added lec 23 and revised stuff
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@ -2,14 +2,14 @@
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### $$\frac{ dy }{ dx } +P(x)y=Q(x)y^n \quad\quad n\in\mathbb{R},\quad n\ne0,1$$
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>I'm calling this #de_b_type1. This is in standard form btw.
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It looks almost like a linear equation! In fact if n=0 it is by definition. We will see further that if n=1 you get a separable equation. So we ignore the cases when $n=0,1$ as these can be solved with prior tools.
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It looks almost like a linear equation! In fact if $n=0$ it is by definition. We will see further that if $n=1$ you get a separable equation. So we ignore the cases when $n=0,1$ as these can be solved with prior tools.
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Bernoulli's equations are important as you will see it in biology and in engineering.
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If y is + then y(x)=0 is a solution to the equation:
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If $y$ is $+$ then $y(x)=0$ is a solution to the equation:
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$\frac{dy}{dx}+0=0\quad\Rightarrow \quad0=0$
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Let's move the y to the LHS:
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$y^{-n}\frac{ dy }{ dx }+P(x)y^{1-n}=Q(x)$
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notice that y(x)=0 is no longer a solution! It was lost due to dividing by zero. So from here on out we will have to remember to add it back in our final answers.
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notice that $y(x)=0$ is no longer a solution! It was lost due to dividing by zero. So from here on out we will have to remember to add it back in our final answers.
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let $y^{1-n}=u$
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Differentiating this with respect to x gives us:
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$(1-n)y^{-n}\frac{ dy }{ dx }=\frac{du}{dx}$
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@ -34,7 +34,7 @@ $y'y^{-2}+y^{-1}=x^2$
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>Note that we lost the y(x)=0 solution here, we will have to add it back in the end.
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let $u=y^{1-n}=y^{-1}$
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Differentiating wrt. x we get: $\frac{du}{dx}=-y^{-2}{\frac{dy}{dx}}$
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Differentiating wrt. $x$ we get: $\frac{du}{dx}=-y^{-2}{\frac{dy}{dx}}$
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$y^{-2}{\frac{dy}{dx}=-\frac{ du }{ dx }}$
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$y^{-2}{\frac{dy}{dx}+y^{-1}=-\frac{ du }{ dx }}+y^{-1}$
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${x^2=-\frac{ du }{ dx }}+y^{-1}$
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@ -40,7 +40,7 @@ $$y'+y-\int _{0} ^t y(v)\sin(t-v) \, dv =-\sin t,\qquad y(0)=1$$
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this is called an integral-differential equation.
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we can convert it to a differential equation by taking the derivative of both sides (wrt to dt.):
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$y''+y'-y\sin(t-v)=-\cos t$
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ew thats a gross second order linear equation. lets do it another way
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ew thats a gross second order linear equation. lets solve using laplace.
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$sY-1+Y-\mathcal{L}\{(y*\sin)(t)\}=-\frac{1}{s^2+1}$
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$\left( s+1-\frac{1}{s^2+1}\right)Y(s)=1-\frac{1}{s^2+1}=\frac{s^2}{s^2+1}$
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@ -87,7 +87,7 @@ $\mathcal{L}\{t^n\}= \frac{n!}{s^{n+1}}$
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Today covers all midterm material. Yay!
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#end of lec 15 #start of lec 16
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He advises us to learn the table of common LT's, however a sheet will be provided for the exam.
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You can find the table of common LT's in the course textbook (I saved it and put it on the index page). There's a big table and a small table, the small table is the one he recommends we learn.
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You can find the table of common LT's in the course textbook (I saved them and left a link on the index page). There's a big table and a small table, the small table is the one he recommends we learn.
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## Examples
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#ex #LT
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@ -122,7 +122,7 @@ $$\mathcal{L}^{-1}\left\{ \frac{1}{s^5}+\frac{3}{(2s+5)^2}+\frac{1}{s^2+4s+8}+ \
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Notice that all these terms approach 0 as s approaches inf.
