revised up to and including method of undetermined coefficients

content/Cauchy-Euler equations (lec 10).md

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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.
+
+# Cauchy-Euler equations
+*if it has a name in it, its very important, if it has 2 names its very important!*
+$ax^2{\frac{d^2y}{dx^2}+bx{\frac{ dy }{ dx }}+cy=f(x)},\ x>0$
+where $a,\ b,\ c$ are still constants and $\in \mathbb{R}$
+note if x=0 is not interesting as the derivative terms disappear.
+how to solve? two approaches:
+textbook only use 2nd method. prof doesn't like this.
+you can find both methods in the profs notes.
+you know Stewart? multimillionaire, he's living in a mansion in Ontario.
+introduce change of variables:
+$x=e^t\Rightarrow t=\ln x$ (x is always +)
+(do $x=-e^t$ if you need it to be negative.)
+find derivatives with respect to t now. y is a function of t which is a function of x.
+$\frac{dy}{dx}=\frac{dy}{dt}{\frac{dt}{dx}}=\frac{ dy }{ dt }{\frac{1}{x}}\Rightarrow \underset{ Impor\tan t }{ x\frac{dy}{dx}=\frac{dy}{dt} }$
+compute 2nd derivative of y wrt to x:
+$\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2}\left( \frac{dt}{dx} \right)^2+\frac{dy}{dt}{\frac{d^2t}{dx^2}}=\frac{1}{x^2}{\frac{d^2y}{dt^2}}-\frac{\frac{1}{x^2}dy}{dt}$
+$\underset{ \mathrm{Im}por\tan t }{ x^22{\frac{d^2y}{dx^2}}=\frac{d^2y}{dt^2}-\frac{dy}{dt} }$
+$$a\frac{d^2y}{dt^2}+(b-a){\frac{dy}{dt}}+cy=f(e^t)$$
+^ this is a constant coefficient equation now! We can solve it now using prior tools.
+
+#ex 
+solve:
+$$x^2{\frac{d^2y}{dx^2}}+3x{\frac{dy}{dx}}+y=x^{-1},\ x>0$$
+$x=e^t$
+transform using the technique we showed just earlier:
+$\frac{d^2y}{dt^2}+2{\frac{dy}{dt}}+y=e^{-t}$
+1) $r^2+2r+1=0$
+$r_{1,2}=-1$
+$y_{h}(t)=c_{1}e^{-t}+c_{2}te^{-t}$
+2) $y_{p}(t)=At^2e^{-t}$ <- using method of undetermined coefficients
+$A=\frac{1}{2}$
+general solution in terms of t:
+$y_{1}(t)=c_{1}e^t+c_{2}te^{-t}+\frac{1}{2}t^2e^{-t}$
+bottom line: solution in terms of t, but we want solution wrt to x:
+$y_{1}(x)=c_{1}e^{-\ln(x)}+c_{2}\ln(x)e^{-\ln(x)}+\frac{1}{2}\ln(x)^2e^{-\ln(x)}$
+$=c_{1}x^-1+c_{2}\ln(x)x^-1+\frac{1}{2}{\ln(x)^2}x^-1$
+#end of lecture 10

content/Free vibrations (lec 11).md

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+#start of lecture 11
+last lecture we did cauchy euler equations:
+$ax^2{\frac{d^2y}{dx^2}+bx{\frac{ dy }{ dx }}+cy=f(x)},\ x>0$
+where $a,\ b,\ c$ are still constants and $\in \mathbb{R}$
+1) $x=e^t$
+$a{\frac{d^2y}{dt^2}}+(b-a){\frac{dy}{dt}}+cy=f(e^t)$ <- lousy notation, the y here isnt quite the same as in the above definition.
+2) $y=x^r$
+$ar^2+(b-a)r+C=0$
+three cases:
+(i) $r_1\ne r_{2}$
+then: $y_{h}(x)=c_{1}x^{r_{1}}+c_{2}x^{r_{2}}$
+(ii) $r_{1}=r_{2}=r$
+then: $y_{h}(x)=c_{1}x^r+c_{2}x^r\ln(x)$
+(iii) $r_{1,2}=\alpha+i\beta$
+then: $y_{h}(x)=x^2(c_{1}\cos(\ln \beta x)+c_{2}\sin \ln(\beta x))$
+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).
+
+not all equations can fall into cauchy euler type.
+$y''+p(x)y'+q(x)y=f(x)$ (1) <- no general solution procudure always
+but, if $y_{1}(x)$ solves $y''+p(x)y'+q(x)y=0$
+then we can find the general solution to the non homogenous equation (1) by guessing it in the form $y(x)=v(x)y_{1}(x)$
+$y'=v'y_{1}+vy_{1}'$
+$y''=v''y_{1}+2v'y_{1}'+vy_{1}''$
+$(v''y_{1}+2v_{1}'y_{1}'+y_{1}''v)+p(x)(v'y_{1}+vy_{1}')+q(x)vy_{1}=f(x)$
+$v\cancelto{ 0 }{ (y_{1}''+p(x)y_{1}'+q(x)y_{1}) }+v''y_{1}+(2y_{1}'+p(x)y_{1})v'=f$
+$y_{1}v''+()$
+$v''+\left( \frac{2y_{1}'}{y_{1}}+p \right)v'=\frac{f}{y_{1}}$
+$v'=u$
+$u'+\left( \frac{2y_{1}'}{y_{1}}+p \right)u=\frac{f}{y_{1}}$<- this is a linear first order equation
+how to solve linear first order equation? we  compute the integrating factor $\mu$
+$\mu=e^{\int(2y_{1}'/y_{1}+p)dx}=e^{\ln(y_{1})^2}e^{\int P(x) \, dx}=y_{1}^2e^{\int p(x) \, dx}$
+isnt this nice? some kind of magic. We made some guesses and we arrived somewhere.
+
+#ex find the general solution to the equation:
+$y''+4xy'+(4x^2+2)y=8e^{-x(x+2)}$
+if $y_{1}(x)=e^{-x^2}$ is one solution.
+therefore were finding the solution of the form: $y(x)=v(x)y_{1}=v(x)e^{-x^2}$
+$v'=u$
+$u'+\left( \frac{2y_{1}'}{y_{1}}+4x \right)u=\frac{8{e^{-x^2}e^{-2x}}}{e^{-x^2}}$ <-(p(x)=4x)
+$u'+\left( \frac{2{e^{-x^2}(-2x)}}{e^{-x^2}}+4x \right)u=8e^{-2x}$
+$u'=8e^{-2x}$
+$u=-4e^{-2x}+c_{1}$
+$v'=u=-4e^{-2x}+c_{1}$
+$v(x)=2e^{-2x}+c_{1}x+c_{2}$
+general solution:
+$$y(x)=v(x)y_{1}(x)=(2e^{-2x}+c_{1}x+c_{2})e^{-x^2}$$
+
+## Free vibrations
+$mr^2+br+k=0$ characteristic polynomail
+(i) $r_{1}\ne r_{2}$ $b^2-4mk>0$
+$y_{h}(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}$
+$r_{1,2}=-\frac{b}{2m}\pm \frac{\sqrt{ b^2-4mk }}{2m}<0$
+then the limit of the homogenous solution is 0 as t->$\infty$ (over damped case)
+(ii) $r_{1}=r_{2}=-\frac{b}{2m}$
+$r_{1}=r_{2}=-\frac{b}{2m}$
+$y_{h}(t)=e^-\frac{b}{2m}+c_{2}te^{-b/2m}t$ limit =0 as t approches inf critically damped

