fix voparam

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

@@ -94,9 +94,9 @@ $-2B\cos t+2A\sin(t)=2\cos t(t)$
 $\implies A=0,\ B=-1$
 $y_{p_{1}}=-\sin(t)$ 
 $y''-2y'+y=e^t\ln(t)$ cant use undetermined coefficients, use variation of parameters
-$y''_{p}(t)=v_{1}y_{1}+v_{2}y_{2}$
+$y_{p}''(t)=v_{1}y_{1}+v_{2}y_{2}$
 $=v_{1}e^t+v_{2}te^t$
-Compute v1 and v2. This time lets do it using the linear system for funzies:
+Compute $v_{1}$ and $v_{2}$. This time let's do it using the linear system for practice:
 eq1) $e^tv_{1}'+te^tv_{2}'=0$
 eq2) $e^tv_{1}'+(te^t+e^t){v_{2}'}=e^t{\ln t}$
 subtract eq1 from eq2: $v_{2}'=\ln(t)$
@@ -107,7 +107,7 @@ $=t\ln(t)-t$ no constant of integration.
 compute $v_{1}$ now:
 $v_{1}'=-tv_{2}'$
 $=-t\ln t$
-integrate to get v_1:
+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):
 $v_{1}=-\frac{1}{2}(t^2\ln t)-\int t^2\frac{1}{t} \, dt$
@@ -115,5 +115,6 @@ $=-\frac{1}{2}\left( t^2\ln t-\frac{t^2}{2} \right)=-\frac{1}{2}t^2\ln t+\frac{1
 $y_{p}''(t)=(\frac{1}{2}t^2\ln t+\frac{1}{4}t^2)e^t+(t\ln t-t)te^t$
 $y_{p}(t)=-\sin(t)+\frac{1}{2}t^2\ln(t)e^t-\frac{3}{4}t^2e^t$
 general solution is produced by adding the homogenous eq with $y_{p}(t)$
-general: 
-$$y(t)=c_{1}e^t+c_{2}te^t+y_{p}(t)$$
+general solution: 
+$$y(t)=c_{1}e^t+c_{2}te^t-\sin(t)+\frac{1}{2}t^2\ln(t)e^t-\frac{3}{4}t^2e^t$$
+We are done.