added lec 14 and added cauchy euler page and revised voparam page
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@ -46,6 +46,7 @@ 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 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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$y_{h}(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}$
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@ -0,0 +1,14 @@
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# Laplace transform
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From now on, LT is short for Laplace Transform
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What is LT? It's denoted as $\mathcal{L}$ and it's an operator defined as the following:
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$$\underset{ =F(s) }{ \mathcal{L}\{f(t)\}(s) }:=\int _{0}^\infty e^{-st}f(t) dt=\lim_{ T \to \infty }\int _{0} ^T e^{-st}f(t)\, dt$$
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This doesn't look like anything useful, but later on we will learn how it is.
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$\mathcal{L}\{0\}=0$ Look at your bank account, integrate 0 you still get 0 :D
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$\mathcal{L}\{1\}=\int_{0}^\infty e^{-st} \, dt=-\frac{1}{s}e^{-st}|_{0}^\infty=\frac{1}{s}$ if s>0
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$\mathcal{L}\{e^{at}\}$=$\int_{0}^{\infty} e^{at}e^{-st}\, dt=\int_{0}^{\infty}e^{-(s-a)t}dt=\frac{1}{s-a}$ if $s-a>0$
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$\mathcal{L}\{\sin bt\}=\int _{0}^{\infty}e^{-st}\sin(bt) \, dt=\frac{b}{s^2+b^2}$ by integration by parts
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similarly can be done for cos, but we have run out of time.
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#end of lec 14
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@ -67,9 +67,9 @@ We wanna solve this IVP! We know from earlier that it must have a unique solutio
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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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lets try $y_{p_{1}}=Ae^{-t}$ Does this work?
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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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$y_{p_{1}}''=Ae^{-t}$ plug in these three and we find that A=2. Yes it works!
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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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@ -0,0 +1,22 @@
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# Variation of parameters
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$ay''+by'+cy=f(t)$
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1) $y_{h}=c_{1}y_{1}(t)+c_{2}y_{2}(t)$ <- h is homogenous, ie: $f(t)=0$
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Lagrange proposed this method to find the particular solution $y_{p}$:
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$y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$ <- btw $y_{1}$ and $y_{2}$ are often called a fundamental pair.
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we put $y_p$ into the equation and make it equal to the RHS. To do so, find the derivatives first:
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$y'_{p}=v_{1}'y_{1}+v_{1}y_{1}'+v_{2}'y_{2}+v_{2}y_{2}'$
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to avoid second derivatives in the equation and problems with uniqueness, Lagrange imposed:
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1) $v_{1}'y_{1}+v_{2}'y_{2}=0$ this simplifies our work down the road as well.
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$y'_{p}=\cancel{ v_{1}'y_{1} }+v_{1}y_{1}'+\cancel{ v_{2}'y_{2} }+v_{2}y_{2}'$
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so $y_{p}''=v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}''$
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now we plug into the second order equation:
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$a(v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}'')+b(v_{1}y_{1}'+v_{2}y_{2}')+c(v_{1}y_{1}+v_{2}y_{2})=f(t)$
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$v_{1}(\cancel{ ay_{1}''+ by_{1}' +cy_{1} })+v_{2}(\cancel{ ay_{2}''+ by_{2}'+cy_{2} })+a(v_{1}'y_{1}'+v_{2}'y_{2}')$
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Both $y_{1}$ and $y_{2}$ are solutions to the homogenous counterpart. So the first two terms above equal to zero.
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2) $v_{1}'y_{1}'+v_{2}'y_{2}'=\frac{f(t)}{a}$
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we now have a system of two equations:
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$\det \begin{pmatrix}y_{1} & y_{2} \\y_{1}'& y_{2}'\end{pmatrix}$ = Wronskian = $W[y_{1},y_{2}]\ne 0$
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by definition $y_1$ and $y_2$ are linearly independent solutions so the above can never be 0!
