added lec 12

content/Free vibrations (lec 11).md

@@ -1,56 +0,0 @@
-#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/Free vibrations (lec 11-12).md

@@ -0,0 +1,120 @@
+#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 polynomial
+(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 approaches inf (critically damped)
+
+#end of lec 11 #start of lec 12 (oct 2 2023)
+![[Drawing 2023-10-02 13.02.06.excalidraw]]
+let $\omega =\frac{\sqrt{ 4mk-b^2 }}{2m}$ (angular frequency)
+then the underdamped case is:
+$y(t)=(c_{1}\cos \omega t+c_{2}\sin \omega t)e^{\frac{-b}{2m}t}$
+we know the trig identity:
+$\sin(\alpha+\beta)=\sin \alpha\cos \beta+\cos \alpha \sin \beta$
+cant make c_1 c_2 sin or cos so what we do?
+do a power transform to convert cartesian into cylindrical coordinates
+$c_{1}=A\sin \phi$
+$c_{2}=A\cos \phi$
+then:
+$Ae^{-bt/2m}(\sin \phi \cos \omega t+\cos \phi \sin \omega t)$
+$=Ae^{-bt/2m}\sin(\omega t+\phi)$ where $\phi$ is the phase shift.
+and $\frac{\omega}{2\pi}$ is the natural frequency
+$\frac{2\pi}{\omega}$ is the period
+but this is all classical mechanics, but beatifully the world of electronic circuits of R L C also has these equations. Biology too. Nature is beautiful and harmonic.
+btw we know $A=\sqrt{ c_{1}^2+c_{2}^2 }$
+and $\tan \phi=\frac{c_{1}}{c_{2}}$
+so we can get A and phi from c_1 and c_2.
+this under damped case also reaches 0 as t->$\infty$
+
+this system in the drawing is in free vibrattion (RHS=0 means no external force=free vibration.)
+#ex 
+$y''+by'+25y=0 \qquad y(0)=1\quad y'(0)=0$
+1) b=0 -> no friction in the system (undamped)
+$b^2-4mk$
+$y(t)=c_{1}\cos 5t+c_{2}\sin 5t$
+$y(0)=c_1=1$
+$y'(0)=0=c_{2}$
+then $\sin 5t\Rightarrow y(t)=\cos(5t)=\sin\left( 5t+\frac{\pi}{2} \right)$ (by trig identity)
+important take away from undamped case: amplitude is constant 1, oscillates forever.
+2) b=6
+compute $b^2-4mk=36-4*25=-64$
+$r_{1,2}=-\frac{6}{2}\pm4i$
+$y(t)=e^{-3t}(c_{1}\cos4t+c_{2}\sin4t)$
+still under damped situation.
+$y(0)=1=c_{1}$
+$y'(0)=0=-3c_{1}+4c_{2}\Rightarrow c_{2}=\frac{3}{4}$
+$A=\frac{5}{4}$
+$\tan \phi=\frac{4}{3}$
+$\phi \approx 0.9273\dots$
+$$y(t)=\frac{5}{4}e^{-3t}\sin(4t+\phi)$$
+"I know engineers loves calculators, I know mathematicians hates calculators, and that's probably the only difference between mathematicians and engineers." -Peter (referring to calculating arctan(4/3) on an exam)
+3) b=10
+$r_{1,2}=-5$
+$y(t)=(c_{1}+c_{2}t)e^{-5t}$
+$y(0)=1=c_{1}$
+$y'(0)=c_{2}-5c_{1}=0$
+$c_{2}=5$
+$y(t)=(1+5t)e^{-5t}\rightarrow0_{as\ t\to\infty}$
+$y(t)=(1+5t)e^{-5t}>0$
+4) b=12
+ $r_{1,2}=-6\pm \sqrt{ 11 }$
+$y(t)=c_{1}e^{(-6\pm \sqrt{ 11 })t}+c_{2}e^{(-6-\sqrt{11 })t}$
+$y(0)=c_{1}+c_{2}=1$
+$y'(0)=(-6+\sqrt{ 11 })c_{1}+(-6-\sqrt{ 11 })c_{2}=0$
+$c_{1}=\frac{11+6\sqrt{ 11 }}{22}$
+$c_{2}=\frac{{11-6\sqrt{ 11 }}}{22}$
+this is an over damped case.
+
+lets look at the graphs: (graphs featuring the three cases shown on projector.)
+#end of lec 12

