revise mouc and cauchy

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

@@ -36,7 +36,7 @@ $A=\frac{1}{2}$
 general solution in terms of t:
 $y(t)=c_{1}e^{-t}+c_{2}te^{-t}+\frac{1}{2}t^2e^{-t}$
 but we want solution in terms of x:
-$y(x)=c_{1}e^{-\ln(x)}+c_{2}\ln(x)e^{-\ln(x)}+\frac{1}{2}\ln(x)^2e^{-\ln(x)}$
+$y(x)=c_{1}e^{-\ln(x)}+c_{2}\ln(x)e^{-\ln(x)}+\frac{1}{2}\ln(x)^2e^{-\ln(x)}$ <- This is rather lousy notation, the y here isn't the same as the y above. Conceptually though, it's all oke doke.
 $$y(x)=c_{1}x^{-1}+c_{2}\ln(x)x^{-1}+\frac{1}{2}{\ln(x)^2}x^{-1}$$
 We are done.
 

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

@@ -102,8 +102,9 @@ 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) $ay''+by'+cy=P_{m}(t)e^{\alpha t}\cos(\beta t)+P_{m}(t)e^{\alpha t}\sin(\beta t)$
+case ii) $ay''+by'+cy=P_{m}(t)e^{\alpha t}\cos(\beta t)+Q_{n}(t)e^{\alpha t}\sin(\beta t)$
-Then we guess the particular solution is of the form: $y_{p}(t)=t^s[(A_{k}t^k+A_{K-1}t^{k-1}+\dots+A_{0})e^{\alpha t}\cos(\beta t)+(B_{k}t^k+B_{k-1}t^{k-1}+\dots+B_{0})e^{\alpha t}\sin(\beta t)]$
+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:
+k=max(m,n)
 s=0 if $\alpha+i\beta$ is not a root,
 s=1 if $\alpha+i\beta$ is a root.