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$v_{1}'=-\frac{{f(t)y_{2}t}}{aW[y_{1},y_{2}]}$; $v_{2}'=\frac{{f(t)y_{1}(t)}}{aW[y_{1},y_{2}]}$ <- integrate both sizes to get $v_{1}$ and $v_{2}$. When integrating, you don't need to add a generic constant.
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Finally, your solution is: $y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$ where you get $y_{1},y_{2}$ from your homogenous solution.
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@ -0,0 +1,83 @@
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#start of lec 13
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He has good news. he's excited to tell us about electric currents! In particular, how a radio uses resonance to selectively listen to a particular frequency:
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# Resonance
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Let's imagine a mass spring system which has an applied force with a forcing frequency of $\gamma$ and an amplitude of $F_{o}$ (a constant):
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$my''+by'+ky=F_{o}\cos(\gamma t)$
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(Think of the driving force being the radio transmitter, and the mass-spring system is an LC tank circuit in a old-style radio)
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In order to study the phenomenon of resonance, we need an underdamped system.
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so we let: $b^2-4mk<0$ (ie: complex roots)
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then the homogenous solution becomes:
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$y_{h}(t)=e^{-bt/2m}\left( c_{1}\cos\left( \frac{\sqrt{ 4mk-b^2 }}{2m} t\right)+c_{2}\sin\left( \frac{\sqrt{ 4mk-b^2 }}{2m} t\right) \right)$
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$y_{h}(t)=Ae^{-bt/2m}\sin(\omega t+\phi)$
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where $\omega$ is called the angular frequency and equals $\frac{\sqrt{ 4mk-b^2 }}{2m}$
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and where $Ae^{-bt/2m}$ is called the transient part of the equation (goes to 0 as t->$\infty$).
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For particular solution, we use method of undetermined coefficients #mouc
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We guess: $y_p=A_{1}\cos(\gamma t)+A_{2}\sin(\gamma t)$ where $\gamma\ne \omega$ because if $\gamma=\omega$ and b=0 then we would have to multiply our guess by t.
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$A_{1}=\frac{F_{o}(k-m\gamma)}{(k-m\gamma^2)^2+b^2\gamma^2}$
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$A_{2}=\frac{F_{o}b\gamma}{(k-m\gamma^2)^2+b^2\gamma^2}$
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$y_{p}(t)=\frac{F_{o}}{(k-m\gamma^2)^2+b^2\gamma^2}((k-m\gamma^2 )\cos(\gamma t)+b\gamma \sin \gamma t)$
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$\sin(\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta$
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$k-m\gamma^2=A\sin \theta$
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$br=A\cos \theta$
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$A=\sqrt{ (k-m\gamma^2)^2+b^2\gamma^2 }$
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$\tan \theta=\frac{k-m\gamma^2}{b\gamma}$
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$y_{p}(t)=\frac{F_{o}}{(k-m\gamma^2)^2+b^2\gamma^2}((k-m\gamma^2 )\cos(\gamma t)+b\gamma \sin \gamma t)=\frac{F_{o}\sqrt{ (k-m\gamma^2)^2+b^2\gamma^2 }}{(k-m\gamma^2)^2+b^2\gamma^2}\sin(\gamma t+\theta)$
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$=\frac{F_{o}}{\sqrt{ (k-m\gamma^2)^2+b^2\gamma^2 }}\sin(\gamma t+\theta)$
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where we define $\mu(\gamma)=\frac{1}{\sqrt{ (k-m\gamma^2)^2+b^2\gamma^2 }}$ called the gain factor
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and the general solution is $y(t)=y_{h}(t)+y_{p}(t)$
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See how $y_h$ goes to zero as time progresses but $y_p$ stays? $y_p$ is the steady state part of the solution. If you were to graph $y(t)$ you would see a "beating" effect due to the sum of the two sins that eventually decays off.
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If we make the value in the denominator of the gain factor small, the amplitude goes to a very high value, higher than $F_{o}$! This is the equivalent of tuning a radio receiving circuit to resonate to a certain transmitted frequency, the amplitude grows to a great value in the receiver when excited with the right frequency. This is resonance! as b (friction) is decreased to zero we get stronger and stronger resonance.