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

@@ -11,7 +11,7 @@ if $r_{1}=r_{2}\Rightarrow y_{h}(t)=c_{1}e^{rt}+c_{2}te^{rt}$ <- **critically da
 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 $f(t)\ne 0$ :
+But what about 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: $y''+p(t)y'+q(t)y=f(t)\quad \text{where}\quad y''(t_{o}),\ y'(t_{o}),\ y(t_{o})\in I$
 ---

content/Solve-any-DE.canvas

@@ -3,7 +3,7 @@
 		{"id":"767241b95828457e","type":"text","text":"Integrate both sides","x":-57,"y":-360,"width":352,"height":80},
 		{"id":"3572d3ccf3a666dc","type":"text","text":"# Solution","x":-43,"y":-520,"width":325,"height":90},
 		{"id":"ac4eb08e6ceeccbd","type":"text","text":"# How to solve (almost) any differential equation:","x":-561,"y":-760,"width":1296,"height":136},
-		{"id":"76786ab85409e204","type":"text","text":"updated Sept 20","x":778,"y":-729,"width":301,"height":74},
+		{"id":"76786ab85409e204","type":"text","text":"updated Oct 1","x":778,"y":-729,"width":301,"height":74},
 		{"id":"c6a536ee57248cfd","type":"file","file":"Math 201/Lectures/Separable equations (lec 1).md","x":-102,"y":-170,"width":441,"height":140},
 		{"id":"df03f8f1000d374a","type":"text","text":"If $y'=G(ax+by)$ \n\nsubstitute u=ax+by\n$\\frac{du}{dx}=a+b\\frac{ dy }{ dx}$","x":-251,"y":121,"width":341,"height":165},
 		{"id":"1044e4c2c0610a1e","type":"text","text":"if $y'=G\\left( \\frac{y}{x} \\right)$ \n\nsubstitute $u=\\frac{y}{x}$\n$\\frac{dy}{dx}=u+x{\\frac{du}{dx}}$","x":150,"y":121,"width":315,"height":165},
@@ -20,11 +20,6 @@
 		{"id":"ece672db8e16ac5a","type":"text","text":"substitute $x=u+k$\n$y=v+l$","x":9,"y":769,"width":242,"height":156},
 		{"id":"120b3008bd08d69a","type":"file","file":"Math 201/Lectures/Linear coefficient equations (lec 4).md","x":-81,"y":1035,"width":422,"height":143},
 		{"id":"bebd67e847df16e1","type":"text","text":"Shortcut:\n$$I(x)=e^{\\int (1-n)P(x) \\, dx }$$\n$$y^{1-n}=\\frac{1}{I(x)}\\left( \\int (1-n)I(x)Q(x) \\, dx  +C\\right)$$","x":-1333,"y":-430,"width":505,"height":260},
-		{"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 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},
 		{"id":"ba1ef733f104894d","type":"text","text":"$M(x,y)=\\frac{\\partial}{\\partial x} \\int N \\, dy+g(x)$\n","x":875,"y":354,"width":340,"height":77},
 		{"id":"3cc6d2966364d19c","type":"text","text":"$F(x,y)=\\int M \\, dx+g(y)$","x":520,"y":497,"width":315,"height":94},
@@ -35,9 +30,28 @@
 		{"id":"d172c6ad7421b458","type":"file","file":"Math 201/Lectures/Exact equations (lec 4-5).md","x":674,"y":903,"width":393,"height":133},
 		{"id":"d3ffeeccd9e88489","type":"text","text":"$M(x,y)dx+N(x,y)dy=0$","x":674,"y":1036,"width":342,"height":83},
 		{"id":"937768cc91c15daa","type":"text","text":"$(a_{1}x+b_{1}y+c_{1})dx+(a_{2}x+b_{2}y+c_{2})dy=0$","x":-81,"y":1178,"width":516,"height":63},
-		{"id":"59f62d39b48e7b57","x":1663,"y":-306,"width":250,"height":60,"type":"text","text":"under-damped"},
+		{"id":"b3e73030feee12da","type":"text","text":"case 2)\ncritically-damped\n$r_{1}=r_{2}=r$","x":1420,"y":177,"width":300,"height":140},