content/Method of undetermined coefficients (lec 8-9).md

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+
+#start of lec 8 (sept 22)
+last lecture we talked about $ay''+b'y+cy=f(t)$
+
+in the case when $f(t)=0$ :
+1) $ay''+b'y+cy=0$
+then $ar^2+br+c=0$ and solve with quadratic formula
+general solutions are: 
+if $r_{1}\ne r_{2}\Rightarrow y_{h}(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}$  <- **overdamped** 
+if $r_{1}=r_{2}\Rightarrow y_{h}(t)=c_{1}e^{rt}+c_{2}te^{rt}$ <- **critically damped**
+if $r_{1,2}\in \mathbb{C}\Rightarrow y_{h}(t)=e^{\alpha t}(c_{1}\cos(\beta t)+c_{2}\sin(\beta t))$ <- **underdamped**
+where h means homogenous, (when $f(t)=0$ the equation is homogenous.)
+
+in the case when $f(t)\ne 0$ :
+2) If $y_{p}(t)$ solves 1) then the general solution to $y(t)$ is $y(t)=y_{h}(t)+y_{p}(t)$
+theorem: if $p(t),\ q(t),\ f(t)$ are continuous on $I$ then the following IVP has a unique solution: $y''+p(t)y'+q(t)y=f(t)\quad \text{where}\quad y''(t_{o}),\ y'(t_{o}),\ y(t_{o})\in I$
+---
+# Method of undetermined coefficients:
+#ex #mouc Find the general solution for:
+$$y''-4y'+4y=3t+9$$
+The equation is certainly non-homogenous.
+First we have to find general solution to the homogenous equation (ie: find $y_{h}(t))$:
+1) $y''-4y'+4y=0$
+characteristic eq: $r^2-4r+4=0$
+$r=2$ (repeated root)
+$y_{h}(t)=c_{1}e^{2t}+c_{2}te^{2t}$
+Good. Now we need $y_{p}(t):$
+Look at the equation again: $y''+{-4}y'+4y=3t+9$
+We are looking for a particular polynomial where the power is not greater than 1. Because if for example $y_{p}(t)=t^2$ then the LHS would be a degree 2 polynomial and yet the RHS is only a degree one polynomial.
+So we guess that the equation will be of the form:
+2) $y_{p}(t)=At+B$
+$y_{p}'=A,\ y_{p}''=0$
+$0-4A+4(At+B)=3t+9$
+$4A=3,\ -4A+4B=9$
+$A=\frac{3}{4},\ B=3$
+$y_{p}(t)=\frac{3}{4}t+3$ <- our guess worked!
+general solution: $$y(t)=c_{1}e^{2t}+c_{2}te^{2t}+\frac{3}{4}t+3$$
+So the big takeaway from this example is if the RHS of the eq is a polynomial of degree u, we try to find a solution as a polynomial of degree u
+
+#ex #second_order_nonhomogenous #mouc
+Find the general solution of the following:
+$$y''-4y'+4y=2e^{2t}$$
+
+1) $y_h(t)=c_{1}e^{2t}+c_{2}te^{2t}$ (computed earlier)
+2) $y_p(t)=\ ?$
+we observe the RHS is some exponential, we need the function + its derivative + its second derivative to equal that, we have no option but suspect that its $Ae^{2t}$
+but then the LHS becomes 0! -> $4Ae^{2t}-4\cdot 2Ae^{2t}+4Ae^{2t}=0$
+so $Ae^{2t}$ is a wrong guess.
+So what do we do? Let's try $Ate^{2t}$ take $c_{2}=A, c_{1}=0$, this does not work again. LHS becomes 0 again -> $A(t4e^{2t}+2e^{2t}+2e^{2t})-4A(t2e^{2t}+e^{2t})+4Ate^{2t}=0$
+so let's try $At^2e^{2t}$: 
+$A(\cancel{ t^24e^{2t} }+2e^{2t}2t+e^{2t}2+2t2e^{2t})-4A(\cancel{ t^22e^{2t} }+e^{2t}2t)+\cancel{ 4At^2e^{2t} }$
+$=8Ate^{2t}+2Ae^{2t}-8Ate^{2t}$
