt/2m}\sin(\omega t+\phi)$ where $\phi$ is the phase shift. and $\frac{\omega}{2\pi}$ is the natural frequency $\frac{2\pi}{\omega}$ is the period -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. +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. btw we know $A=\sqrt{ c_{1}^2+c_{2}^2 }$ and $\tan \phi=\frac{c_{1}}{c_{2}}$ so we can get A and phi from c_1 and c_2. diff --git a/content/Reduction of order (lec 11).md b/content/Reduction of order (lec 11).md new file mode 100644 index 0000000..685c47a --- /dev/null +++ b/content/Reduction of order (lec 11).md @@ -0,0 +1,41 @@ +# Reduction of order +#reduction_of_order +Consider the equation: +$$y''+p(x)y'+q(x)y=f(x)$$this equation (1) does not have a general solution procedure always. +But, if $y_{1}(x)$ solves the homogenous counterpart: $y''+p(x)y'+q(x)y=0$ +then we can find the general solution to the non homogenous equation (1) by guessing it in the form: $y(x)=v(x)y_{1}(x)$ +let's calculate the derivatives wrt. x: +$y'=v'y_{1}+vy_{1}'$ +$y''=v''y_{1}+2v'y_{1}'+vy_{1}''$ +plugging in: +$(v''y_{1}+2v_{1}'y_{1}'+y_{1}''v)+p(x)(v'y_{1}+vy_{1}')+q(x)vy_{1}=f(x)$ +$v\cancelto{ 0 }{ (y_{1}''+p(x)y_{1}'+q(x)y_{1}) }+v''y_{1}+(2y_{1}'+p(x)y_{1})v'=f(x)$ +$y_{1}v''+(2y_{1}'+p(x)y_{1})=f(x)$ +$v''+\left( \frac{2y_{1}'}{y_{1}}+p(x) \right)v'=\frac{f(x)}{y_{1}}$ +substitute $v'=u$ +$u'+\left( \frac{2y_{1}'}{y_{1}}+p(x) \right)u=\frac{f(x)}{y_{1}}$<- This is now a linear first order equation #de_L_type2 +This can be solved with prior tools now, We compute the integrating factor $\mu$ +$\mu=e^{\int(2y_{1}'/y_{1}+p)dx}=e^{\ln(y_{1}^2)}e^{\int P(x) \, dx}=y_{1}^2\cdot e^{\int p(x) \, dx}$ +From there, continue on as you would with any linear first order equation. +Isn't this nice? some kind of magic. We made some guesses and we arrived somewhere. + +#ex #second_order_nonhomogenous #reduction_of_order +Find the general solution to the equation: +$$y''+4xy'+(4x^2+2)y=8e^{-x(x+2)}$$ +if $y_{1}(x)=e^{-x^2}$ is one solution. +> Ouch look at those x terms. And the exponent on the RHS. This isn't even in Cauchy Euler form! + +we guess the general solution will be in the form: $y(x)=v(x)y_{1}=v(x)e^{-x^2}$ +substitute: $v'=u$ +plug into formula derived above: +$u'+\left( \frac{2y_{1}'}{y_{1}}+4x \right)u=\frac{8{e^{-x^2}e^{-2x}}}{e^{-x^2}}$ <-(note: $p(x)=4x$) +$u'+\underbrace{ \left( \frac{2{e^{-x^2}(-2x)}}{e^{-x^2}}+4x \right) }_{ =0 }u=8e^{-2x}$ +$u'=8e^{-2x}$ +> Lucky us! This is just a separable equation. No need to treat it like a linear equation. + +$u=-4e^{-2x}+c_{1}$ +$v'=u=-4e^{-2x}+c_{1}$ +$v(x)=2e^{-2x}+c_{1}x+c_{2}$ +general solution: +$$y(x)=v(x)y_{1}(x)=(2e^{-2x}+c_{1}x+c_{2})e^{-x^2}$$ +We are done. \ No newline at end of file diff --git a/content/_index.md b/content/_index.md index f1b9844..ee5eeb1 100644 --- a/content/_index.md +++ b/content/_index.md @@ -10,7 +10,8 @@ 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) -[Cauchy-Euler equations (lec 10)](cauchy-euler-equations-lec-10.html) +[Cauchy-Euler equations (lec 10-11)](cauchy-euler-equations-lec-10-11.html) +[Reduction of order (lec 11)](reduction-of-order-lec-11.html) [Free vibrations (lec 11-12)](free-vibrations-lec-11-12.html) (raw notes, not reviewed or revised yet.) [Resonance & AM (lec 13-14)](resonance-am-lec-13-14.html) [Laplace transform (lec 14-16)](laplace-transform-lec-14-16.html) (raw notes, not reviewed or revised yet.)