diff --git a/content/Cauchy-Euler equations (lec 10).md b/content/Cauchy-Euler equations (lec 10).md
index d1e14af..1776995 100644
--- a/content/Cauchy-Euler equations (lec 10).md	
+++ b/content/Cauchy-Euler equations (lec 10).md	
@@ -1,39 +1,43 @@
 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!*
-$ax^2{\frac{d^2y}{dx^2}+bx{\frac{ dy }{ dx }}+cy=f(x)},\ x>0$
+*If it has a name in it, its very important, if it has 2 names, its very very important!*
+#cauchy-euler equations are equations in the form:
+$$ax^2{\frac{d^2y}{dx^2}+bx{\frac{ dy }{ dx }}+cy=f(x)},\ x>0$$
 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.
+Note: x=0 is not interesting as the derivative terms disappear.
+How to solve? There are two approaches:
+Textbook only use 2nd method, prof doesn't like this. You can find both methods in the profs notes. Btw, do 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.
-$\frac{dy}{dx}=\frac{dy}{dt}{\frac{dt}{dx}}=\frac{ dy }{ dt }{\frac{1}{x}}\Rightarrow \underset{ Impor\tan t }{ x\frac{dy}{dx}=\frac{dy}{dt} }$
+$\frac{dy}{dx}=\frac{dy}{dt}{\frac{dt}{dx}}=\frac{ dy }{ dt }{\frac{1}{x}}\Rightarrow \underset{ \text{Important} }{ x\frac{dy}{dx}=\frac{dy}{dt} }$
 compute 2nd derivative of y wrt to x:
-$\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2}\left( \frac{dt}{dx} \right)^2+\frac{dy}{dt}{\frac{d^2t}{dx^2}}=\frac{1}{x^2}{\frac{d^2y}{dt^2}}-\frac{\frac{1}{x^2}dy}{dt}$
-$\underset{ \mathrm{Im}por\tan t }{ x^22{\frac{d^2y}{dx^2}}=\frac{d^2y}{dt^2}-\frac{dy}{dt} }$
+$\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2} \frac{dt}{dx}\cdot \frac{dt}{dx}+\frac{dy}{dt}{\frac{d^2t}{dx^2}}=\frac{1}{x^2}{\frac{d^2y}{dt^2}}-\frac{\frac{1}{x^2}dy}{dt}$
+$\underset{ \text{Important} }{ x^2{\frac{d^2y}{dx^2}}=\frac{d^2y}{dt^2}-\frac{dy}{dt} }$
+plugging those derivatives in we get:
 $$a\frac{d^2y}{dt^2}+(b-a){\frac{dy}{dt}}+cy=f(e^t)$$
 ^ this is a constant coefficient equation now! We can solve it now using prior tools.
 
