forked from Sasserisop/MATH201
added lec 16, inverse LT stuff
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we know how to solve second order equations where a,b,c are constants. Even if they're not constant some can be expressed as a linear equation. But not always will they be solvable. However, there is one class of second order equations with non constant coefficients that are always solvable.
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We know how to solve second order equations where a, b, c are constants. Even if they're not constant some can be expressed as a linear equation. But not always will they be solvable. However, there is one class of second order equations with non constant coefficients that are always solvable.
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# Cauchy-Euler equations
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*If it has a name in it, its very important, if it has 2 names, its very very important!*
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@ -44,4 +44,4 @@ $g(x)=3x+C_{1}$
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$F(x,y)=(x^2-1)y+g(x)\Rightarrow F(x,y)=(x^2-1)y+3x=C_{2}-C_{1}=C$
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We are done:
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$$(x^2-1)y+3x=C$$
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there is also another method to solve exact equations (see Wikipedia article, but the prof says this method is easier, I believe him)
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there is also another method to solve exact equations (see Wikipedia article, but the prof says this method is easier, I believe him)
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@ -3,17 +3,26 @@ 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 isnt quite the same as in the above definition.
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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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three cases:
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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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then: $y_{h}(x)=c_{1}x^r+c_{2}x^r\ln(x)$
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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: $y_{h}(x)=x^2(c_{1}\cos(\ln \beta x)+c_{2}\sin \ln(\beta x))$
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now find one particular solution for a non homogenous soultion, using variation of parameters, combine the y_h and y_p to get y(x).
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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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@ -1,63 +0,0 @@
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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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Here are some trivial example computations:
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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 #start of lec 15
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compute the LT of this funny function:
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$f(t)=\begin{pmatrix}1, & 0\leq t\leq 1 \\ 2, & 1<t\leq 2 \\0, & 2<t \end{pmatrix}$
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$F(s)=\int _{0}^1 e^{-st}\, dt+2\int _{1}^2 e^{-st}\, dt+0$
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$=-\frac{1}{s}e^{-st}|_{t=0}^{t=1}-\frac{2}{s}e^{-st}|_{t=1}^{t=2}=-\frac{1}{s}(e^{-s}-1)-\frac{2}{5}(e^{-2s}-e^{-s})$
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We have shown how to compute the LT of a choppy function.
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$\mathcal{L}\{\alpha f(t)+\beta g(t)\}=\alpha \mathcal{L}\{f\}+\beta y\mathcal{L}\{g\}$
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the LT is a linear operator. as shown above. Proof:
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$\mathcal{L}\{\alpha f(t)+\beta g(t)\}=\alpha \int_{0}^{\infty}f(t)e^{-st} \, dt+\beta \int_{0}^{\infty}g(t)e^{-st} \, dt \quad \Box$
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Def: $f(t)$ is piecewise continuous on an interval $I$ if $f(t)$ is continuous on $I$, except possibly at a /finite/ number of points of /jump/ discontinuity
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what is continuouity? the limit exists and equals the value at that point.
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![[Drawing 2023-10-11 13.17.32.excalidraw]]
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Def: f(t) is of exponential order $\alpha$ if there $\exists\ T,\ M, T\geq0$ such that $\forall\ t>T : f(t)\leq Me^{\alpha t}$
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this is important so that f(t)e^{-st} doesn't go into infinity,
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this can be proven, but not necessary
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Theorem: If $f(t)$ is piecewise continuous and of an exponential order $\alpha$ on $[0, \infty)$, then $\mathcal{L}\{f\}$ exists $s>\alpha$
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and we are guarenteed that $s>\alpha$
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$e^{\alpha t}$ is this of exponential order? Of course, $M=1,\alpha=\alpha$
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what about $\sin(t)$? yes, sin is bounded b/w -1 and 1
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What about $e^{t^2}$? No, t^2 always outgrows $\alpha t$ eventually. This function does not have a LT.