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Lets find the maximum of the amplitude (resonance point)
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take the derivative of $\mu$ wrt to $\gamma$:
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$\mu'(\gamma)=-\frac{{2m^2\gamma\left( \gamma^2-\left( \frac{k}{m}-\frac{b^2}{2m^2} \right) \right)}}{[(k-m\gamma^2)^2+b^2\gamma^2]^{3/2}}=0$
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case one: $\gamma=0$ not interesting, because then the force applied would just a constant force.
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case two:
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$\gamma_{r}=\sqrt{ \frac{k}{m}-\frac{b^2}{2m^2} }$ where the r means resonance
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By plugging in $\gamma_{r}$ into $\mu()$ we can know the maximum amplitude at resonance: $\mu_{max}(\gamma_{r})=\mu(\gamma_{r})=\frac{2m}{b\sqrt{ 4mk-b^2 }}$
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if $b^2\geq 4mk>2mk$ (overly damped/critically damped case) (no resonance, imaginary numbers) $\gamma=0$
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if $b^2<2mk<4mk$ (assumed from the very beginning above) then we get a resonant frequency: $\gamma_{r}=\sqrt{\frac{ 2mk-b^2}{2m^2}}$
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what if b=0? (no resistance):
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$my''+ky=F_{o}\cos(\gamma t)$
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$y_{h}(t)=c_{1}\cos \omega t+c_{2}\sin \omega t$ , $\omega=\sqrt{ \frac{k}{m} }$
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$=A\sin(\omega t+\phi)$
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$y_{p}(t)=A_{1}\cos(\gamma t)+A_{2}\sin(\gamma t)$
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assume $\gamma=\omega$ with zero resistance we get:
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$y_{p}(t)=\frac{F_{o}}{2m\omega}t\sin \omega t \underset{ t\to \infty }{ \to }\infty$ (your circuit blows up! Or equivalently, your bridge collapses.)
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#end of lec 13
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#start of lecture 14
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# Amplitude modulation
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Last lecture we showed we can selectively listen to a specific signal by using resonance.
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He has something else he's excited to show us; amplitude modulation! Lets recall from last lecture:
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$my''+ky=F_{o}\cos(\gamma t)$ (undamped driven mass-spring system)
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solving characteristic equation:
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$\omega=\sqrt{ \frac{k}{m} }$
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$y_{h}(t)=c_{1}\cos(\omega t)+c_{2}\sin(\omega t)$ ; $\gamma\ne\omega$
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$y_{p}=A\cos(\gamma t)$ we can guess the particular solution is a constant times $\cos$. There will be no $\sin$ term on the LHS as there's no first derivative (aka no friction)
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$\Rightarrow A=\frac{F_{o}}{m(\omega^2-\gamma^2)}\cos(\gamma t)$
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$y(t)=c_{1}\cos(\omega t)+c_{2}\sin(\omega t)+\frac{F_{o}}{m(\omega^2-\gamma^2)}\cos(\gamma t)$
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assume $y(0)=0=y'(0)$
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$c_{2}=0$ $c_{1}=-\frac{F_{o}}{m(\omega^2-\gamma)}$
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$y(t)=\frac{F_{o}}{m(\omega^2-\gamma^2)}(\cos(\gamma t)-\cos(\omega t))$
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Hmm, It's not very easy to visualize a cos minus cos term.
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Use trig identity to make equation easier to visualize: $2\sin(\alpha)\sin(\beta)=\cos\left( \frac{{\alpha-\beta}}{2} \right)-\cos\left( \frac{{\alpha+\beta}}{2} \right)$
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$2\gamma t=\alpha-\beta$
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$2\omega t=\alpha+\beta$
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$\Rightarrow \alpha=\frac{(\omega+\gamma)t}{2}$ $\beta=\frac{(\omega-\gamma)t}{2}$
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$$y(t)=\frac{F_{o}}{m(\omega^2-\gamma^2)}\sin\frac{(\omega-\gamma)t}{2}\sin\frac{(\omega+\gamma)t}{2}$$
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This is an amplitude modulated signal! also can be seen as a "beating frequency".