+$=2Ae^{2t}=2e^{2t},\ A=1$ This one works!
+we know the homogenous solution and the particular solution. Sum them together to get the general solution:
+$$y(t)=c_{1}e^{2t}+c_{2}te^{2t}+t^2e^{2t}$$
+Moral of story? if RHS is constant times $e^{2t}$ we guess with an exponent with a constant, if its homogenous we multiply by t, if still not a valid solution then we multiply by t again.
+#ex #IVP #second_order_nonhomogenous #mouc
+$$y''+2y'+2y=2e^{-t}+5\cos t \qquad y(0)=3,\ y'(0)=1$$
+We wanna solve this IVP! We know from earlier that it must have a unique solution.
+1) set RHS to 0: $r^2+2r+2=0$
+ $r_{1,2}=-1\pm i$ 
+ >someone mentions sqrt(i). sqrt(i) is interesting, but not the topic for today.
+ 
+ $y_{h}(t)=e^{-t}(c_{1}\cos(t)+c_{2}\sin(t))$
+ 2) $y_{p}(t)=\ ?$
+ RHS is much more complicated, sum of 2 functions. Lets use principle of super position:
+ $y_{p}(t)=y_{p_{1}}(t)+y_{p_{2}}(t)$
+ where $y_{p_{1}}$ solves $y''+2y'+2y=2e^{-t}$
+ $y_{p_{2}}$ solves $y''+2y'+2y=5\cos (t)$
+ 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.
+ $y_{p_{1}}'=-Ae^{-t}$
+ $y_{p_{1}}''=Ae^{-t}$ plug in these three and we find that A=2
+ 
+second equation, not so easy:
+solution of $\cos(t)$ doesn't quite work because the LHS will obtain a $\sin$ term. Lets try this instead:
+$y_{p_{2}}=A\cos(t)+B\sin(t)$
+$y_{p_{2}}'=-A\sin(t)+B\cos (t)$
+$y_{p_{2}}''=-A\cos t-B\sin t$
+$(A+2B)\cos(t)+(-2A+B)\sin(t)=5\cos(t)$
+$A+2B=5$
+$-2A+B=0$ -> solving the system of linear equations yields: A=1, B=2
+but $y_{p_{1}}\ne y_{p_{2}}$ because of the $e^{-t}$ term.
+The general solution is:
+$y(t)=c_{1}e^{-t}\cos(t)+c_{2}e^{-t}\sin t+2e^{-t}+\cos t+2\sin t$
+now we solve the IVP:
+$y(0)=3=c_{1}+3=3\implies c_{1}=0$
+$y'(t)=c_{2}e^{-t}\cos(t)+\sin(t)(-1)e^{-t}-2e^t-\sin(t)+2\cos(t)$
+
+$y'(0)=c_{2}+0-2+0+2$
+$y'(0)=1=c_{2}$
+final solution to IVP:
+$$y(t)=e^{-t}(\sin t+2)+\cos t+2\sin t$$
+
+#end of lec 8 #start of lec 9
+*remember in a previous example when we had to guess that $y_{p}=At^2e^{2t}$? Here is a generalized algorithm that can find $y_{p}$ when the RHS falls under the following form. Reducing the guess work to zero:*
+# Generalized guesses for undetermined coefficients:
+case i) $ay''+by'+cy=P_m(t)e^{rt}$
+where $P_{m}(t)=a_{m}t^m+a_{m-1}t^{m-1}+ \dots +a_{0}$ *i.e. P is a polynomial degree m.*
+Then we guess the particular solution is of the form: $y_{p}(t)=t^s(b_{m}t^m+b_{m-1}t^{m-1}+\dots+b_{0})e^{rt}$
+where:
+s=0, if r is not a root,
+s=1 if r is a single root,
+s=2 if r is a double root.
+
+case ii) $ay''+by'+cy=P_{m}(t)e^{\alpha t}\cos(\beta t)+P_{m}(t)e^{\alpha t}\sin(\beta t)$
+Then we guess the particular solution is of the form: $y_{p}(t)=t^s[(A_{k}t^k+A_{K-1}t^{k-1}+\dots+A_{0})e^{\alpha t}\cos(\beta t)+(B_{k}t^k+B_{k-1}t^{k-1}+\dots+B_{0})e^{\alpha t}\sin(\beta t)]$
+where:
+s=0 if $\alpha+i\beta$ is not a root,
+s=1 if $\alpha+i\beta$ is a root.