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![[Drawing 2023-10-11 13.21.18.excalidraw]]
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Theorem if $F(s)$ is a LT then $\lim_{ s \to \infty }F(s)=0$
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properties:
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assume LT of f(t) exists:
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first property: $\mathcal{L}\{e^{\alpha t}f(t)\}=\int _{0}^\infty e^{\alpha t}e^{-st}\, dt=\int _{0}^\infty e^{-(s-a)t}f(t)\, dt=F(s-a)$
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$\mathcal{L}\{e^{\alpha t}\cdot 1\}=\frac{1}{s-\alpha}$
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$\mathcal{L}\{e^{\alpha t}\sin(bt)\}=\frac{b}{(s-a)^2+b^2}$ these properties are essential for the midterm.
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using integration by parts:
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$\mathcal{L}\{f'(t)\}(s)=\int _{0}^\infty e^{-st}f(t)\, dt=e^{-st}f(t)|_{t=0}^{t\to \infty}+s \underbrace{ \int e^{-st}f(t) \, dt }_{ F(s) }$$=sF(s)-f(0)$
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$\mathcal{L}{f''}=s^2F(s)-sf(0)-f'(0))$
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using proof by induction (try at home!):
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$\mathcal{L}\{f^{(m)}\}=s^mF(s)-s^{m-1}f(0)-\dots-f^{m-1}(0)$
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trig stuff:
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$\mathcal{L}\{\sin bt\}=\frac{b}{s^2+b^2}$
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$\mathcal{L}\{\cos bt\}=\frac{s}{s^2+b^2}$
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$\cos(bt)=\frac{1}{b}(\sin(bt))'$
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$\frac{dF}{ds}(s)=\frac{d}{ds}\int _{0 } ^\infty e^{-st}f(t)\, dt=\int _{0} ^\infty f(t) \frac{d}{ds}(e^{-st}) \, dt$
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$=-\int _{0} ^\infty e^{-st}tf(t)\, dt$
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$-\frac{dF}{ds}=\mathcal{L}\{tf(t)\}$; again we can use induction to prove:
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$\mathcal{L}\{t''f(t)\}=(-1)^{n} \frac{d^nF}{ds^n}$
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$\mathcal{L}\{t 1\}=-(s^-1)^1=s^-2$; $\mathcal{L}\{t^n\}= \frac{n!}{s^{n+1}}$ (can be proven by induction)
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today covers all midterm material. Yay!
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#end of lec 15
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@ -0,0 +1,152 @@
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# Laplace transform
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From now on, LT is short for Laplace Transform #LT
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What is LT? It's denoted as $\mathcal{L}$ and it's an operator defined as the following:
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$${\underbrace{ \mathcal{L}\{f(t)\}(s) }_{ =F(s) }}:=\int _{0}^\infty e^{-st}f(t) dt=\lim_{ T \to \infty }\int _{0} ^T e^{-st}f(t)\, dt$$
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This doesn't look like anything useful, but later on we will learn how it is.
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Btw, an operator is something that takes in a function and spits out a new function. For example, integration is a kind of operator.
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Here are some trivial example computations:
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$\mathcal{L}\{0\}=0$ Look at your bank account, integrate 0 you still get 0 :D
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$\mathcal{L}\{1\}=\int_{0}^\infty e^{-st} \, dt=-\frac{1}{s}e^{-st}|_{0}^\infty=\frac{1}{s}$ if s>0
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$\mathcal{L}\{e^{at}\}$=$\int_{0}^{\infty} e^{at}e^{-st}\, dt=\int_{0}^{\infty}e^{-(s-a)t}dt=\frac{1}{s-a}$ if $s-a>0$
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$\mathcal{L}\{\sin bt\}=\int _{0}^{\infty}e^{-st}\sin(bt) \, dt=\frac{b}{s^2+b^2}$ by integration by parts
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similarly can be done for cos, but we have run out of time. (Don't worry, you will find these properties and many more on the table provided in the next lectures)
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#end of lec 14 #start of lec 15
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#ex #LT
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compute the LT of this funny function:
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$f(t)=\begin{cases}1 &\text{if } 0\leq t\leq 1 \\ 2 &\text{if } 1<t\leq 2 \\0 & \text{if } 2<t \end{cases}$
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$F(s)=\int _{0}^1 e^{-st}\, dt+2\int _{1}^2 e^{-st}\, dt+0$
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$F(s)=-\frac{1}{s}e^{-st}|_{t=0}^{t=1}-\frac{2}{s}e^{-st}|_{t=1}^{t=2}$
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$$F(s)=-\frac{1}{s}(e^{-s}-1)-\frac{2}{5}(e^{-2s}-e^{-s})$$
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We have shown how to compute the LT of a choppy function.