-#ex 
-solve:
+## Example:
+#ex #second_order #second_order_nonhomogenous #cauchy-euler
+Find the general solution for:
 $$x^2{\frac{d^2y}{dx^2}}+3x{\frac{dy}{dx}}+y=x^{-1},\ x>0$$
-$x=e^t$
+substitute: $x=e^t$
 transform using the technique we showed just earlier:
 $\frac{d^2y}{dt^2}+2{\frac{dy}{dt}}+y=e^{-t}$
-1) $r^2+2r+1=0$
+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:
-$y_{1}(x)=c_{1}e^{-\ln(x)}+c_{2}\ln(x)e^{-\ln(x)}+\frac{1}{2}\ln(x)^2e^{-\ln(x)}$
-$=c_{1}x^-1+c_{2}\ln(x)x^-1+\frac{1}{2}{\ln(x)^2}x^-1$
+$y(t)=c_{1}e^{-t}+c_{2}te^{-t}+\frac{1}{2}t^2e^{-t}$
+but we want solution in terms of x:
+$y(x)=c_{1}e^{-\ln(x)}+c_{2}\ln(x)e^{-\ln(x)}+\frac{1}{2}\ln(x)^2e^{-\ln(x)}$
+$$y(x)=c_{1}x^{-1}+c_{2}\ln(x)x^{-1}+\frac{1}{2}{\ln(x)^2}x^{-1}$$
+We are done.
+
 #end of lecture 10
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diff --git a/content/Free vibrations (lec 11-12).md b/content/Free vibrations (lec 11-12).md
index 96a2a98..44da303 100644
--- a/content/Free vibrations (lec 11-12).md	
+++ b/content/Free vibrations (lec 11-12).md	
@@ -99,7 +99,7 @@ $A=\frac{5}{4}$
 $\tan \phi=\frac{4}{3}$
 $\phi \approx 0.9273\dots$
 $$y(t)=\frac{5}{4}e^{-3t}\sin(4t+\phi)$$
-"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)
+"I know engineers love calculators, I know mathematicians hate 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
 $r_{1,2}=-5$
 $y(t)=(c_{1}+c_{2}t)e^{-5t}$
diff --git a/content/Resonance & AM (lec 13-14).md b/content/Resonance & AM (lec 13-14).md
index 7daa9fd..4844fa0 100644
--- a/content/Resonance & AM (lec 13-14).md	
+++ b/content/Resonance & AM (lec 13-14).md	
@@ -1,5 +1,5 @@
 #start of lec 13
-He has good news. he's excited to tell us about electric currents! In particular, how a radio uses resonance to selectively listen to a particular frequency:
+He has good news. he's excited to tell us about electric currents! In particular, how a radios use resonance to selectively listen to a particular frequency:
 # Resonance
 Let's imagine a mass spring system which has an applied force with a forcing frequency of $\gamma$ and an amplitude of $F_{o}$ (a constant):
 $my''+by'+ky=F_{o}\cos(\gamma t)$
@@ -35,10 +35,11 @@ take the derivative of $\mu$ wrt to $\gamma$:
 $\mu'(\gamma)=-\frac{{2m^2\gamma\left( \gamma^2-\left( \frac{k}{m}-\frac{b^2}{2m^2} \right) \right)}}{[(k-m\gamma^2)^2+b^2\gamma^2]^{3/2}}=0$
 case one: $\gamma=0$ not interesting, because then the force applied would just a constant force.
 case two:
-$\gamma_{r}=\sqrt{ \frac{k}{m}-\frac{b^2}{2m^2} }$ where the r means resonance
-By plugging in $\gamma_{r}$ into $\mu()$ we can know the maximum amplitude at resonance: $\mu_{max}(\gamma_{r})=\mu(\gamma_{r})=\frac{2m}{b\sqrt{ 4mk-b^2 }}$
+$$\gamma_{r}=\sqrt{ \frac{k}{m}-\frac{b^2}{2m^2} }$$
+where the r means resonance.
+By plugging in $\gamma_{r}$ into $\mu()$ we can know the maximum amplitude at resonance: $$\mu_{max}(\gamma_{r})=\mu(\gamma_{r})=\frac{2m}{b\sqrt{ 4mk-b^2 }}$$
 if $b^2\geq 4mk>2mk$ (overly damped/critically damped case) (no resonance, imaginary numbers) $\gamma=0$
-if $b^2<2mk<4mk$ (assumed from the very beginning above) then we get a resonant frequency: $\gamma_{r}=\sqrt{\frac{  2mk-b^2}{2m^2}}$
+if $b^2<2mk<4mk$ (assumed from the very beginning above) then we get a resonant frequency $\gamma_{r}$ which can be calculated as shown above.
 
 what if b=0? (no resistance):
 $my''+ky=F_{o}\cos(\gamma t)$
@@ -48,8 +49,7 @@ $=A\sin(\omega t+\phi)$
 $y_{p}(t)=A_{1}\cos(\gamma t)+A_{2}\sin(\gamma t)$
 assume $\gamma=\omega$ with zero resistance we get:
 $y_{p}(t)=\frac{F_{o}}{2m\omega}t\sin \omega t \underset{ t\to \infty }{ \to }\infty$ (your circuit blows up! Or equivalently, your bridge collapses.)
-#end of lec 13
-#start of lecture 14
+#end of lec 13 #start of lecture 14
 # Amplitude modulation
 Last lecture we showed we can selectively listen to a specific signal by using resonance.
 He has something else he's excited to show us; amplitude modulation! Lets recall from last lecture:
@@ -74,7 +74,7 @@ This is an amplitude modulated signal! also can be seen as a "beating frequency"
 To recap, taking a frictionless mass spring system which is initially at rest, and applying a driving frequency that is strictly different from the system's natural frequency, results in a displacement that follows an AM modulated signal. How cool is that! 
 >Sometimes I feel this effect when I'm sitting in the back of the bus, where the bus is stopped at a red light. I wonder if this modulated vibration pattern is attributed to this exact phenomenon.
 