content/More second order stuff (lec 8).md

@@ -1,83 +0,0 @@
-
-#start of lec 8 (sept 22)
-last lecture we talked about $ay''+b'y+cy=f(t)$
-
-in the case when $f(t)=0$ :
-1) $ay''+b'y+cy=0$
-then $ar^2+br+c=0$ and solve with quadratic formula
-general solution is: $y_{h}(t)=c_{1}e^{r_{1}(t)}+c_{2}e^{r_{2}t}$ where h means homogenous, ( because when =0 its homogenous)
-
-if $r_{1}=r_{2}$ then $y_{h}(t)=c_{1}e^{r(t)}+c_{2}e^{rt}$
-if imaginary roots:
-$y_{h}(t)=e^{\alpha t}(c_{1}\cos(\beta t)+c_{2}\sin(\beta t))$
-2) If $y_{p}(t)$ solves 1) then its general solution is $y(t)=y_{h}(t)+y_{p}(t)$
-theorem: if $p(t),\ g(t),\ f(t)$ are continuous on $I$ then the IVP $y''+p(t)y'+q(t)y=f(t), y(t_{o}),\ y'(t_{o})=y_{1} t_{o}\in I$ has a unique solution
-method of undetermined coeffecients:
-#ex
-$y''\pm_{4}y'+4y=3t+9$ lets find general solution, its centainly non homogenous.
-first we have to find general solution to the homogenous equation:
-1) $y''-4y'+4y=0$
-characteristic eq: $r^2-4r+4=0$ what are the roots?
-$r=2$ (repeated solution)
-$y_{h}(t)=c_{1}e^{2t}+c_{2}te^{2t}$
-we are looking for a particular polynomial where the power is not greater than 1 (?)
-2) $y_{p}(t)=At+B$
-$y_{p}'=A,\ y_{p}''=0$
-$-4A+4(At+B)=3t+9$
-$4A=3,\ -4A+4B=9$
-$A=\frac{3}{4},\ B=3$
-$y(t)=\frac{3}{4}t+3$
-general solution: $$y(t)=c_{1}e^{2t}+c_{2}te^{2t}+\frac{3}{4}t+3$$
-so big takeaway is if the RHS of eq is a polynomial of  degree u, we try to find a solution as a polynomial of degree u
-
-#ex 
-$$y''-4y'+4y=e^{2t}$$
-find general solution.
-1) $y_h(t)=c_{1}e^{2t}+c_{2}te^{2t}$ (computed earlier)
-2) $y_p(t)$ 
-we observe the RHS is some exponential, we need the derivative + its second derivative to equal that, we have no option but suspect that its $Ae^{2t}$
-but then the LHS becomes 0!
-so $Ae^{2t}$ is a wrong guess.
-so what do we do? try $Ate^{2t}$ take $c_{2}=A, c_{1}=0$, this does not work again. LHS becomes 0 again
-so try $At^2e^{2t}$
-$2Ae^{2t}=2e^{2t},\ A=1$ This one works!
-we know the homogeenous solution.
-$y(t)=c_{1}e^{2t}+c_{2}te^{2t}+t^2e^{2t}$ is the general solution
-moral of sotry? if RHS is constant times $e^2t$ we guess with an exponent with a constant, if its homogenous we multiply by t, if still not a valid solution then we multiply by t again.
-Ex:
-$y''+2y'+2y=2e^{-t}+\cos t,\ y(0)=3,\ y'(0)=1$ I wanna solve this IVP! it must have a unique solution.
-1) set RHS to 0: $r^2+2r+2=0$
- $r_{1,2}=-1\pm i$ sqrt(i) is interesting, but not the topic for today.
- $y_{h}(t)=e^{-t}(c_{1}\cos(t)+c_{2}\sin(t))$
- 2) $y_{p}(t)=$
- RHS is much more complicated, sum of 2 functions. Lets use principle of super position
- $y_{p}(t)=y_{p_{1}}+y_{p_{2}}$
- where $y_{p_{1}}$ solves $y''+2y'+2y=2e^{-t}$
- $y_{p_{2}}$ solves $y''+2y'+2y=5\cos (t)$
- 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.
- $y_{p_{1}}'=-Ae^{-t}$
- $y_{p_{1}}''=Ae^-t$ plug in these three and we find that A=2
- 
-second equation, not so easy:
-solution of cos t doenst quite work
-$y_{p_{2}}=A\cos(t)+B\sin(t)$
-$y_{p_{2}}'=-A\sin(t)+b\cos (t)$
-$y_{p_{2}}''=-A\cos t-B\sin t$
-$(A+2B)\cos(t)+(-2A+B)\sin(t)=5\cos(t)$
-$A+2B=0$
-$-2A+B=0$ -> A=1, B=2
-but $y_{p_{1}}\ne y_{p_{2}}$ because of the $e^{-t}$ term
-$y(t)=c_{1}e^{-t}\cos(t)+c_{2}e^{-t}\sin t+2e^{-t}+\cos t+2\sin t$
-$y(0)=3=c_{1}+3=3\implies c_{1}=0$
-$y'(0)=1=c_{2}$
-final solution $y(t)=e^{-t}(\sin t+2)+\cos t+2\sin t$
-
-If we have an equation of the from:
-1) $ay''+by'+cy=P_{m}(t)e^{rt}$
-where $p_{m}(t)=a_{m}t^m+a_{m-1}t^{m-1}+ \dots +a_{0}$
-then the guess is: $y_{p}(t)=t^s(b_{mt}t^m+b_{m-1}t^{m-1}+\dots+b_{0})e^{rt}$
-(i) s=0 if r is not a characteristic polynomial
-(ii)) s=1 if r is a single root
-(iii) s=2 if r is a double root
-we will talk about this more in the coming lecture.
-#end of lec 8

content/Second order homogenous linear equations (lec 5-7).md

@@ -16,9 +16,9 @@ The equation $my''+by'+ky=0$ is a homogenous second order equation. (in this cas
 It's called second order because we have second derivative in the equation.
 
 
-#ex #second_order 
+#ex #second_order_homogenous
 ### $$y''-4y'+3y=0$$
-(This is obviously homogenous as stated by prof, although I don't understand why that is.)
+(This equation is homogenous as the RHS is equal to 0)
 Imagine there's no y' (meaning no friction) you kind want the derivates to equal itself, an exponential!
 We guess the solution is of the form $y(t)=e^{rt}$
 $y(t)=e^{rt}$
@@ -38,10 +38,10 @@ and we're done.
 ---
 #end of lec 5 #start of lec 6
 
-#ex #IVP #second_order Same equation from last lecture, but now an IVP:
+#ex #IVP #second_order_homogenous Same equation from last lecture, but now an IVP:
 
 $$y(t)=c_{1}e^{t}+c_{2}e^{3t} \quad c_{1},c_{2}\in\mathbb{R} \quad\text{let } y(0)=0,\ y'(0)=4\quad \text{ What is } c_{1}, c_{2}?$$
-> Lemma: $y'(t)=c_{1}y_{1}+c_{2}y_{2}$
+> Lemma: if $y(t)=c_{1}e^{t}+c_{2}e^{3t}$ then $y'(t)=c_{1}y_{1}+c_{2}y_{2}$
 > proof: let $y_{1}=e^{r_{1}t}\qquad y_{2}=e^{r_{2}t}$
 > $y(t)=c_{1}e^t+c_{2}e^{3t}$
 > $y'(t)=c_{1}r_{1}e^{r_{1}t}+r_{2}c_{2}e^{r_{2}t}$
@@ -60,11 +60,12 @@ Solving the linear system of equations gives: $c_{1}=-2,\ c_{2}=2$ which gives t
 $$y'(t)=-2e^t+2e^{3t}$$
 