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$\mathcal{L}\{\alpha f(t)+\beta g(t)\}=\alpha \mathcal{L}\{f\}+\beta y\mathcal{L}\{g\}$
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the LT is a linear operator. as shown above. Proof:
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$\mathcal{L}\{\alpha f(t)+\beta g(t)\}=\alpha \int_{0}^{\infty}f(t)e^{-st} \, dt+\beta \int_{0}^{\infty}g(t)e^{-st} \, dt \quad \Box$
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Def: $f(t)$ is piecewise continuous on an interval $I$ if $f(t)$ is continuous on $I$, except possibly at a /finite/ number of points of /jump/ discontinuity
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what is continuouity? the limit exists and equals the value at that point.
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![[Drawing 2023-10-11 13.17.32.excalidraw]]
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Def: f(t) is of exponential order $\alpha$ if there $\exists\ T,\ M, T\geq0$ such that $\forall\ t>T : f(t)\leq Me^{\alpha t}$
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this is important so that f(t)e^{-st} doesn't go into infinity,
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this can be proven, but not necessary
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Theorem: If $f(t)$ is piecewise continuous and of an exponential order $\alpha$ on $[0, \infty)$, then $\mathcal{L}\{f\}$ exists $s>\alpha$
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and we are guarenteed that $s>\alpha$
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$e^{\alpha t}$ is this of exponential order? Of course, $M=1,\alpha=\alpha$
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what about $\sin(t)$? yes, sin is bounded b/w -1 and 1
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What about $e^{t^2}$? No, t^2 always outgrows $\alpha t$ eventually. This function does not have a LT.
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![[Drawing 2023-10-11 13.21.18.excalidraw]]
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Theorem if $F(s)$ is a LT then $\lim_{ s \to \infty }F(s)=0$
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properties:
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assume LT of f(t) exists:
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first property: $\mathcal{L}\{e^{\alpha t}f(t)\}=\int _{0}^\infty e^{\alpha t}e^{-st}\, dt=\int _{0}^\infty e^{-(s-a)t}f(t)\, dt=F(s-a)$
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$\mathcal{L}\{e^{\alpha t}\cdot 1\}=\frac{1}{s-\alpha}$
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$\mathcal{L}\{e^{\alpha t}\sin(bt)\}=\frac{b}{(s-a)^2+b^2}$ these properties are essential for the midterm.
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What if we calculate the LT of $f'$ ?
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using integration by parts:
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$\mathcal{L}\{f'(t)\}(s)=\int _{0}^\infty e^{-st}f(t)\, dt=e^{-st}f(t)|_{t=0}^{t\to \infty}+s \underbrace{ \int e^{-st}f(t) \, dt }_{ F(s) }$$=sF(s)-f(0)$
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$\mathcal{L}\{f''\}=s^2F(s)-sf(0)-f'(0)$
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in general, we can use proof by induction to show the following (try at home!):
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$\mathcal{L}\{f^{(m)}\}=s^mF(s)-s^{m-1}f(0)-\dots-f^{m-1}(0)$
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trig stuff:
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$\mathcal{L}\{\sin bt\}=\frac{b}{s^2+b^2}$
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$\mathcal{L}\{\cos bt\}=\frac{s}{s^2+b^2}$
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$\cos(bt)=\frac{1}{b}(\sin(bt))'$
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$F'(s)=\frac{dF}{ds}(s)=\frac{d}{ds}\int _{0 } ^\infty e^{-st}f(t)\, dt=\int _{0} ^\infty f(t) \frac{d}{ds}(e^{-st}) \, dt$
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$=-\int _{0} ^\infty e^{-st}tf(t)\, dt$
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so that means:
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$-\frac{dF}{ds}=\mathcal{L}\{tf(t)\}$; again we can use induction to prove:
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$\mathcal{L}\{t''f(t)\}=(-1)^{n} \frac{d^nF}{ds^n}$
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$\mathcal{L}\{t 1\}=-(s^{-1})^1=s^{-2}$
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$\mathcal{L}\{t^n\}= \frac{n!}{s^{n+1}}$ (can be proven by induction)
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today covers all midterm material. Yay!
