revise mouc and cauchy

This commit is contained in:
Sasserisop 2023-10-10 17:52:23 -06:00
parent e6ef4f1229
commit 81b0363ce5
2 changed files with 4 additions and 3 deletions

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@ -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.

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