 ---
+# Principle of super position
 Remember from the example above where I said the "general solution is the sum of the two possibilities"? Let's explore and see why that is:
 Recap: suppose we have an equation of the form $ay''+by'+cy=0$
 $y(t)=e^{rt}$
 then $ar^2+br+c=0$
-case i) $r_{1},r_{2}=\frac{{-b\pm \sqrt{ b^2-4ac }}}{2a}, {r_{1}}\ne r_{2}$
+case i) $r_{1},r_{2}=\frac{{-b\pm \sqrt{ b^2-4ac }}}{2a}, {r_{1}}\ne r_{2}$ (over damped case, there are 3 possible cases, see below.)
 $y_{1}(t)=e^{r_{1}t}\qquad y_{2}(t)=e^{r_{2}t}$
 $y(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}$ is also a solution. But why? Principle of super position.
 ## Principle of super position:
@@ -84,9 +85,9 @@ $y_2(t)=e^{r_{2}t}$ solves $ay''+by'+cy=0$
 $f_{1}(t)=f_{2}(t)=0$
 This following can be concluded:
 $y(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}$ must solve $ay''+by'+cy=0$ by principle of super position.
-> Yay! Note this is only true when $f_{1}(t)=f_{2}(t)=0$ aka your RHS in the second order equation must be 0.
+> Yay! Note this is only true when $f_{1}(t)=f_{2}(t)=0$ aka your RHS in the second order equation must be 0. aka the equation is homogenous.
 
-case ii) $r_{1}=r_{2}=\frac{-b}{2a}$ if $b^2-4ac=0$
+case ii) $r_{1}=r_{2}=\frac{-b}{2a}$ if $b^2-4ac=0$ (critically damped)
 if we assume $y_{1}=e^{r_{1}t}, y_{2}=e^{r_{1}t}$ like before then we get:
 $y(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{1}t}=ce^{r_{1}t}$ <- this doesn't seem like it works! We need two integration constants for a second order equation.
 $y_{1}(t)=e^{-bt/2a}, y_{2}(t)=te^{-bt/2a}$ for time being we take this as true, we can prove it later.
@@ -96,83 +97,94 @@ we can check later at home, but also, how was the idea for this found? He will t
 ### linear algebra 101: linear independence makes unit vectors, which forms a basis.
 definition: if $y_{1}, y_{2}$ are solutions to $a(t)y''+b(t)y+c(t)=0$ on some interval $I_{1}$
 then they are called linearly independent if none of them is a constant multiple of the other.
-Theorem: If $y_{1}(t), y_{2}(t)$ are linearly independent solutions to $ay''+by'+cy=0$ then any other solution can be written as $y(t)=c_{1}y_{1}(t)+c_{2}y_{2}(t)$
+Theorem: If $y_{1}(t), y_{2}(t)$ are linearly independent solutions to $ay''+by'+cy=0$ then any other solution can be written as $y(t)=c_{1}y_{1}(t)+c_{2}y_{2}(t)$ (due to super position)
 
 how do we know the two solutions are linearly independent? Test for linear independence:
 $y_{1}, y_{2}$ are solutions to $a(t)y''+b(t)y+c(t)y=0$ on some interval $I_{1}$
-then they are called linearly independent iff 
+then they are called linearly independent iff:
 $W(y_{1},y_{2})(t)=\det \begin{pmatrix}y_{1} & y_{2} \\y_{1}' & y_{2}'\end{pmatrix}\ne 0$
-(i) $b^2-4ac>0 \Rightarrow r_{1}\ne r_{2}$
+case i) $b^2-4ac>0 \Rightarrow r_{1}\ne r_{2}$
 $y_{1}=e^{r_{1}t}, y_{2}=e^{r_{2}}t$
 $W(y_{1},y_{2})=\det\begin{pmatrix}e^{r_{1}t} & e^{r_{2}t} \\r_{1}e^{r_{1}t} & r_{2}e^{r_{2}t}\end{pmatrix}$
-$=e^{t(r_{1}+r_{2})}(r_{2}-r_{1})\ne 0$
+$=e^{t(r_{1}+r_{2})}(r_{2}-r_{1})\ne 0$ (because this can only be 0 when $r_{1}= r_{2}$ but we know $r_{1}\ne r_{2}$)
-(ii) $b^2-4ac=0\Rightarrow$
+case ii) $b^2-4ac=0\Rightarrow$
 $r_{1}=r_{2}=-\frac{b}{2a}=r$
 $y_{1}(t)=e^{rt}, y_{2}(t)=te^{rt}$
 $W(y_{1},y_{2})=\det\begin{pmatrix}e^{rt} & te^{rt} \\re^{rt} & e^{rt}(1+rt)\end{pmatrix}$
-$=e^{2rt}(1+rt)-rte^{rt}e^{2rt}$
+$=e^{2rt}(1+rt)-rte^{2rt}$
 $=e^{2rt}\ne 0$
+> But what about case iii) ? it wasn't covered here. But now we proven that for case i and ii their two respective solutions are linearly independent and therefore we know we can safely apply the principle of super position on them and obtain their general solutions.
 