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To recap, taking a frictionless mass spring system which is initially at rest, and applying a driving frequency that is strictly different from the system's natural frequency, results in a displacement that follows an AM modulated signal. How cool is that!
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>Sometimes I feel this effect when I'm sitting in the back of the bus, where the bus is stopped at a red light. I wonder if this modulated vibration pattern is attributed to this exact phenomenon.
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# Shortcut for solving DE of a mass spring system
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![[Drawing 2023-10-06 13.24.11.excalidraw]]
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$my''+by'+ky=mg+F$
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move $mg$ to LHS and replace $y$ with $y_{new}$ (remember, the $\frac{mg}{k}$ is a constant, its derivative is 0):
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$m\left( y-\frac{mg}{k} \right)''+b\left( y-\frac{mg}{k} \right)'+k\left( y-\frac{mg}{k} \right)=F$
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$my_{new}''+by_{new}'+ky_{new}=F$
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This simplifies our approach to solving a mass spring system. We could do it without this rearrangement, but it's more complex as the RHS has a sum of two terms. Either way works though, pick what you like.
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@ -1,41 +0,0 @@
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#start of lec 13
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He has good news. he's excited to tell us about electric currents!
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$my''+by'+ky=F_{o}\cos(\gamma t)$ the system has an applied force with a forcing frequency of $\gamma$ and an amplitude of $F_{o}$ (a constant)
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$b^2-4mk<0$ (complex roots)
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$y_{h}(t)=e^{-bt/2m}\left( c_{1}\cos\left( \frac{\sqrt{ 4mk-b^2 }}{2m} t\right)+c_{2}\sin\left( \frac{\sqrt{ 4mk-b^2 }}{2m} t\right) \right)$
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$y_{h}(t)=Ae^{-bt/2m}\sin(\omega t+\phi)$ where $\omega$ is angular frequency and equals $\frac{\sqrt{ 4mk-b^2 }}{2m}$
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where $Ae^{-bt/2m}$ is called the transient part of the equation (makes equation go to 0 as t->$\infty$)
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We guess: $y_p=A_{1}\cos(\gamma t)+A_{2}\sin(\gamma t)$ where $\gamma\ne \omega$ because if $\gamma=\omega$ and b=0 then we would have to multiply our guess by t.
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$A_{1}=\frac{F_{o}(k-m\gamma)}{(k-m\gamma^2)^2+b^2\gamma^2}$
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$A_{2}=\frac{F_{o}b\gamma}{(k-m\gamma^2)^2+b^2\gamma^2}$
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$y_{p}(t)=\frac{F_{o}}{(k-m\gamma^2)^2+b^2\gamma^2}((k-m\gamma^2 )\cos(\gamma t)+b\gamma \sin \gamma t)$
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$\sin(\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta$
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$k-m\gamma^2=A\sin \theta$
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$br=A\cos \theta$
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$A=\sqrt{ (k-m\gamma^2)^2+b^2\gamma^2 }$
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$\tan \theta=\frac{k-m\gamma^2}{b\gamma}$
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$y_{p}(t)=\frac{F_{o}}{(k-m\gamma^2)^2+b^2\gamma^2}((k-m\gamma^2 )\cos(\gamma t)+b\gamma \sin \gamma t)=\frac{F_{o}\sqrt{ (k-m\gamma^2)^2+b^2\gamma^2 }}{(k-m\gamma^2)^2+\beta^2\gamma^2}\sin(\gamma t+\theta)$
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$=\frac{F_{o}}{\sqrt{ (k-m\gamma^2)^2+b^2\gamma^2 }}\sin(\gamma t+\theta)$
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where we define $\mu(\gamma)=\frac{1}{(k-m\gamma^2)^2+b^2\gamma^2}$ called the gain factor
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and the general solution is $y(t)=y_{h}(t)+y_{p}(t)$
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see how y_h goes to zero but y_p stays? y_p is the steady state part of the solution. if you were to graph it you would see a wah wah effect that decays to zero.
