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revised reduction of order and cauchy

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content/Cauchy-Euler equations (lec 10).md
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@ -18,8 +18,8 @@ compute 2nd derivative of y wrt to x:
$\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2} \frac{dt}{dx}\cdot \frac{dt}{dx}+\frac{dy}{dt}{\frac{d^2t}{dx^2}}=\frac{1}{x^2}{\frac{d^2y}{dt^2}}-\frac{\frac{1}{x^2}dy}{dt}$ $\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2} \frac{dt}{dx}\cdot \frac{dt}{dx}+\frac{dy}{dt}{\frac{d^2t}{dx^2}}=\frac{1}{x^2}{\frac{d^2y}{dt^2}}-\frac{\frac{1}{x^2}dy}{dt}$
$\underset{ \text{Important} }{ x^2{\frac{d^2y}{dx^2}}=\frac{d^2y}{dt^2}-\frac{dy}{dt} }$ $\underset{ \text{Important} }{ x^2{\frac{d^2y}{dx^2}}=\frac{d^2y}{dt^2}-\frac{dy}{dt} }$
plugging those derivatives in we get: plugging those derivatives in we get:
$$a\frac{d^2y}{dt^2}+(b-a){\frac{dy}{dt}}+cy=f(e^t)$$ $$a\frac{d^2y}{dt^2}+(b-a){\frac{dy}{dt}}+cy(t)=f(e^t)$$
^ this is a constant coefficient equation now! We can solve it now using prior tools. ^ this is a constant coefficient second order non-homogenous equation now! We can solve it now using prior tools.
## Example: ## Example:
#ex #second_order #second_order_nonhomogenous #cauchy-euler #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. $y(x)=c_{1}e^{-\ln(x)}+c_{2}\ln(x)e^{-\ln(x)}+\frac{1}{2}\ln(x)^2e^{-\ln(x)}$ <- This is rather lousy notation, the y here isn't the same as the y above. Conceptually though, it's all oke doke.
$$y(x)=c_{1}x^{-1}+c_{2}\ln(x)x^{-1}+\frac{1}{2}{\ln(x)^2}x^{-1}$$ $$y(x)=c_{1}x^{-1}+c_{2}\ln(x)x^{-1}+\frac{1}{2}{\ln(x)^2}x^{-1}$$
We are done. We are done.
#end of lecture 10 #start of lecture 11
#end of lecture 10 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.

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content/Free vibrations (lec 11-12).md
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@ -1,60 +1,4 @@
#start of lecture 11 # Free vibrations
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 are when there are no externally applied forces acting upon an oscillatory system. RHS=0. Free vibrations are when there are no externally applied forces acting upon an oscillatory system. RHS=0.
$mr^2+br+k=0$ characteristic polynomial $mr^2+br+k=0$ characteristic polynomial
(i) $r_{1}\ne r_{2}$ $b^2-4mk>0$ (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. $=Ae^{-bt/2m}\sin(\omega t+\phi)$ where $\phi$ is the phase shift.
and $\frac{\omega}{2\pi}$ is the natural frequency and $\frac{\omega}{2\pi}$ is the natural frequency
$\frac{2\pi}{\omega}$ is the period $\frac{2\pi}{\omega}$ 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 $A=\sqrt{ c_{1}^2+c_{2}^2 }$ btw we know $A=\sqrt{ c_{1}^2+c_{2}^2 }$
and $\tan \phi=\frac{c_{1}}{c_{2}}$ and $\tan \phi=\frac{c_{1}}{c_{2}}$
so we can get A and phi from c_1 and c_2. so we can get A and phi from c_1 and c_2.

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content/Reduction of order (lec 11).md Normal file
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# 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.

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content/_index.md
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@ -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) [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) [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) [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.) [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) [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.) [Laplace transform (lec 14-16)](laplace-transform-lec-14-16.html) (raw notes, not reviewed or revised yet.)