-		{"id":"7089887c01722e83","x":1880,"y":-173,"width":250,"height":60,"type":"text","text":"over-damped"},
+		{"id":"59f62d39b48e7b57","type":"text","text":"case 3)\nunder-damped\n$r_{1,2}=\\alpha\\pm \\beta i$","x":1760,"y":177,"width":280,"height":140},
-		{"id":"b3e73030feee12da","x":1843,"y":33,"width":287,"height":50,"type":"text","text":"critically-damped"}
+		{"id":"7089887c01722e83","type":"text","text":"case 1)\nover-damped $r_{1}\\ne r_{2}$","x":1140,"y":177,"width":240,"height":140},
+		{"id":"fb1b8ea1fa7e997b","type":"text","text":"general solution:\n$y_{h}(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}$","x":1079,"y":-83,"width":281,"height":103},
+		{"id":"21dddcaf48c1f19e","type":"text","text":"general solution:\n$y_{h}(t)=c_{1}e^{rt}+c_{2}te^{rt}$","x":1410,"y":-83,"width":320,"height":90},
+		{"id":"e85838ac18ea4a1e","type":"text","text":"general solution:\n$y_{h}(t)=e^{\\alpha t}(c_{1}\\cos\\beta t+c_{2}\\sin\\beta t)$","x":1760,"y":-83,"width":380,"height":103},
+		{"id":"cd7490f8cce0b6e0","type":"file","file":"Math 201/Lectures/Second order homogenous linear equations (lec 5-7).md","x":1336,"y":808,"width":465,"height":190},
+		{"id":"e063ab92aef817e4","type":"text","text":"obtain characteristic equation","x":1420,"y":619,"width":298,"height":78},
+		{"id":"d9167b6da862a39f","type":"file","file":"Math 201/Lectures/Free vibrations (lec 11-12).md","x":1467,"y":-875,"width":433,"height":146},
+		{"id":"9d6ed2c739b0615e","type":"file","file":"Math 201/Lectures/Variation of parameters (lec 9-10).md","x":1467,"y":-655,"width":530,"height":153},
+		{"id":"2b3a2c1a51539afd","type":"text","text":"idk how to solve these yet","x":1139,"y":-802,"width":241,"height":102},
+		{"id":"3f081acda4f30a27","type":"text","text":"solve for $r_{1}$ & $r_{2}$ using quadratic formula $\\frac{{-b\\pm \\sqrt{ b^2-4ac }}}{2a}$","x":1365,"y":419,"width":408,"height":125},
+		{"id":"50996e52b63ee5e0","type":"text","text":"$ay''+by'+cy=0$","x":1336,"y":998,"width":250,"height":60},
+		{"id":"174f52023f715bf2","type":"file","file":"Math 201/Lectures/Method of undetermined coefficients (lec 8-9).md","x":2240,"y":1143,"width":498,"height":196},
+		{"id":"0daa567ae142c34f","type":"text","text":"split into the homogenous solution and particular solution","x":2313,"y":927,"width":352,"height":131},
+		{"id":"e5369edab4d8a607","type":"text","text":"find $y_{h}(t)$\n(the easy part)","x":1911,"y":895,"width":229,"height":98},
+		{"id":"208cafba1dc71798","type":"text","text":"Use intuition and guess and check. (Practice improves intuition)","x":2068,"y":247,"width":332,"height":132},
+		{"id":"eded7f0b26c3ebc5","type":"text","text":"use generalized algorithm:\ncase 1) $ay''+by'+cy=P_{m}(t)e^{rt}$\nthen: $y_{p}(t)=t^s(b_{m}t^m+b_{m-1}t^{m-1}+\\dots+b_{0})e^{rt}$\nwhere:\ns=0, if r is not a root,\ns=1 if r is a single root,\ns=2 if r is a double root\n\ncase 2)\n$ay''+by'+cy=P_{m}(t)e^{\\alpha t}\\cos(\\beta t)+P_{m}(t)e^{\\alpha t}\\sin(\\beta t)$\nthen: $y_{p}(t)=t^s[(A_{k}t^k+A_{K-1}t^{k-1}+\\dots+A_{0})e^{\\alpha t}\\cos(\\beta t)+(B_{k}t^k+B_{k-1}t^{k-1}+\\dots+B_{0})e^{\\alpha t}\\sin(\\beta t)]$\nwhere:\ns=0 if $\\alpha+i\\beta$ is not a root,\ns=1 if $\\alpha+i\\beta$ is a root.","x":2489,"y":-156,"width":1054,"height":575},