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Using linearity:
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$\mathcal{L}^{-1}\left\{ \frac{1}{s^5} \right\}+\mathcal{L}^{-1}\left\{ \frac{3}{\left( 2s+5 \right)^2} \right\}+\mathcal{L}^{-1}\left\{ \frac{1}{s^2+4s+8} \right\}+\mathcal{L}^{-1}\left\{ \frac{{s+1}}{s^2+2s+10} \right\}$
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Now we use the table! Um, but a lot of these terms don't really fit the table.
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Now we use the table! But a lot of these terms don't really fit the table.
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You'll see how computing an inverse LT is kinda like solving a puzzle, or cooking imo. You look at the terms you are given and you try to manipulate it to fit something in the table. Kinda like how you stare at your pantry and think of what food you can cook with it. It's kinda fun!
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First term: take out an $\frac{1}{4!}$. This is possible due to linearity:
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$\frac{1}{4!}\mathcal{L}^{-1}\left\{ \frac{4!}{s^5} \right\}$ <- Now we can use $\mathcal{L}\{t^n\}=\frac{n!}{s^{n+1}}$
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@ -135,7 +135,6 @@ $+\mathcal{L}^{-1}\left\{ \frac{{s+1}}{(s+1)^2+3^2} \right\}$ <- We can use $\ma
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All in all we get:
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$$\mathcal{L}^{-1}\left\{ \frac{1}{s^5}+\frac{3}{(2s+5)^2}+\frac{1}{s^2+4s+8}+ \frac{{s+1}}{s^2+2s+10} \right\}=\frac{1}{4!}t^4+\frac{3}{4}te^{-5t/2}+\frac{1}{2}\sin(2t)e^{-2t}+e^{-t}\cos(3t)$$
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Isn't that fun!
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Well, every good thing must come to an end, we are done.
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#ex #inv_LT #partial_fractions
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Find the inverse LT of:
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@ -1,73 +0,0 @@
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#start of lec 22
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Finished chapter 7 of the course textbook, Let's begin chapter 8!
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# Power series
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A power series is defined by:
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$$\sum_{n=0}^\infty a_{n}(X-X_{0})^n=a_{0}+a_{1}(x-x_{0})+a_{2}(x-x_{0})^2+\dots$$
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It is convergent if:
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$$\sum_{n=0} ^ \infty a_{n}(x-x_{0})^n<\infty \text{ at a given x}$$
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Otherwise, it is divergent.
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If $\sum_{n=0}^\infty \mid a_{n}(x-x_{0})^n\mid$ is convergent
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$\implies\sum_{n=0}^\infty a_{n}(x-x_{0})^n$ is absolutely convergent
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Just because something is absolutely convergent doesn't mean it is conditionally convergent. think of the harmonic series. It is absolutely convergent but also divergent (conditionally divergent).
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Theorem: With each $\sum_{n=0}^{\infty}a_{n}(x-x_{0})^n$ we can associate $0\leq \rho\leq \infty$ such that
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$\sum_{n=0} ^\infty a_{n}(x-x_{0})^n$ is absolutely convergent
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for all x such that $\mid x-x_{0}\mid<\rho$, divergent for all x where $\mid x-x_{0}\mid>\rho$
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"Who keeps stealing the whiteboard erases? (jokingly) It's a useless object, anyways"
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![[Drawing 2023-10-30 13.12.57.excalidraw.png]]
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how can we find $\rho$?
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Ratio test: If $\lim_{ n \to \infty }\mid \frac{a_{n+1}}{a_{n}}\mid=L$
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then $\rho=\frac{1}{L}$
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## Examples:
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#ex
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is this convergent? Divergent? and where so?
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$\sum_{n=0}^\infty \frac{2^{-n}}{n+1}(x-1)^n$
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determine the convergent set.
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Use ratio test:
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$\lim_{ n \to \infty }\mid \frac{a_{n+1}}{a_{n}}\mid=\lim_{ n \to \infty } \frac{2^{-(n+1)}}{n+2} \frac{n+1}{2^{-n}}=\frac{1}{2}\implies \rho=2$
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so it's convergent on $-1<x<3$, divergent on $\mid x-1\mid>2$
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But what about on the points $-1$ and $3$?
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plug in $x_{0}=-1$
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$\sum_{n=0}^{\infty} \frac{2^{-n}}{n+1}(-2)^n=\sum_{n=0}^\infty \frac{(-1)^n}{n+1}<\infty$ <- That is the alternating harmonic series, it is convergent.