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if we make the value in the denominator of the gain factor small the amplitude goes to a very high value, higher than $F_{o}$ this is equivalent to tuning a radio circuit to resonate to a certain transmitted frequency, the amplitude grows to a great value in the receiver when excited with the right frequency. This is resonance! as b (friction) decreases to zero we get stronger and stronger resonance.
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lets find the maximum of the amplitude (resonance point)
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$\mu'(\gamma)=-\frac{{2m^2\gamma\left( \gamma^2-\left( \frac{k}{m}-\frac{b^2}{2m^2} \right) \right)}}{[(k-m\gamma^2)^2+b^2\gamma^2]^{3/2}}=0$
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cases: $\gamma=0$ not interesting, beacuse then the force applied is a constant force.
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$\gamma_{r}=\sqrt{ \frac{k}{m}-\frac{b^2}{2m^2} }$ where the r means resonance
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by taking second derivative$\mu_{max}(\gamma_{r})=\frac{2m}{b\sqrt{ 4mk-b^2 }}$
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if $b^2\geq 4mk>2mk$ (overly damped/critically damped case) (no resonance, imaginary numbers) $\gamma=0$
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if $b^2<2mk<4mk$ (assumed from the beginning above) then we get a resonant frequency $\gamma_{r}=\sqrt{\frac{ 2mk-b^2}{2m^2}}$
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what if b=0? (no resistance):
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$my''+ky=F_{o}\cos(\gamma t)$
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$y_{h}(t)=c_{1}\cos \omega t+c_{2}\sin \omega t$ , $\omega=\sqrt{ \frac{k}{m} }$
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$=A\sin(\omega t+\phi)$
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$y_{p}(t)=A_{1}\cos(\gamma t)+A_{2}\sin(\gamma t)$
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assume $\gamma=\omega$ with zero resistance we get:
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$y_{p}(t)=\frac{F_{o}}{2m\omega}t\sin \omega t \underset{ t\to \infty }{ \to }\infty$ (your circuit blows up! Or equivalently, your bridge collapses.)
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#end of lec 13
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@ -1,90 +1,105 @@
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# Variation of parameters
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$ay''+by'+cy=f(t)$
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1) $y_{h}=c_{1}y_{1}(t)+c_{2}y_{2}t$ <- h is homogenous, ie: $f(t)=0$
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Lagrange proposed: find a particular solution of $y_{p}$
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$y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$ <- btw $y_{1}$ and $y_{2}$ are often called a fundamental pair.
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we put $y_p$ into the equation and make it equal to the RHS. To do so, find the derivatives first:
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$y'_{p}=v_{1}'y_{1}+v_{1}y_{1}'+v_{2}'y_{2}+v_{2}y_{2}'$
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to avoid second derivatives in the equation and problems with uniqueness, Lagrange imposed:
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1) $v_{1}'y_{1}+v_{2}'y_{2}=0$ this simplifies our work down the road as well.
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$y'_{p}=\cancel{ v_{1}'y_{1} }+v_{1}y_{1}'+\cancel{ v_{2}'y_{2} }+v_{2}y_{2}'$
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so $y_{p}''=v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}''$
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now we plug into the second order equation:
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$a(v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}'')+b(v_{1}y_{1}'+v_{2}y_{2}')+c(v_{1}y_{1}+v_{2}y_{2})=f(t)$
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$v_{1}(\cancel{ ay_{1}''+ by_{1}' +cy_{1} })+v_{2}(\cancel{ ay_{2}''+ by_{2}'+cy_{2} })+a(v_{1}'y_{1}'+v_{2}'y_{2}')$
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Both $y_{1}$ and $y_{2}$ are solutions to the homogenous counterpart. So the first two terms above equal to zero.