+#ex #IVP #second_order_homogenous
+$$y''-2y'+y=0 \qquad y(0)=1,\ y'(0)=0$$
 
-#ex #IVP #second_order
+$e^{rt}({ r^2-2r+1 })=0$
-$$y''-2y'+y=0, y(0)=1, y'(0)=0$$
+$(r-1)^2=0\Rightarrow r_{1}=r_{2}=1$ (this is case ii, critically damped)
-
-$e^{rt}(\underset{ = }{ r^2-2r+1 })=0$
-$(r-1)^2=0\Rightarrow r_{1}=r_{2}=1$
 $y(t)=c_{1}e^t+c_{2}te^t$
 $y(0)=c_{1}=1$
 $y(t)=e^t+c_{2}te^t$
-$y'(0)=0=1+c_{2}\Rightarrow c_{2}=1$
+
+
+$y'(0)=0$
+$=e^0+c_{2}(0e^0+1\cdot e^0)$
+$=1+c_{2}$
+$\Rightarrow c_{2}=-1$
 $$y(t)=e^t-te^t$$
-Wow, that was a lot today, my notes look like a mess. I'll have to clean this up and understand what's going on later.
+#end of lecture 6 #start of lecture 7 (sept 20)
-#end of lecture 6
 
-
+Last lecture we covered case i) and case ii).
-#start of lecture 7 (sept 20)
-from last class: if $r_{1}=r_{2}=-\frac{b}{2a}=r$
-$y_{1}=e^{rt}, \quad y_{2}=te^{rt}$
-general solution:
-$$y(t)=c_{1}y_{1}(t)+c_{2}y_{2}(t)$$
 however there's a third option:
 $b^2-4ac<0$
-then we have complex roots
+Then we have complex roots:
-$r_{1,2}=-\frac{b}{2a}\pm\frac{i\sqrt{ 4ac-b^2 }}{2a}=\alpha+i\beta$ <- Complex conjugates. And due to fundamental theorem of algebra, there are only 2 roots.
+$r_{1,2}=-\frac{b}{2a}\pm\frac{i\sqrt{ 4ac-b^2 }}{2a}=\alpha\pm i\beta$ <- Complex conjugates. And due to fundamental theorem of algebra, there are only 2 roots in this degree 2 polynomial equation.
 
 $e^{r_{1}t}=e^{(\alpha+i\beta)t}=e^{\alpha t}+e^{i\beta t}$
 side note: there are no numbers that are more than two components that are "useful", even quaternions
-$e^{i\beta t}=e^{i\theta}$
+let $e^{i\beta t}=e^{i\theta}$
 expand into power series:
-$=1+\frac{i\theta}{1!}+\frac{{(i\theta)^2}}{2!}=\frac{{(i\theta)^3}}{3!}\dots$
+>Recall from math 101: $e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$
-$=1+\frac{i\theta}{1!}-\frac{\theta^2}{2!}-\frac{i\theta^3}{3!}+\frac{\theta^4}{4!}+\frac{i\theta^5}{5!}+\dots$
-$=\left( 1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\dots \right)+i\left( \frac{\theta}{1!}-\frac{\theta^3}{2!}+\frac{\theta^5}{3!}\dots\right)$
-$e^{i\theta}=\cos(\theta)+i\sin(\theta) \quad \Box$ We have proven the Euler formula 
 
-$y(t)=e^{rt}=e^{\alpha t}(\cos \beta t+i\sin \beta t)$
+$=1+\frac{i\theta}{1!}+\frac{{(i\theta)^2}}{2!}+\frac{{(i\theta)^3}}{3!}+\dots$
+$=1+\frac{i\theta}{1!}-\frac{\theta^2}{2!}-\frac{i\theta^3}{3!}+\frac{\theta^4}{4!}+\frac{i\theta^5}{5!}-\dots$
+$=\left( 1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\dots \right)+i\left( \frac{\theta}{1!}-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\dots\right)$
+>Recall from math 101: $\cos(x)=\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}\quad\text{and}\quad \sin(x)=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}$
+
+$e^{i\theta}=\cos(\theta)+i\sin(\theta) \quad \Box$ We have proven the Euler formula.
+
+we guess the solution is of the form: $y(t)=e^{rt}=e^{(\alpha+i\beta)t}=e^{\alpha t}(\cos \beta t+i\sin \beta t)$
 Lemma: If $u(t)+iv(t)$ solves $ay''+by'+cy=0$ then $u(t),\ v(t)$ are also solutions.
 Proof:
 $a(u+iv)''+b(u+iv)'+c(u+iv)=0$
-$\underbrace{ { (au''+bu'+cu) } }_{ =0 }+i\underbrace{ (av''+bv'+cv) }_{ =0 }=0$
+$\underbrace{ { (au''+bu'+cu) } }_{ =0 }+i\underbrace{ (av''+bv'+cv) }_{ =0 }=0$ <- since the RHS is zero, both the real and imaginary parts must also equal zero.
 $y_{1}(t)=e^{\alpha t}\cos(\beta t),\ y_{2}(t)=e^{\alpha t}\sin(\beta t)$
 $\alpha=-\frac{b}{2a},\quad \beta={\frac{\sqrt{ 4ac-b^2 }}{2a}}$
+then by principle of super position:
 $y(t)=c_{1}y_{1}+c_{2}y_{2}$
 now we have to test the two solutions are linearly independent
-$W[y_{1},y_{2}]=\det\begin{pmatrix}y_{1}  & y_{2}  \\ y_{1}' & y_{2}'\end{pmatrix}\ne0$ <- remember to do/show this at home
+$W[y_{1},y_{2}]=\det\begin{pmatrix}y_{1}  & y_{2}  \\ y_{1}' & y_{2}'\end{pmatrix}\ne0$
+$=\det \begin{pmatrix}e^{\alpha t}\cos(\beta t)& e^{\alpha t}\sin(\beta t)\\ -\beta\sin(\beta t)e^{\alpha t}+\cos(\beta t)\alpha e^{\alpha t} & \beta\cos(\beta t)e^{\alpha t}+\sin(\beta t)\alpha e^{\alpha t}\end{pmatrix}$
+$=e^{2\alpha t}[\cos\beta t(\beta \cos(\beta t)+\alpha\sin(\beta t))-\sin \beta t(-\beta \sin(\beta t)+\alpha\cos(\beta t))]$
+$=e^{2\alpha t}(\beta \cos^2(\beta t)+\beta \sin^2(\beta t))$
+$=e^{2\alpha t}\beta\ne 0$
+> Yay! We have shown the two solutions are indeed always linearly independent.
 