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plug in $x_{0}=3$:
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$\sum_{n=0}^\infty \frac{2^{-n}}{n+1}2^n=\sum_{n=0}^\infty \frac{1}{n+1}>\infty$ <- harmonic series, this diverges.
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so the power series is convergent on $[-1,3)$ divergent otherwise.
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$$\text{ converges only on: } [-1,3)$$
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Assume that $\sum_{n=0}^\infty a_{n}(x-x_{0})^n$ and $\sum_{n=0}^\infty b_{n}(x-x_{0})^n$ are converget with $\rho>0$
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Then:
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1.) $\sum_{n=0}^\infty a_{n}(x-x_{0})^n+\sum_{n=0}^{\infty}b_{n}(x-x_{0})^n=\sum_{n=0}^\infty(a_{n}+b_{n})(x-x_{0})^n$
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That has a radius of convergence of at least $\rho$.
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2.) $\left( \sum_{n=0}^\infty a_{n}(x-x_{0})^n \right)\left( \sum_{n=0}^\infty b_{n}(x-x_{0})^n \right) \qquad c_n=\sum_{k=0}^n a_{k}b_{n-k}$(Cauchy)
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$=(a_{0}+a_{1}(x-x_{0})+a_{2}(x-x_{0})^2+\dots)(b_{0}+b_{1}(x-x_{0})+b_{2}(x-x_{0})^2+\dots)$
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$=a_{0}b_{0}+(a_{0}b_{1}+a_{1}b_{0})(x-x_{0})+(a_{0}b_{2}+a_{1}b_{1}+a_{2}b_{1})(x-x_{0})^2+\dots$ (Cauchy multiplication)
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more Definitions of power series:
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If $\sum_{n=0}^{\infty}a_{n}(x-x_{0})^n$ is convergent with $\rho>0$
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$\mid x-x_{0}\mid<\rho$
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we can differentiate this infinite sum and get:
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$\implies y'(x)=\sum_{n=1}^\infty a_{n}n(x-x_{0})^{n-1}$
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$y''(x)=\sum_{n=2}^\infty a_{n}n(n-1)(x-x_{0})^{n-2}$
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Theorem: If $y(x)$ is infinitely many times differentiable on some interval: $\mid x-x_{0}\mid<\rho$
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then: $\sum_{n=0}^\infty \frac{y^{(n)}(x_{0})}{n!}(x-x_{0})^n$ (Taylor series)
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"believe me, taylor series is the most important theorem in engineering."
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"I mean engineering is all about approximations, do you know how your calculator computes ...? Taylor series!"
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"Applied mathematics is all about approximating and then measuring how good your approximation is, it's what engineering is all about." -Prof (loosy quotes, can't keep up with how enthusiastic he is!)
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Definition: If $y(x)$ can be represented with a power series on $\mid x-x_{0}\mid$ then $y(x)$ is an analytic function on $(x_{0}-\rho,x_{0}+\rho)$
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btw analytic functions are very important in complex calculus MATH301. (i don't have that next term)
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$f(x)=\sum_{n=0}^\infty a_{n}(x-x_{0})^n<\infty$
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$f'(x)=\sum_{n=1}^\infty a_{n}n(x-x_{0})^{n-1}$
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$f(x)+f'(x)=\sum_{n=0}^\infty a_{n}(x-x_{0})^n+\sum_{n=1}^\infty a_{n}n(x-x_{0})^{n-1}$
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let $n-1=k$
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$=\sum_{n=0}^\infty a_{n}(x-x_{0})^n+\sum_{k=0}^\infty a_{n}(k+1)(x-x_{0})^{k}$
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$=\sum_{n=0}^\infty(a_{n}+a_{n}(n+1))(x-x_{0})^n$
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Last theorem fo' da day:
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If $\sum_{n=0}^\infty a_{n}(x-x_{0})^n=0$ for all x$\in(x_{0}-\rho,x_{0}+\rho)$ where $\rho>0$
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$\implies a_{n}=0$, $n=0,1,2,\dots$
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#end of lec 22
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@ -0,0 +1,156 @@
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#start of lec 22
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Finished chapter 7 of the course textbook, Let's begin chapter 8!