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2) $v_{1}'y_{1}'+v_{2}'y_{2}'=\frac{f(t)}{a}$
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$\det \begin{pmatrix}y_{1} & y_{2} \\y_{1}'& y_{2}'\end{pmatrix}$ = Wronskian = $W[y_{1},y_{2}]\ne 0$
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by definition $y_1$ and $y_2$ are linearly independent solutions so the above can never be 0!
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$v_{1}'=\frac{{f(t)y_{2}t}}{aW[y_{1},y_{2}]}$; $v_{2}'=-\frac{{f(t)y_{1}(t)}}{aW[y_{1},y_{2}]}$ <- integrate both sizes to get $v_{1}$ and $v_{2}$. When integrating, you don't need to add a generic constant.
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# Variation of parameters
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#voparam
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>[Professors definition/derivations during lecture](prof-variation-of-parameters.html) <- I found this to be too big brain for me. Here is a simplified definition:
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#ex #second_order #IVP
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## Variation of parameters is a method to solve:
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## $$ay''+by'+cy=f(t)$$
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First, find the homogenous solution:
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$y_{h}=c_{1}y_{1}(t)+c_{2}y_{2}(t)$
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Now we need the particular solution, let $y_{p}$ be in the following form:
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$y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$ <- btw $y_{1}$ and $y_{2}$ are often called a fundamental pair. They are obtained from your homogenous solution.
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Impose the following:
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1) $v_{1}'y_{1}+v_{2}'y_{2}=0$
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Compute the derivatives and simplify:
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$y'_{p}=v_{1}y_{1}'+v_{2}y_{2}'$
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$y_{p}''=v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}''$
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Now we plug those into the second order equation and simplify:
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2) $v_{1}'y_{1}'+v_{2}'y_{2}'=\frac{f(t)}{a}$
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We now have a system of two equations (1 and 2). Now we can solve for $v_{1}$ and $v_{2}$:
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Using Cramer's rule, we can solve for the system of equations and obtain the solutions:
|
||||
$v_{1}'=-\frac{{f(t)y_{2}(t)}}{aW[y_{1},y_{2}]}$; $v_{2}'=\frac{{f(t)y_{1}(t)}}{aW[y_{1},y_{2}]}$ <- integrate both sizes to get $v_{1}$ and $v_{2}$. When integrating, you don't need to add a generic constant.
|
||||
also, $W[y_1,y_{2}]$ is the Wrońskian, and it equals to: $\det \begin{pmatrix}y_{1}&y_{2}\\ y_{1}' &y_{2}'\end{pmatrix}=y_{1}y_{2}'-y_{2}y_{1}'$
|
||||
Finally, the general solution is:
|
||||
$$y(t)=y_{h}+y_{p}\qquad \text{where}\qquad y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$$
|
||||
## What you need to remember
|
||||
So, what do you need to commit to memory? I believe memorizing these three is a good tradeoff between memory allocated and speed for when you're solving a #voparam problem:
|
||||
1) $y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$
|
||||
2) $v_{1}'=-\frac{{f(t)y_{2}(t)}}{aW[y_{1},y_{2}]}$
|
||||
3) $v_{2}'=\frac{{f(t)y_{1}(t)}}{aW[y_{1},y_{2}]}$
|
||||
Alternatively, you could memorize the system of equations and solve for $v_{1}'$ and $v_{2}'$. Ie:
|
||||
1) $y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$
|
||||
2) $v_{1}'y_{1}+v_{2}'y_{2}=0$
|
||||
3) $v_{1}'y_{1}'+v_{2}'y_{2}'=\frac{f(t)}{a}$
|
||||
This is what the prof likes. I love you Dr. Minev, but this I personally disagree with. I think I'll stick with the formulas.
|
||||
---
|
||||
#ex #second_order #IVP #voparam #mouc
|
||||
$y''+4y=2\tan(2t)-e^t \qquad y(0)=0 \qquad y'(0)=\frac{4}{5}$
|
||||
can we use undetermined coefficients? yes and no
|
||||
find general solution to homogenous 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
|
||||
$y_{p}^1(t)=Ae^{t}$
|
||||
Can we use undetermined coefficients? Yes and no. We can use it on the $e^t$ term. However, guessing and checking $y_{p}$ for the $2\tan(2t)$ term might take a really, really long time.