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#end of lec 15
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#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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#ex #LT
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lets try to compute the LT of:
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$\mathcal{L}\{t\cos (t)e^t\}= ?$
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Look at the table, which one would be useful? (the bottom one?)
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We know $\mathcal{L}\{t^nf(t)\}=(-1)^n \frac{d^nF}{ds^n}$ this property was shown last lecture, can be proven by induction. It's good to learn this by heart.
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$\mathcal{L}\{t\cdot\underbrace{ \cos (t)e^t }_{ f(t) }\}=-\frac{d}{ds}\mathcal{L}\{e^t\cos t\}$
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how do we compute that $\mathcal{L}$? Well now we can use that last row in the table!
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$=-\frac{d}{ds}\left( \frac{{s-1}}{(s-1)^2+1^2} \right)$
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$$=\frac{{(s-1)^2-1}}{((s-1)^2+1)^2}$$
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Doing this normally would be hard, with the table its a piece of cake!
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# Inverse Laplace Transform
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we know LT is a operator, we can also define an inverse transform! The inverse of LT. #inv_LT
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Here's the definition:
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$$\mathcal{L}^{-1}\{F(s)\}=f(t)$$
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Recall the forward LT was: $\mathcal{L}\{f(t)\}=F(s)$
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Next time hopefully, we will see how this will all be useful for solving DE!
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We know the forward transform is linear, the inverse is also linear.
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$\mathcal{L}^{-1}\{\alpha F(s)+\beta G(s)\}=\alpha \mathcal{L}^{-1}\{F\}+\beta \mathcal{L}^{-1}\{G\}$
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This can be proven rather easily due to the linearity of the forward transform. (wasn't done in class unfortunately.)
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#ex
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$\mathcal{L}^{-1}\left\{ \frac{1}{s^5}+\frac{3}{(2s+5)^2}+\frac{1}{s^2+4s+8}+ \frac{{s-1}}{s^2+2s+10} \right\}$ notice that all these terms approach 0 as s approaches inf.
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using linearity:
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second term doesn't look like its in the table, 3rd term looks like second last row
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$\mathcal{L}^{-1}\left\{ \frac{1}{s^5} \right\}+\frac{3}{4}\mathcal{L}^{-1}\left\{ \frac{1}{\left( s+\frac{5}{2} \right)^2} \right\}+\frac{1}{2}\mathcal{L}^{-1}{\frac{2}{(s+2)^2+2^2}}+\mathcal{L}^{-1}\left\{ \frac{{s+1}}{(s+1)^2+3^2} \right\}$
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multiply the first term by 4!, divide outside by 4!
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=$\frac{1}{4!}t^4+\frac{3}{4}te^{-5t/2}+\frac{1}{2}\sin(2t)e^{-2t}+e^{-t}\cos(3t)$
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it's kinda fun, try to match each term with something in the table, like a puzzle.
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We are done.
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#ex #inv_LT
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$\mathcal{L}^{-1}\{\frac{1}{(s-3)(s^2+2s+2)}\}$ notice the numerator is at least one degree lower than the denominator. The limit of the overall term is zero as s->inf
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it doesn't look like anything in the table, can we factor the denominator? not really, they have complex solutions. So maybe split the terms using partial fractions!