-# Shortcut for solving DE of a mass spring system
+# Shortcut for solving DE of a mass-spring system
 ![[Drawing 2023-10-06 13.24.11.excalidraw]]
 $my''+by'+ky=mg+F$
 move $mg$ to LHS and replace $y$ with $y_{new}$ (remember, the $\frac{mg}{k}$ is a constant, its derivative is 0):
diff --git a/content/Separable equations (lec 1).md b/content/Separable equations (lec 1).md
index 1a1c19e..86c7a43 100644
--- a/content/Separable equations (lec 1).md	
+++ b/content/Separable equations (lec 1).md	
@@ -44,4 +44,4 @@ $C=-1$
 So, the answer is:
 $$y=\arctan(x^2+1)$$
 
-#end of Lecture 1
+#end of Lecture 1
\ No newline at end of file
diff --git a/content/Variation of parameters (lec 9-10).md b/content/Variation of parameters (lec 9-10).md
index a6c30cd..65f9167 100644
--- a/content/Variation of parameters (lec 9-10).md	
+++ b/content/Variation of parameters (lec 9-10).md	
@@ -1,9 +1,10 @@
 # Variation of parameters 
-#voparam
 >[Professors definition/derivations during lecture](prof-variation-of-parameters.html) <- I found this to be too big brain for me. Here is a simplified definition:
 
 ## Variation of parameters is a method to solve:
 ## $$ay''+by'+cy=f(t)$$
+#voparam
+Variation of parameters is generally known to be a more powerful replacement for method of undetermined coefficients.
 First, find the homogenous solution:
 $y_{h}=c_{1}y_{1}(t)+c_{2}y_{2}(t)$
 Now we need the particular solution, let $y_{p}$ be in the following form:
@@ -34,7 +35,8 @@ Alternatively, you could memorize the system of equations and solve for $v_{1}'$
 This is what the prof likes. I love you Dr. Minev, but this I personally disagree with. I think I'll stick with the formulas.
 ---
 #ex #second_order #IVP #voparam #mouc
-$y''+4y=2\tan(2t)-e^t \qquad y(0)=0 \qquad y'(0)=\frac{4}{5}$
+Solve the 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. We can use it on the $e^t$ term. However, guessing and checking $y_{p}$ for the $2\tan(2t)$ term might take a really, really long time.
 First, find general solution to homogenous counterpart:
 $y''+4y=0$ -> $r^2+4=0$ -> $r_{1,2}=\pm 2i$
@@ -77,6 +79,7 @@ $$y(t)=\frac{1}{5}\cos(2t)+\sin(2t)-\frac{1}{5}e^t+v_{1}(t)\cos(2t)+v_{2}(t)\sin
 
 #end of lecture 9 #start of lecture 10
 #ex #second_order #voparam #mouc
+Find the general solution for:
 $$y''-2y'+y=e^t\ln(t)+2\cos(t)$$
 Find homogenous solution first:
 $r^2-2r+1=0$
diff --git a/content/_index.md b/content/_index.md
index 232c6fc..e7f7409 100644
--- a/content/_index.md
+++ b/content/_index.md
@@ -10,11 +10,11 @@ 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) 
-[Cauchy-Euler equations (lec 10)](cauchy-euler-equations-lec-10.html) (raw notes, not reviewed or revised yet.)
+[Cauchy-Euler equations (lec 10)](cauchy-euler-equations-lec-10.html)
 [Free vibrations (lec 11-12)](free-vibrations-lec-11-12.html) (raw notes, not reviewed or revised yet.)
 [Resonance & AM (lec 13-14)](resonance-am-lec-13-14.html)
 [Laplace transform (lec 14)](laplace-transform-lec-14.html)
 
 </br>
 [How to solve any DE, a flow chart](Solve-any-DE.png)
-</br>
+</br>
\ No newline at end of file