|
||||
First, find general solution to homogenous counterpart:
|
||||
$y''+4y=0$ -> $r^2+4=0$ -> $r_{1,2}=\pm 2i$
|
||||
$y_{h}(t)=c_{1}\cos(2t)+c_{2}\sin(2t)$ Done. Easy peasy.
|
||||
For $-e^t$ lets use method of undetermined coefficients:
|
||||
$y''+4y=-e^t$
|
||||
$y_{p_{1}}(t)=Ae^{t}$
|
||||
$5Ae^t=-e^t$
|
||||
$A=-\frac{1}{5}$
|
||||
$y_{p}^1(t)=-\frac{1}{5}e^t$
|
||||
(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:
|
||||
we get a system of eq:
|
||||
$\cos(2t)v_{1}'+\sin(2t)v_{2}'=0$
|
||||
$-2\sin(2t)v_{1}'+2\cos(2t)v_{2}'=2\tan(2t)$
|
||||
> we know these two will give a unique solution.
|
||||
>to solve system of eq multiply each by:
|
||||
>$2\cos(2t)$
|
||||
>$\sin(2t)$
|
||||
$y_{p_{1}}(t)=-\frac{1}{5}e^t$
|
||||
|
||||
$2(\sin^2(2t)+\cos^2(2t))v_{2}'=2\tan(2t)\cos(2t)$
|
||||
$v_{2}'=\sin(2t)$
|
||||
$v_{2}(t)=-\frac{1}{2}\cos(2t)$ no constant of integration, we want one solution only
|
||||
$v_{1}'=-{\frac{\sin^2(2t)}{\cos(2t)}}$
|
||||
Now for $2\tan(2t)$, we cannot realistically use method of undetermined coefficients.
|
||||
Let's use variation of parameters instead:
|
||||
$y''+4y=2\tan(2t)$
|
||||
$y_{p_{2}}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$
|
||||
where $y_{1}=\cos(2t)$, $y_{2}=\sin(2t)$
|
||||
recall:
|
||||
$v_{1}'=-\frac{f(t)y_{2}}{aW[y_{1},y_{2}]}$
|
||||
$v_{2}'=\frac{f(t)y_{1}}{aW[y_{1},y_{2}]}$
|
||||
plugging in:
|
||||
$v_{1}'=-\frac{2\tan(2t)\sin(2t)}{(1)(\cos2t(2)\cos2t-\sin2t(-2)\sin2t}=-\frac{2\tan(2t)\sin(2t)}{2\cos^22t+2\sin^22t}=-\tan(2t)\sin(2t)$
|
||||
$v_{2}'=\frac{2\tan(2t)\cos(2t)}{2}=\sin(2t)$ <- isn't it nice how we can reuse our computation for the denominator? :D
|
||||
Now we integrate.
|
||||
$v_{2}=-\frac{1}{2}\cos(2t)$ <- Don't add a constant of integration, we want one solution only.
|
||||
$v_{1}=-\int \frac{\sin^2(2t)}{\cos(2t)} \, dt$
|
||||
$v_{1}=-\int \frac{{1-\cos^2(2t)}}{\cos(2t)} \, dt$
|
||||
$v_1=-\int sec(2t) \, dt+\int \cos(2t) \, dt$
|
||||
$v_{1}(t)=-\frac{1}{2}\ln\mid sec(2t)+\tan(2t)\mid+\frac{1}{2}\sin(2t)$
|
||||
$y_{p}^2(t)=v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)$
|
||||
$y(t)=y_{h}(t)+y_{p}^1(t)+y_{p}^2(t)$
|
||||
$y_{p_{2}}(t)=v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)$
|
||||
> ^Now you can start to see how guessing $y_{p_{2}}$ would take a really, really long time.