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$=\mathcal{L}^{-1}\left\{ \frac{A}{{s-3}}+\frac{{Bs+C}}{s^2+2s+2} \right\}$
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$=\mathcal{L}^{-1}\left\{ \frac{{As^2+2As+2A+Bs^3+3Bs+Cs-3C}}{(s-3)(s^2+2s+2)} \right\}$
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we get a linear system of equations:
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$\begin{bmatrix}A+B&=0 \\2A-3B+C&=0 \\ 2A-3C&=1\end{bmatrix}$
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skipping some computations:
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$A=\frac{1}{17},\ B=-\frac{1}{17},\ C=-\frac{5}{17}$
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$=\frac{1}{17}\mathcal{L}^{-1}\left\{ \frac{1}{(s-3)} \right\}-\frac{1}{17}\mathcal{L}^{-1}\left\{ \frac{s}{(s+1)^2+1^2} \right\}-\frac{5}{17}\mathcal{L}^{-1}\left\{ \frac{1}{(s+1)^2+1^2} \right\}$
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$=\frac{1}{17}e^{3t}-\frac{1}{17}$
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for second term we use an important identity: $s=s+1-1$:
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$-\frac{1}{17}\mathcal{L}^{-1}\left\{ \frac{{s+1-1}}{(s+1)^2+1^2} \right\}$
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final answer:
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$$\frac{1}{17}e^{3t}-\frac{1}{17}e^{-t}\cos(t)-\frac{4}{17}\sin(t)e^{-t}$$
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partial fractions, hopefully you remember from math 101:
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For each term $(s+a)^k$
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we include $\frac{A_{1}}{s+a}+\frac{A_{2}}{(s+a)^2}+\dots + \frac{A_{k}}{(s+a)^k}$
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for each term $(s^2+as+b)^k$
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we include: $\frac{B_{1}s+c_{1}}{(s^2+as+b)}+\frac{{B_{2}s+c_{2}}}{(s^2+as+b)^2}+\dots+\frac{{B_{k}s+c_{k}}}{(s^2+as+b)^k}$
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#ex #inv_LT
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$\mathcal{L}^{-1}\left\{ \frac{{3s^2+5s+3}}{s^4+s^3} \right\}$
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factor out s^3 (s+1)s^3
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$\mathcal{L}^{-1}\left\{ \frac{A}{s}+\frac{B}{s^2}+\frac{C}{s^3}+\frac{D}{s+1} \right\}$
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$=\mathcal{L}^{-1}\left\{ \frac{{As^2(s+1)+Bs(s+1)+C(s+1)+Ds^3}}{(s+1)s^3} \right\}$
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we get a linear system:
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$\begin{matrix}A+D=0 \\A+B=3 \\B+C=5 \\C=3\end{matrix}$
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after some computation for solving linear system:
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$A=1,\ B=2,\ C=3,\ D=-1$
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so:
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$\mathcal{L}^{-1}\left\{ \frac{1}{s}+\frac{2}{s^2}+\frac{3}{s^3}+\frac{-1}{s+1} \right\}$
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$=1+2t+\frac{3}{2}t^2-e^{-t}$
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is the final answer.
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#ex #inv_LT
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$\mathcal{L}^{-1}\left\{ \ln \frac{{s^2+9}}{s^2+1} \right\}$
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how do we get rid of the ln? work with the derivative.
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property: $\mathcal{L}\{tf(t)\}=-\frac{dF}{ds}$
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$\mathcal{L}^{-1}\left\{ \frac{d}{ds}\ln \frac{{s^2+9}}{s^2+1} \right\}=-tf(t)$
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partial fractions:
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$\mathcal{L}^{-1}\left\{ -\frac{16s}{(s^2+1)(s^2+9)} \right\}=2\mathcal{L}^{-1}\left\{ \frac{s}{{s^2+9}}-\frac{s}{{s^2+1}} \right\}=2(\cos 3t-\cos t)$
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$f(t)=-\frac{2}{t}(\cos 3t-\cos t)$
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#end of lec 16
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@ -10,7 +10,9 @@ $a(x)y'+a'(x)y=f(x)=(ay)'$
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$ay=\int f(x) \, dx$ <-yay! We can find the solutions to y.
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we can rewrite the linear equation in what's called standard form:
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$$\frac{ dy }{ dx }+P(x)y=Q(x)$$ (I'm calling this #de_L_type2 )
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$$\frac{ dy }{ dx }+P(x)y=Q(x)$$
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>I'm calling this #de_L_type2 )
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we will define a function $\mu(x)$ called the integration factor, also expressed as $I(x)$
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Multiply both sides by $\mu(x)$
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$\mu(x) \frac{ dy }{ dx }+\underbrace{ \mu(x)P(x) }_{ \mu'(x) }y=\mu(x)Q(x)$
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@ -13,7 +13,7 @@ I have written these notes for myself, I thought it would be cool to share them.
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[Cauchy-Euler equations (lec 10)](cauchy-euler-equations-lec-10.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 14-15)](laplace-transform-lec-14-15.html) (raw notes, not reviewed or revised yet.)
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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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</br>
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[How to solve any DE, a flow chart](Solve-any-DE.png)
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Reference in New Issue