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# Power series
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A power series is defined by:
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$$\sum_{n=0}^\infty a_{n}(X-X_{0})^n=a_{0}+a_{1}(x-x_{0})+a_{2}(x-x_{0})^2+\dots$$
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It is convergent if:
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$$\sum_{n=0} ^ \infty a_{n}(x-x_{0})^n<\infty \text{ at a given x}$$
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Otherwise, it is divergent.
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If $\sum_{n=0}^\infty \mid a_{n}(x-x_{0})^n\mid$ is convergent
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$\implies\sum_{n=0}^\infty a_{n}(x-x_{0})^n$ is absolutely convergent
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Just because something is absolutely convergent doesn't mean it is conditionally convergent. think of the harmonic series. It is absolutely convergent but also divergent (conditionally divergent).
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Theorem: With each $\sum_{n=0}^{\infty}a_{n}(x-x_{0})^n$ we can associate $0\leq \rho\leq \infty$ such that
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$\sum_{n=0} ^\infty a_{n}(x-x_{0})^n$ is absolutely convergent
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for all x such that $\mid x-x_{0}\mid<\rho$, divergent for all x where $\mid x-x_{0}\mid>\rho$
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"Who keeps stealing the whiteboard erases? (jokingly) It's a useless object, anyways"
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![[Drawing 2023-10-30 13.12.57.excalidraw.png]]
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how can we find $\rho$?
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Ratio test: If $\lim_{ n \to \infty }\mid \frac{a_{n+1}}{a_{n}}\mid=L$
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then $\rho=\frac{1}{L}$
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## Examples:
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#ex
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is this convergent? Divergent? and where so?
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$\sum_{n=0}^\infty \frac{2^{-n}}{n+1}(x-1)^n$
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determine the convergent set.
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Use ratio test:
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$\lim_{ n \to \infty }\mid \frac{a_{n+1}}{a_{n}}\mid=\lim_{ n \to \infty } \frac{2^{-(n+1)}}{n+2} \frac{n+1}{2^{-n}}=\frac{1}{2}\implies \rho=2$
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so it's convergent on $-1<x<3$, divergent on $\mid x-1\mid>2$
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But what about on the points $-1$ and $3$?
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plug in $x_{0}=-1$
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$\sum_{n=0}^{\infty} \frac{2^{-n}}{n+1}(-2)^n=\sum_{n=0}^\infty \frac{(-1)^n}{n+1}<\infty$ <- That is the alternating harmonic series, it is convergent.
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plug in $x_{0}=3$:
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$\sum_{n=0}^\infty \frac{2^{-n}}{n+1}2^n=\sum_{n=0}^\infty \frac{1}{n+1}>\infty$ <- harmonic series, this diverges.
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so the power series is convergent on $[-1,3)$ divergent otherwise.
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$$\text{ converges only on: } [-1,3)$$
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Assume that $\sum_{n=0}^\infty a_{n}(x-x_{0})^n$ and $\sum_{n=0}^\infty b_{n}(x-x_{0})^n$ are converget with $\rho>0$
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Then:
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1.) $\sum_{n=0}^\infty a_{n}(x-x_{0})^n+\sum_{n=0}^{\infty}b_{n}(x-x_{0})^n=\sum_{n=0}^\infty(a_{n}+b_{n})(x-x_{0})^n$
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That has a radius of convergence of at least $\rho$.