|
||||
|
||||
$y(t)=y_{h}(t)+y_{p_{1}}(t)+y_{p_{2}}(t)$
|
||||
=$c_{1}\cos(2t)+c_{2}\sin(2t)+v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)-\frac{1}{5}e^t$
|
||||
is 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: $y'(0)=2c_{2}+y_{p}'(0)=2c_{2}+v_{1}'(0)+2v_{2}(0)-\frac{1}{5}=\frac{4}{5}\Rightarrow c_{2}=1$
|
||||
$y(0)=0=c_{1}-y_{p}(0)=c_{1}-\frac{1}{5} \implies c_{1}=\frac{1}{5}$
|
||||
$y'(0)=\frac{4}{5}=2c_{2}+v_{1}'(0)+2v_{2}(0)-\frac{1}{5} \implies c_{2}=1$
|
||||
$$y(t)=\frac{1}{5}\cos(2t)+\sin(2t)-\frac{1}{5}e^t+v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)$$
|
||||
|
||||
#end of lecture 9
|
||||
#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
|
||||
|
||||
#ex #second_order #voparam #mouc
|
||||
$$y''-2y'+y=e^t\ln(t)+2\cos(t)$$
|
||||
i) $y_{h}(t)=?$
|
||||
Find homogenous solution first:
|
||||
$r^2-2r+1=0$
|
||||
$r_{1,2}=1$
|
||||
$r_{1,2}=1$ (repeated root)
|
||||
$y_{h}(t)=c_{1}e^t+c_{2}te^t$
|
||||
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.)
|
||||
let's use method of undetermined coefficients:
|
||||
$y_{p_{1}}=A\cos(t)+B\sin(t)$ is our guess
|
||||
$y_{p_{1}}'=-A\sin t+B\cos t$
|
||||
$y_{p_{1}}''=-A\cos t-B\sin t$
|
||||
$-A\cos t-B\sin t+2A\sin t-2B\cos t+A\cos t+B\sin t=2\cos t$
|
||||
$-2B\cos t+2A\sin(t)=2\cos t(t)$
|
||||
$\implies A=0,\ B=-1$
|
||||
$y_{p_{1}}=-\sin(t)$
|
||||
$y''-2y'+y=e^t\ln(t)$ 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) $e^t+v_{1}'+te^tv_{2}'=0$
|
||||
Compute v1 and v2. This time lets do it using the linear system for funzies:
|
||||
eq1) $e^tv_{1}'+te^tv_{2}'=0$
|
||||
eq2) $e^tv_{1}'+(te^t+e^t){v_{2}'}=e^t{\ln t}$
|
||||
subtract eq1 from eq2 $v_{2}'=\ln(t)$
|
||||
subtract eq1 from eq2: $v_{2}'=\ln(t)$
|
||||
$v_{2}(t)=\int \ln(t) \, dt$
|
||||
integrate by parts
|
||||
$=t\ln(t)-\int t\frac{1}{t} \, dt$
|
||||
|
|
|
|||
|
|
@ -9,9 +9,12 @@ I have written these notes for myself, I thought it would be cool to share them.
|
|||
[Exact equations (lec 4-5)](exact-equations-lec-4-5.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)
|
||||
[Variation of parameters (lec 9-10)](variation-of-parameters-lec-9-10.html) (raw notes, not reviewed or revised yet.)
|
||||
[Variation of parameters (lec 9-10)](variation-of-parameters-lec-9-10.html)
|
||||
[Cauchy-Euler equations (lec 10)](cauchy-euler-equations-lec-10.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 in free vibrations (lec 13)](resonance-in-free-vibrations-lec-13.html) (raw notes, not reviewed or revised yet.)
|
||||
[Resonance in free vibrations (lec 13-14)](resonance-in-free-vibrations-lec-13.html)
|
||||
[Laplace transform (lec 14)](laplace-transform-lec-14.html)
|
||||
|
||||
</br>
|
||||
[How to solve any DE, a flow chart](Solve-any-DE.png)
|
||||
</br>
|
||||
Loading…
Reference in New Issue