-#ex #IVP 
+#ex #IVP #second_order_homogenous  
 $$y''-2y'+5y=0 \quad y(0)=0 \quad y'(0)=2$$
 $r^2-2r+5=0$<-characteristic equation
+$r_{1,2}=\frac{{-b\pm \sqrt{ b^2-4ac }}}{2a}$
 $r_{1,2}={1\pm \frac{\sqrt{ -4b }}{2}}=1\pm2i$
 $y_{1}=e^t\cos(2t)$ $y_{2}=e^t\sin(2t)$
-general solution: $y(t)=e^{\alpha t}(c_{1}\cancel{ \cos 2 t }+c_{2}\sin 2 t)$
+general solution: $y(t)=e^t(c_{1} \cos 2 t +c_{2}\sin 2 t)$
-$y(0)=0=c_{1}$
+$y(0)=0=c_{1}$ <- this helps us calculate y' easier as we can cross out the $\cos(2t)$ term before taking the derivative.
-$y'(0)=c_{2}(e^t)$
+$y'(t)=\frac{d}{dt}(e^tc_{2}\sin(2t))=e^tc_{2}2\cos(2t)+c_{2}\sin(2t)e^t$
-*I missed stuff here that he erased*
+$y'(0)=2=c_{2}2$
-general solution is:
+therefore: $c_{1}=0,\ c_{2}=1$ and the solution to the IVP is:
 $$y(t)=e^t\sin(2t)$$
-it has a nice graph, where if it was a circuit it would blow up
+The solution has a nice graph, where if it was a circuit it would blow up,
-or if it was a bridge it would collapse
+or if it was a bridge it would oscillate and eventually collapse.
 
-## Something more difficult now:
+## In the next lecture:
+something more difficult now:
 $ay''+by'+cy=f(t)$ Again, a mass-spring system without any external force.
 if f(t)=0 we can find the solution easily and use superposition to get the general solution
 $ay''+by'+cy=0$
@@ -185,4 +197,4 @@ $y(t)=c_{1}y_{1}(t)+c_{2}y_{2}(t)+y_{p}(t)$ must solve $ay''+by'+cy=f(t)$
 
 Theorem: If $a(t),\ b(t),\ c(t)$ are continuous on $I$ , then IVP: $a(t)y''+b(t)y'+c(t)y=f(t)$ ; $y(t_{o})=y_{o}$ \ , $y'(t_{o})=y_{1}$ has a unique solution.
 we will do the proofs next class.
-#end of lecture 7
+#end of lecture 7

content/Separable equations (lec 1).md

@@ -44,4 +44,8 @@ $C=-1$
 So, the answer is:
 $$y=\arctan(x^2+1)$$
 
-#end of Lecture 1
+#end of Lecture 1
+
+$\int \frac{1}{x^5} \, dx$
+$a\in\mathbb{C}$
+$\Rightarrow$

content/Solve-any-DE.canvas

@@ -22,7 +22,7 @@
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 		{"id":"3f081acda4f30a27","type":"text","text":"solve for $r_{1}$ & $r_{2}$ using quadratic formula","x":1315,"y":183,"width":274,"height":125},
 		{"id":"4ffaa5c9a7e8d22b","type":"text","text":"use principle of super position","x":1329,"y":29,"width":245,"height":96},
-		{"id":"cd7490f8cce0b6e0","type":"file","file":"Math 201/Lectures/Second order linear equations (lec 5-7).md","x":1215,"y":780,"width":474,"height":145},
+		{"id":"cd7490f8cce0b6e0","type":"file","file":"Math 201/Lectures/Second order homogenous linear equations (lec 5-7).md","x":1215,"y":780,"width":474,"height":145},
 		{"id":"cd31ca74652b6936","type":"text","text":"substitute $y(t)=e^{rt}$ and its derivatives in the equation","x":1293,"y":556,"width":317,"height":122},
 		{"id":"e063ab92aef817e4","type":"text","text":"divide both sides of equation by $e^{rt}$","x":1322,"y":382,"width":260,"height":110},
 		{"id":"d98da52cb7139c25","type":"text","text":"$N(x,y)=\\frac{\\partial}{\\partial y} \\int M \\, dx+g(y)$\n","x":514,"y":354,"width":328,"height":77},