+		{"id":"d327b2b2e59fc4cc","type":"text","text":"general solution will be $y(t)=y_{h}(t)+y_{p}(t)$","x":2108,"y":-261,"width":336,"height":105},
+		{"id":"b9455e68aa57c635","type":"text","text":"$ay''+by'+cy=f(t)$","x":2239,"y":1339,"width":283,"height":59},
+		{"id":"5ef326cd80c96f9a","type":"text","text":"if $f(t)$ is a sum of two functions, split it up and find $y_{p_{1}}$, $y_{p_{2}}, \\dots$ separately, then $y_{p}$ will be the sum of the all the $y_{p_{n}}$'s you find.","x":2276,"y":658,"width":426,"height":188},
+		{"id":"ac0242af88518588","type":"text","text":"find $y_{p}(t)$ (hard part)","x":2389,"y":520,"width":200,"height":99}
 	],
 	"edges":[
 		{"id":"7f7a9f03d9ef74ef","fromNode":"c6a536ee57248cfd","fromSide":"top","toNode":"767241b95828457e","toSide":"bottom"},
@@ -55,13 +69,10 @@
 		{"id":"bbc151d3f261f179","fromNode":"5263b3c4f5b28ec6","fromSide":"top","toNode":"3572d3ccf3a666dc","toSide":"left"},
 		{"id":"473fd48634427ca8","fromNode":"a662aed47cf28581","fromSide":"top","toNode":"ba1ef733f104894d","toSide":"bottom"},
 		{"id":"c3fd9e04ae24009a","fromNode":"3cc6d2966364d19c","fromSide":"top","toNode":"d98da52cb7139c25","toSide":"bottom"},
-		{"id":"6f5e1fd90c03ff8e","fromNode":"cd7490f8cce0b6e0","fromSide":"top","toNode":"cd31ca74652b6936","toSide":"bottom"},
 		{"id":"256d32160fde6619","fromNode":"c280d80abc2ea256","fromSide":"top","toNode":"df03f8f1000d374a","toSide":"bottom"},
 		{"id":"06197d071305053e","fromNode":"df03f8f1000d374a","fromSide":"top","toNode":"c6a536ee57248cfd","toSide":"bottom"},
 		{"id":"2930b4b645b64fe6","fromNode":"e71a5b824ac543f8","fromSide":"top","toNode":"c6a536ee57248cfd","toSide":"left"},
-		{"id":"3b69d186f470909b","fromNode":"cd31ca74652b6936","fromSide":"top","toNode":"e063ab92aef817e4","toSide":"bottom"},
 		{"id":"f99e2001e302efe7","fromNode":"e063ab92aef817e4","fromSide":"top","toNode":"3f081acda4f30a27","toSide":"bottom"},
-		{"id":"5b2a75e3d7143938","fromNode":"4ffaa5c9a7e8d22b","fromSide":"top","toNode":"3572d3ccf3a666dc","toSide":"right"},
 		{"id":"499d02da2174d229","fromNode":"120b3008bd08d69a","fromSide":"top","toNode":"ece672db8e16ac5a","toSide":"bottom"},
 		{"id":"c71b296bcb5e22a4","fromNode":"ece672db8e16ac5a","fromSide":"top","toNode":"109d20a863d93116","toSide":"bottom"},
 		{"id":"c03033b857ed5370","fromNode":"d172c6ad7421b458","fromSide":"top","toNode":"90a401b5afaf25e6","toSide":"bottom"},
@@ -72,7 +83,28 @@
 		{"id":"85f3f204e5750054","fromNode":"5dc72ba2fdd9b5af","fromSide":"top","toNode":"dc3f64df05e215d4","toSide":"bottom"},
 		{"id":"87edf0438b210dcd","fromNode":"dc3f64df05e215d4","fromSide":"top","toNode":"3572d3ccf3a666dc","toSide":"right"},
 		{"id":"e54d4010a7ef62c8","fromNode":"bebd67e847df16e1","fromSide":"right","toNode":"3572d3ccf3a666dc","toSide":"left"},