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2.) $\left( \sum_{n=0}^\infty a_{n}(x-x_{0})^n \right)\left( \sum_{n=0}^\infty b_{n}(x-x_{0})^n \right) \qquad c_n=\sum_{k=0}^n a_{k}b_{n-k}$(Cauchy)
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$=(a_{0}+a_{1}(x-x_{0})+a_{2}(x-x_{0})^2+\dots)(b_{0}+b_{1}(x-x_{0})+b_{2}(x-x_{0})^2+\dots)$
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$=a_{0}b_{0}+(a_{0}b_{1}+a_{1}b_{0})(x-x_{0})+(a_{0}b_{2}+a_{1}b_{1}+a_{2}b_{1})(x-x_{0})^2+\dots$ (Cauchy multiplication)
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more Definitions of power series:
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If $\sum_{n=0}^{\infty}a_{n}(x-x_{0})^n$ is convergent with $\rho>0$
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$\mid x-x_{0}\mid<\rho$
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we can differentiate this infinite sum and get:
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$\implies y'(x)=\sum_{n=1}^\infty a_{n}n(x-x_{0})^{n-1}$
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$y''(x)=\sum_{n=2}^\infty a_{n}n(n-1)(x-x_{0})^{n-2}$
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Theorem: If $y(x)$ is infinitely many times differentiable on some interval: $\mid x-x_{0}\mid<\rho$
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then: $\sum_{n=0}^\infty \frac{y^{(n)}(x_{0})}{n!}(x-x_{0})^n$ (Taylor series)
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"believe me, taylor series is the most important theorem in engineering."
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"I mean engineering is all about approximations, do you know how your calculator computes ...? Taylor series!"
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"Applied mathematics is all about approximating and then measuring how good your approximation is, it's what engineering is all about." -Prof (loosy quotes, can't keep up with how enthusiastic he is!)
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Definition: If $y(x)$ can be represented with a power series on $\mid x-x_{0}\mid$ then $y(x)$ is an analytic function on $(x_{0}-\rho,x_{0}+\rho)$
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btw analytic functions are very important in complex calculus MATH301. (i don't have that next term)
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$f(x)=\sum_{n=0}^\infty a_{n}(x-x_{0})^n<\infty$
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$f'(x)=\sum_{n=1}^\infty a_{n}n(x-x_{0})^{n-1}$
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$f(x)+f'(x)=\sum_{n=0}^\infty a_{n}(x-x_{0})^n+\sum_{n=1}^\infty a_{n}n(x-x_{0})^{n-1}$
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let $n-1=k$
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$=\sum_{n=0}^\infty a_{n}(x-x_{0})^n+\sum_{k=0}^\infty a_{n}(k+1)(x-x_{0})^{k}$
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$=\sum_{n=0}^\infty(a_{n}+a_{n}(n+1))(x-x_{0})^n$
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Last theorem fo' da day:
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If $\sum_{n=0}^\infty a_{n}(x-x_{0})^n=0$ for all x$\in(x_{0}-\rho,x_{0}+\rho)$ where $\rho>0$
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$\implies a_{n}=0$, $n=0,1,2,\dots$
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#end of lec 22 #start of lec 23
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Mid terms are almost done being marked!
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## Examples
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Let's start using power series to start solving DE!
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No magic formulas we need to memorize when solving equations using power series (Yay!)
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#ex
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$$y'-2xy=0 \qquad x_{0}=0$$
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note this is separable and linear, so we can already solve this. This time we do it with power series
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y should be an analytic function (meaning, infinitely many times differentiable)
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so we should expect we can represent it as a power series
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$y(x)=\sum_{n=0}^\infty a_{n}x^n$
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$y'(x)=\sum_{n=1}^\infty a_{n}nx^{n-1}$
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plug these into the equation:
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$\sum_{n=1}^\infty a_{n}nx^{n-1}-\sum_{n=0}^\infty 2a_{n}x^{n+1}=0$
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if the entire interval is zero, we should expect all the coefficients to equal 0
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we need to combine the summations.
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shift the index!
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$k=n-1,\ k=n+1$
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$\sum_{k=0}^\infty a_{k+1}(k+1)x^{k}-\sum_{k=1}^\infty 2a_{k-1}x^k=0$
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$a_{1}+\sum_{k=1}^\infty (\underbrace{ a_{k+1}(k+1)-2a_{k-1} }_{ =0 })x^k=0$
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The whole series equals zerro,
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so $a_{1}=0$ is the first observation
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second observation:
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$a_{k+1}=\frac{2}{k+1}a_{k-1}$ where $k=1,2,3,\dots$ This is called a recursive relation. (if we know one index we can produce some other index recursively)
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from this equation:
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$a_{1}, a_{3}, a_{5}, \dots=0$
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$a_{2k+1}=0, k=0,1,2,\dots$
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this means half of our power series disappears!