content/Variation of parameters (lec 9-10).md

@@ -1,20 +1,7 @@
-#start of lec 9
+# Variation of parameters
-1) $ay''+by'+cy=P_m(t)e^{rt}$
-$y_{p}(t)=t^s(b_{m}t^m+b_{m-1}t^{m-1}+\dots+b_{0})e^{rt}$
-s=0, if r is not a root
-s=1 if r is a single root
-s=2 if r is a double root
-where P is a polynomial degree m.
-
-2) $ay''+by'+cy=P_{m}(t)e^{\alpha t}\cos(\beta t)+P_{m}(t)e^{\alpha t}\sin(\beta t)$
-3) $y_{p}(t)=t^s[(A_{k}t^k+A_{K-1}t^{k-1}+\dots+A_{0})e^{\alpha t}\cos(\beta t)+(B_{k}t^k+B_{k-1}t^{k-1}+\dots+B_{0})e^{\alpha t}\sin(\beta t)]$
-s=0 if $\alpha+i\beta$ is not a root
-s=1 if $\alpha+i\beta$ is a root
-
-variation of parameters:
 $ay''+by'+cy=f(t)$
 1) $y_{h}=c_{1}y_{1}(t)+c_{2}y_{2}t$ <- h is homogenous, ie: $f(t)=0$
-lagrange proposed: find a particular solution of y_p
+Lagrange proposed: find a particular solution of $y_{p}$
 $y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$ <- btw $y_{1}$ and $y_{2}$ are often called a fundamental pair.
 we put y_p into the equation and make it equal to the RHS
 $y'_{p}=v_{1}y_{1}+v_{1}y_{1}'+v_{2}'y_{2}+v_{2}y_{2}'$
@@ -24,14 +11,14 @@ so $y_{p}''=v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}''$
 $a(v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}'')+b(v_{1}y_{1}'+v_{2}y_{2}')+c(v_{1}y_{1}+v_{2}y_{2})=f(t)$
 $v_{1}(ay_{1}''+\cancelto{ 0 }{ by_{1}' }+cy_{1})+v_{2}(ay_{2}''+\cancelto{ 0 }{ by_{2}'C }+cy_{2})+a(v_{1}'y_{1}'+v_{2}'y_{2}')$
 2) $v_{1}'y_{1}'+v_{2}'y_{2}'=\frac{f(t)}{a}$
-$\det \begin{pmatrix}y_{1} & y_{2} \\y_{1}'& y_{2}'\end{pmatrix}$ = rronsky = $W[y_{1},y_{2}]\ne 0$ this can never be 0!
+$\det \begin{pmatrix}y_{1} & y_{2} \\y_{1}'& y_{2}'\end{pmatrix}$ = Wronskian = $W[y_{1},y_{2}]\ne 0$ 
-by definition $y_1$ and $y_2$ are linearly independant solutions so the above can never be 0!
+by definition $y_1$ and $y_2$ are linearly independent solutions so the above can never be 0!
 $v_{1}'=\frac{{f(t)y_{2}t}}{aW[y_{1},y_{2}]}$; $v_{2}'=-\frac{{f(t)y_{1}(t)}}{aW[y_{1},y_{2}]}$ <- integrate both sizes to get v1,2. When integrating, you don't need to add a generic constant.
 
 #ex #second_order #IVP 
 $y''+4y=2\tan(2t)-e^t \qquad y(0)=0 \qquad y'(0)=\frac{4}{5}$
 can we use undetermined coefficients? yes and no
-find general solution to homogenous countepart
+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
@@ -68,4 +55,47 @@ $y(0)=0=c_{1}-y_{p}(0)=c_{1}-\frac{1}{5}\Rightarrow c_{1}=\frac{1}{5}$
 skipping some differentiation: $y'(0)=2c_{2}+y_{p}'(0)=2c_{2}+v_{1}'(0)+2v_{2}(0)-\frac{1}{5}=\frac{4}{5}\Rightarrow c_{2}=1$
 $y(t)=\frac{1}{5}\cos(2t)+\sin(2t)-\frac{1}{5}e^t+v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)$
 
-#end of lecture 9
+#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
+$$y''-2y'+y=e^t\ln(t)+2\cos(t)$$
+i) $y_{h}(t)=?$
+$r^2-2r+1=0$
+$r_{1,2}=1$
+$y_{h}(t)=c_{1}e^t+c_{2}te^t$
+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.)
+$y''-2y'+y=e^t\ln(t)$ 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) $e^t+v_{1}'+te^tv_{2}'=0$
+eq2) $e^tv_{1}'+(te^t+e^t){v_{2}'}=e^t{\ln t}$
+subtract eq1 from eq2 $v_{2}'=\ln(t)$
+$v_{2}(t)=\int \ln(t) \, dt$
+integrate by parts
+$=t\ln(t)-\int t\frac{1}{t} \, dt$
+$=t\ln(t)-t$ no constant of integration.
+compute $v_{1}$ now:
+$v_{1}'=-tv_{2}'$
+$=-t\ln t$
+integrate to get v_1:
+$v_{1}=-\int t\ln t \, dt$
+integrate by parts (btw integration by parts will be the most important integration technique in this course):
+$v_{1}=-\frac{1}{2}(t^2\ln t)-\int t^2\frac{1}{t} \, dt$
+$=-\frac{1}{2}\left( t^2\ln t-\frac{t^2}{2} \right)=-\frac{1}{2}t^2\ln t+\frac{1}{4}t^2$
+$y_{p}''(t)=(\frac{1}{2}t^2\ln t+\frac{1}{4}t^2)e^t+(t\ln t-t)te^t$
+$y_{p}(t)=-\sin(t)+\frac{1}{2}t^2\ln(t)e^t-\frac{3}{4}t^2e^t$
+general solution is produced by adding the homogenous eq with $y_{p}(t)$
+general: 
+$$y(t)=c_{1}e^t+c_{2}te^t+y_{p}(t)$$

content/_index.md

@@ -7,9 +7,12 @@ I have written these notes for myself, I thought it would be cool to share them.
 [Bernoulli equations (lec 3)](bernoulli-equations-lec-3.html)
 [Linear coefficient equations (lec 4)](linear-coefficient-equations-lec-4.html)
 [Exact equations (lec 4-5)](exact-equations-lec-4-5.html)
-[Second order linear equations (lec 5-7)](second-order-linear-equations-lec-5-7.html) (raw notes, not reviewed or revised yet.)
+[Second order homogenous linear equations (lec 5-7)](second-order-homogenous-linear-equations-lec-5-7.html) 
-[More second order stuff (lec 8)](more-second-order-stuff-lec-8.html) (raw notes, not reviewed or revised yet.)
+[Method of undetermined coefficients (lec 8-9)](method-of-undetermined-coefficients-lec-8-9.html)
-[Undetermined coefficients (lec 9)](undetermined-coefficients-lec-9.html) (raw notes, not reviewed or revised yet.)
+[Variation of parameters (lec 9-10)](variation-of-parameters-lec-9-10.html) (raw notes, not reviewed or revised yet.)
+[Free vibrations (lec 11)](free-vibrations-lec-11) (raw notes, not reviewed or revised yet.)
 </br>
 [How to solve any DE, a flow chart](Solve-any-DE.png)
 </br>
+I'd like to add a search by tag feature. I'm also thinking of hosting the source code for all this on a git server. That way, people can contribute and fix my notes for me :P
+It would also allow people to contribute or host their own notes, which would be pretty cool. (Side note: I am against sharing instructor materials without their approved consent.)