-		{"id":"d24c3bef53a69a99","fromNode":"3f081acda4f30a27","fromSide":"top","toNode":"4ffaa5c9a7e8d22b","toSide":"bottom"},
+		{"id":"7a3e93142d63ca9f","fromNode":"70316d4131dc52c3","fromSide":"left","toNode":"bebd67e847df16e1","toSide":"bottom"},
-		{"id":"7a3e93142d63ca9f","fromNode":"70316d4131dc52c3","fromSide":"left","toNode":"bebd67e847df16e1","toSide":"bottom"}
+		{"id":"1e6a766d01de63d3","fromNode":"cd7490f8cce0b6e0","fromSide":"top","toNode":"e063ab92aef817e4","toSide":"bottom"},
+		{"id":"76868def5db04e7e","fromNode":"3f081acda4f30a27","fromSide":"top","toNode":"59f62d39b48e7b57","toSide":"bottom"},
+		{"id":"41c1dbd214b8df89","fromNode":"3f081acda4f30a27","fromSide":"top","toNode":"b3e73030feee12da","toSide":"bottom"},
+		{"id":"cf3a2b3df9d5ddb6","fromNode":"3f081acda4f30a27","fromSide":"top","toNode":"7089887c01722e83","toSide":"bottom"},
+		{"id":"d57dcc30c832573e","fromNode":"7089887c01722e83","fromSide":"top","toNode":"fb1b8ea1fa7e997b","toSide":"bottom"},
+		{"id":"a729a47c00ac9e7b","fromNode":"b3e73030feee12da","fromSide":"top","toNode":"21dddcaf48c1f19e","toSide":"bottom"},
+		{"id":"78b5da4563e7e81e","fromNode":"59f62d39b48e7b57","fromSide":"top","toNode":"e85838ac18ea4a1e","toSide":"bottom"},
+		{"id":"32cb3e88211c6942","fromNode":"fb1b8ea1fa7e997b","fromSide":"top","toNode":"3572d3ccf3a666dc","toSide":"right"},
+		{"id":"d3e77c44bc263a1f","fromNode":"21dddcaf48c1f19e","fromSide":"top","toNode":"3572d3ccf3a666dc","toSide":"right"},
+		{"id":"0df1bc4ae125efb4","fromNode":"e85838ac18ea4a1e","fromSide":"top","toNode":"3572d3ccf3a666dc","toSide":"right"},
+		{"id":"e32c61f90abd18c2","fromNode":"9d6ed2c739b0615e","fromSide":"left","toNode":"2b3a2c1a51539afd","toSide":"right"},
+		{"id":"19652a56d16ff5a9","fromNode":"d9167b6da862a39f","fromSide":"left","toNode":"2b3a2c1a51539afd","toSide":"right"},
+		{"id":"daab7bfdf38808c0","fromNode":"174f52023f715bf2","fromSide":"top","toNode":"0daa567ae142c34f","toSide":"bottom"},
+		{"id":"25bc6fcf96f56e4c","fromNode":"0daa567ae142c34f","fromSide":"left","toNode":"e5369edab4d8a607","toSide":"right"},
+		{"id":"cb40bc53f34e134d","fromNode":"e5369edab4d8a607","fromSide":"left","toNode":"cd7490f8cce0b6e0","toSide":"right"},
+		{"id":"b5a960b34f3ba8f6","fromNode":"eded7f0b26c3ebc5","fromSide":"top","toNode":"d327b2b2e59fc4cc","toSide":"right"},
+		{"id":"8f310881de0ef642","fromNode":"208cafba1dc71798","fromSide":"top","toNode":"d327b2b2e59fc4cc","toSide":"bottom"},
+		{"id":"cc48a0286f85f502","fromNode":"d327b2b2e59fc4cc","fromSide":"left","toNode":"3572d3ccf3a666dc","toSide":"right"},
+		{"id":"d7c92d38eca2f187","fromNode":"0daa567ae142c34f","fromSide":"top","toNode":"5ef326cd80c96f9a","toSide":"bottom"},
+		{"id":"15ef69cc4ff38f75","fromNode":"5ef326cd80c96f9a","fromSide":"top","toNode":"ac0242af88518588","toSide":"bottom"},
+		{"id":"875dc8da7b797eb2","fromNode":"ac0242af88518588","fromSide":"top","toNode":"208cafba1dc71798","toSide":"bottom"},
+		{"id":"a6a70cacfc405270","fromNode":"ac0242af88518588","fromSide":"top","toNode":"eded7f0b26c3ebc5","toSide":"bottom"}
 	]
 }