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what happens with the other half?
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$a_{2}$ is related to $a_{0}$ from the above formula
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$a_{2}=\frac{2}{2}a_{0}$ ($k=1$)
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$a_{4}=\frac{2}{3+1}a_{2}=\frac{a_{0}}{2}$ ($k=2$)
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$a_{6}=\frac{2}{5+1} \frac{a_{0}}{2}=\frac{a_{0}}{6}$ ($k=3$)
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$a_{8}=\frac{1}{4} \frac{a_{0}}{6}=\frac{a_{0}}{24}$ ($k=4$)
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you might start noticing a factorial-y pattern:
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$a_{2k}=\frac{1}{k!}a_{0}$ where $k=0,1,2,\dots$
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$y(x)=a_{0}\sum_{k=0} ^\infty \frac{1}{k!}x^{2k}=a_{0}\sum_{k=0} ^\infty \frac{1}{k!}(x^2)^k$
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Does this look like something from math 101?
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Yes! it looks like the taylor series of $e^{x^2}$
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so:
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$$y(x)=a_{0}e^{x^2}$$
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"if we are correct--the same is not true in general in real life--but in mathematics if we are correct we should end up with the same solution" -Prof
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#ex
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$$z''-x^2z'-xz=0 \qquad \text{about } x_{0}=0$$
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using regular methods will be problematic,
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if you use laplace transform you will have problems as well.
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>"you try the simplest thing you know, if you know anything :D" (referring to answering a question about how do we know what method to use?)
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lets use power series:
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assume solution is analytic:
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$z(x)=\sum_{n=0}^\infty a_{n}nx^{n-1}$
|
||||
$z''(x)=\sum_{n=2}^\infty a_{n}n(n-1)x^{n-2}$
|
||||
$\underset{ n-2=k }{ \sum_{n=2}^\infty a_{n}n(n-1)x^{n-2} }-\underset{ n+1=k }{ \sum_{n=1}^\infty a_{n} nx^{n+1} }-\underset{ n+1=k }{ \sum_{n=0}^\infty a_{n}x^{n+1} }=0$
|
||||
|
||||
shift the index to equalize the powers.
|
||||
>loud clash of clans log in sound, class giggles, "whats so funny?" :D "im not a dictator" something about you are not forced to sit through and watch the lecture if you don't like to, "I dont think everybody should like me."
|
||||
|
||||
$\sum_{k=0}^\infty a_{k+2}(k+2)(k+1)x^k-\sum_{k=2}^\infty a_{k-1}x^k-\sum_{k=1}^\infty a_{k-1}x^k=0$
|
||||
$2a_{2}+6a_{3}+\sum_{k=2}^\infty a_{k+2}(k+2)(k+1)x^k-\sum_{k=2}^\infty a_{k-1}(k-1)x^k-a_{0}x-\sum_{k=2}^\infty a_{k-1}x^k=0$
|
||||
$6a_{2}+(6a_{3}-a_{0})x+\sum_{k=2}^\infty (a_{k+2}(k+1)(k+2)-a_{k-1}k)x^k=0$
|
||||
$a_{2}=0 \qquad a_{3}=\frac{a_{0}}{6}$
|
||||
$a_{k+2}=\frac{k}{(k+1)(k+2)}a_{k-1}$ where $k=2,3,4,\dots$
|
||||
Finally, a recursive relation!
|
||||
it should be clear that each step of 3 starting from $a_{2}$ should all equal 0.
|
||||
$a_{2}, a_{5}, a_{8}, \dots=0$
|
||||
$a_{3k-1}=0$ where $k=1,2,\dots$
|
||||
$a_{4}=\frac{2}{3\cdot 4}a_{1}$
|
||||
$a_{7}=\frac{5}{6\cdot 7} \frac{2}{3\cdot 4}a_{1}$
|
||||
realize if we multiply here by 5 and 2:
|
||||
$a_{7}=\frac{5^2}{5\cdot6\cdot 7} \frac{2^2}{2\cdot3\cdot 4}a_{1}$
|
||||
$a_{7}=\frac{(2\\dot{c} 5)^2}{7!}a_{1}$
|
||||
$a_{4}=\frac{2^2}{4!}a_{1}$
|
||||
the pattern leads us to:
|
||||
$a_{3k+1}=\frac{(2\cdot 5 \dots(3k-1))^2}{(3k+1)!}$ where k=1,2,3, ...