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

@@ -3,17 +3,20 @@ $ay''+by'+cy=f(t)$
 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 $y_{p}$
 $y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$ <- 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
+we put $y_p$ into the equation and make it equal to the RHS. To do so, find the derivatives first:
-$y'_{p}=v_{1}y_{1}+v_{1}y_{1}'+v_{2}'y_{2}+v_{2}y_{2}'$
+$y'_{p}=v_{1}'y_{1}+v_{1}y_{1}'+v_{2}'y_{2}+v_{2}y_{2}'$
-to avoid second derivatives in the equation and problems with uniqueness lagrange imposed:
+to avoid second derivatives in the equation and problems with uniqueness, Lagrange imposed:
-1) $v_{1}y_{1}+v_{2}'y_{2}=0$ this simplifies our work down the road as well.
+1) $v_{1}'y_{1}+v_{2}'y_{2}=0$ this simplifies our work down the road as well.
+$y'_{p}=\cancel{ v_{1}'y_{1} }+v_{1}y_{1}'+\cancel{ v_{2}'y_{2} }+v_{2}y_{2}'$
 so $y_{p}''=v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}''$
+now we plug into the second order equation:
 $a(v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}'')+b(v_{1}y_{1}'+v_{2}y_{2}')+c(v_{1}y_{1}+v_{2}y_{2})=f(t)$
-$v_{1}(ay_{1}''+\cancelto{ 0 }{ by_{1}' }+cy_{1})+v_{2}(ay_{2}''+\cancelto{ 0 }{ by_{2}'C }+cy_{2})+a(v_{1}'y_{1}'+v_{2}'y_{2}')$
+$v_{1}(\cancel{ ay_{1}''+ by_{1}' +cy_{1} })+v_{2}(\cancel{ ay_{2}''+ by_{2}'+cy_{2} })+a(v_{1}'y_{1}'+v_{2}'y_{2}')$
+Both $y_{1}$ and $y_{2}$ are solutions to the homogenous counterpart. So the first two terms above equal to zero.
 2) $v_{1}'y_{1}'+v_{2}'y_{2}'=\frac{f(t)}{a}$
 $\det \begin{pmatrix}y_{1} & y_{2} \\y_{1}'& y_{2}'\end{pmatrix}$ = Wronskian = $W[y_{1},y_{2}]\ne 0$ 
 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.
+$v_{1}'=\frac{{f(t)y_{2}t}}{aW[y_{1},y_{2}]}$; $v_{2}'=-\frac{{f(t)y_{1}(t)}}{aW[y_{1},y_{2}]}$ <- integrate both sizes to get $v_{1}$ and $v_{2}$. When integrating, you don't need to add a generic constant.
 
 #ex #second_order #IVP 
 $y''+4y=2\tan(2t)-e^t \qquad y(0)=0 \qquad y'(0)=\frac{4}{5}$
@@ -21,11 +24,11 @@ can we use undetermined coefficients? yes and no
 find general solution to homogenous counterpart
 1) $y''+4y=0$ -> $r^2+4=0$ -> $r_{1,2}=\pm 2i$
 $y_{h}(t)=c_{1}\cos(2t)+c_{2}\sin(2t)$
-2 $y''+4y=-e^t$ <- use method of undetermined coefficients
+2) $y''+4y=-e^t$ <- use method of undetermined coefficients
-$y_{p}'(t)=Ae^{t}$
+$y_{p}^1(t)=Ae^{t}$
 $5Ae^t=-e^t$
 $A=-\frac{1}{5}$
-$y_{p}'(t)=-\frac{1}{5}e^t$
+$y_{p}^1(t)=-\frac{1}{5}e^t$
 (ii) $y''+4y=2\tan(2t)$ <- cant use method of undetermined coefficients
 $y_{p}^2(t)=v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)$
 plugging in:
@@ -41,19 +44,19 @@ $2(\sin^2(2t)+\cos^2(2t))v_{2}'=2\tan(2t)\cos(2t)$
 $v_{2}'=\sin(2t)$
 $v_{2}(t)=-\frac{1}{2}\cos(2t)$ no constant of integration, we want one solution only
 $v_{1}'=-{\frac{\sin^2(2t)}{\cos(2t)}}$
-$v_{1}=-\int \frac{\sin^2(2t)}{\cos(2t)} \, dx$
+$v_{1}=-\int \frac{\sin^2(2t)}{\cos(2t)} \, dt$
 $v_{1}=-\int \frac{{1-\cos^2(2t)}}{\cos(2t)} \, dt$
-$v_1=-\int sec(2t) \, dx+\int \cos(2t) \, dt$
+$v_1=-\int sec(2t) \, dt+\int \cos(2t) \, dt$
 $v_{1}(t)=-\frac{1}{2}\ln\mid sec(2t)+\tan(2t)\mid+\frac{1}{2}\sin(2t)$
 $y_{p}^2(t)=v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)$
-$y(t)=y_{h}(t)+y_{p}'(t)+y_{p}^2(t)$
+$y(t)=y_{h}(t)+y_{p}^1(t)+y_{p}^2(t)$
 =$c_{1}\cos(2t)+c_{2}\sin(2t)+v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)-\frac{1}{5}e^t$
-is our general answer.
+is the general answer.
 
 IVP solution:
 $y(0)=0=c_{1}-y_{p}(0)=c_{1}-\frac{1}{5}\Rightarrow c_{1}=\frac{1}{5}$
 skipping some differentiation: $y'(0)=2c_{2}+y_{p}'(0)=2c_{2}+v_{1}'(0)+2v_{2}(0)-\frac{1}{5}=\frac{4}{5}\Rightarrow c_{2}=1$
-$y(t)=\frac{1}{5}\cos(2t)+\sin(2t)-\frac{1}{5}e^t+v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)$
+$$y(t)=\frac{1}{5}\cos(2t)+\sin(2t)-\frac{1}{5}e^t+v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)$$
 
 #end of lecture 9
 #start of lecture 10

content/_index.md

@@ -10,7 +10,7 @@ I have written these notes for myself, I thought it would be cool to share them.
 [Second order homogenous linear equations (lec 5-7)](second-order-homogenous-linear-equations-lec-5-7.html) 
 [Method of undetermined coefficients (lec 8-9)](method-of-undetermined-coefficients-lec-8-9.html)
 [Variation of parameters (lec 9-10)](variation-of-parameters-lec-9-10.html) (raw notes, not reviewed or revised yet.)
-[Free vibrations (lec 11)](free-vibrations-lec-11.html) (raw notes, not reviewed or revised yet.)
+[Free vibrations (lec 11-12)](free-vibrations-lec-11-12) (raw notes, not reviewed or revised yet.)
 </br>
 [How to solve any DE, a flow chart](Solve-any-DE.png)
 </br>