|
||||
$a_{3k}=\frac{(1\cdot 4\cdot \dots(3k-2))^2}{(3k)!}a_{0}$ k=1,2,...
|
||||
$z(x)=a_{0}\left( 1+\sum_{k=1}^\infty \frac{(1*4*\dots(3k-2))^2}{(3k)!}x^{3k} \right)$
|
||||
$a_{1}\left( x+\sum_{k=1}^\infty \frac{(2\cdot 5\cdot \dots(3k-1))^2}{(3k+1)!} x^{3k+1}\right)$
|
||||
there we go, $z$ is a linear combination of those two expressions
|
||||
class done at 1:56 (a lil late but the journey is worth it)
|
||||
#end of lec 23
|
||||
|
|
@ -1,56 +1,50 @@
|
|||
|
||||
#start of lec 17
|
||||
summary of all the equations (separable linear exact, ...) available on eclass
|
||||
(homogenous (2 forms), Bernoulli equation, linear coefficients,)
|
||||
But first, midterm review!
|
||||
He spends most of the time giving us a summary of all the equations (separable, linear, homogenous, exact, ...) The review is available on the eclass page. 25 minutes spent in total covering what will be on the midterm!
|
||||
|
||||
second order equations:
|
||||
put it in standard form
|
||||
solve the homogenous eq,
|
||||
with constant coefficients
|
||||
3 cases: overdamped critically damped under damped
|
||||
now find particular solution to get general solution $y=y_{h}+y_{p}$
|
||||
either use #mouc or #voparam
|
||||
|
||||
we also have Cauchy-Euler equations (two methods, either guess y=x^r or use the general formula)
|
||||
|
||||
reduction of order is a technique when you're given a solution and you have to find the other
|
||||
|
||||
mechanical systems $my''+by'+ky=F_{o}\cos(\gamma t)$ (he recommends solving with mouc) that gives you the transient solution.
|
||||
the particular solution is the steady part.
|
||||
recall $\mu$=amplitude $\gamma_{r}$=resonance freq $\mu(\gamma_{_{r}})$=amplitude at resonance
|
||||
|
||||
Laplace transforms: definition and 4 important properties
|
||||
That's all that will be covered on the midterm. 25 minutes spent covering what will be on the midterm!
|
||||
|
||||
So, why did we learn all this stuff about laplace transforms? We will now see how its useful:
|
||||
|
||||
#ex #LT #second_order_nonhomogenous
|
||||
So, why did we learn all this stuff about Laplace transforms? We will now see how its useful:
|
||||
# Solving IVP's with Laplace
|
||||
## Example:
|
||||
#ex #LT #second_order_nonhomogenous #IVP
|
||||
Solve the following IVP using Laplace transforms:
|
||||
$$y''+y=t^2+2 \qquad y(0)=1 \qquad y'(0)=-1$$
|
||||
LHS=RHS, so the Laplace transforms of each side must also be equal:
|
||||
LHS=RHS, so the Laplace transforms of each side must also be equal.
|
||||
We take the Laplace transform of both sides:
|
||||
$\mathcal{L}\{y''+y\}=\mathcal{L}\{t^2+2\}$
|
||||
applying linearity:
|
||||
$\mathcal{L}\{y''\}+\mathcal{L}\{y\}=\mathcal{L}\{t^2\}+\mathcal{L}\{2\}$
|
||||
using properties:
|
||||
$s^2Y(s)-sy(0)-y'(0)+Y(s)=\frac{2!}{s^3}+\frac{2}{s}$
|
||||
$s^2Y(s)-s+1+Y(s)=\frac{2!}{s^3}+\frac{2}{s}$
|
||||
factor out Y(s)
|
||||
Now we need to isolate $Y(s)$.
|
||||
So we add $s-1$ to both sides and factor out $Y(s)$:
|
||||
$Y(s)(s^2+1)=\frac{2(s^2+1)}{s^3}+s-1$
|
||||
Divide both sides by $(s^2+1)$ :
|
||||
$Y(s)=\frac{2}{s^3}+\frac{s}{s^2+1}-\frac{1}{s^2+1}$
|
||||
now we take the inverse LT to obtain y(s)
|
||||
Now we take the inverse LT of both sides to obtain $y(t)$
|
||||
$$y(t)=t^2+\cos(t)-\sin(t)$$
|
||||
Done!
|
||||
Done. That was pretty fast!
|
||||
|
||||
second example:
|
||||
#ex #LT #second_order_nonhomogenous
|
||||
Solve the following IVP using Laplace transforms:
|
||||
$$y''-y=(t-2)e^{t-2} \qquad y(2)=0 \qquad y'(2)=0$$
|
||||
we need y at 0 but its as 2.
|
||||
so we make a substitution: $x=t-2$
|
||||
$\frac{d^2y}{dx^2}-y=xe^x$ <- notice the y here is not the same as the y above, lousy notation again :)
|
||||
We are provided $y(2)$ but usually we need it in the form $y(0)$ What do we do?
|
||||
We make a substitution: $x=t-2$
|
||||
$\frac{d^2y}{dx^2}-y=xe^x$ <- notice the y here is not the same as the y above, lousy notation. This graph is shifted 2 units to the left.
|
||||
where $y(0)=0 \qquad \frac{dy}{dx}(0)=0$
|
||||
hit it with the LT!
|
||||
notice 1/s^2 is LT of x so we have to shift it
|
||||
Hit it with the LT!
|
||||
$\frac{1}{s^2}$ is LT of $x$. Using the shifting property, $\frac{1}{(s-\alpha)^2}$ is the LT of $xe^{\alpha x}$
|
||||
$s^2Y(s)-Y(s) =\frac{1}{(s-1)^2}$
|
||||
Isolate $Y(s)$ :
|
||||
$(s^2-1)Y(s)=\frac{1}{(s-1)^2}$
|
||||
$Y(s)=\frac{1}{(s-1)^2(s^2-1)}$
|
||||
$Y(s)=\frac{1}{(s-1)^2(s-1)(s+1)}$
|
||||
Partial fraction time:
|
||||
$Y(s)=\frac{1}{(s-1)^3(s+1)}=\frac{A}{s-1}+\frac{B}{(s-1)^2}+\frac{C}{(s-1)^3}+\frac{D}{s+1}$
|
||||
$\frac{{A(s-1)^2(s+1)+B(s-1)(s+1)+C(s+1)+D(s-1)^3}}{(s-1)^3(s+1)}$
|
||||
$\begin{matrix}A+D=0 \\-A-B-3D=0 \\ -A+c+3D=0 \\ A-B+C-D=1\end{matrix}$
|
||||
$\begin{matrix}A+D=0 \\A-2A+B-3D=0 \\ A-2A+B-B+C+3D=0 \\ A-B+C-D=1\end{matrix}$
|
||||
$A=\frac{1}{8}$
|
||||
$B=-\frac{1}{4}$
|
||||
$C=\frac{1}{2}$
|
||||
|
|
|
|||
|
|
@ -22,7 +22,7 @@ Good luck on midterms! <3 -Oct 18 2023
|
|||
[Periodic functions (lec 19)](periodic-functions-lec-19.html) (raw notes, not reviewed or revised yet.)
|
||||
[Convolution (lec 19-20)](convolution-lec-19-20.html) (raw notes, not reviewed or revised yet.)
|
||||
[Dirak δ-function (lec 21)](dirak-δ-function-lec-21.html) (raw notes, not reviewed or revised yet.)
|
||||
[Power series (lec 22)](power-series-lec-22.html) (raw notes, not reviewed or revised yet.)
|
||||
[Power series (lec 22-23)](power-series-lec-22-23.html) (raw notes, not reviewed or revised yet.)
|
||||
</br>
|
||||
[How to solve any DE, a flow chart](Solve-any-DE.png) (Last updated Oct 1st, needs revision. But it gives a nice overview.)
|
||||
[Big LT table (.png)](drawings/bigLTtable.png)
|
||||
|
|
|
|||
Loading…
Reference in New Issue