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fix voparam

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Sasserisop 2023-10-08 00:22:10 -06:00
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content/Variation of parameters (lec 9-10).md
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@ -94,9 +94,9 @@ $-2B\cos t+2A\sin(t)=2\cos t(t)$
$\implies A=0,\ B=-1$ $\implies A=0,\ B=-1$
$y_{p_{1}}=-\sin(t)$ $y_{p_{1}}=-\sin(t)$
$y''-2y'+y=e^t\ln(t)$ cant use undetermined coefficients, use variation of parameters $y''-2y'+y=e^t\ln(t)$ cant use undetermined coefficients, use variation of parameters
$y''_{p}(t)=v_{1}y_{1}+v_{2}y_{2}$ $y_{p}''(t)=v_{1}y_{1}+v_{2}y_{2}$
$=v_{1}e^t+v_{2}te^t$ $=v_{1}e^t+v_{2}te^t$
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) $e^tv_{1}'+te^tv_{2}'=0$ eq1) $e^tv_{1}'+te^tv_{2}'=0$
eq2) $e^tv_{1}'+(te^t+e^t){v_{2}'}=e^t{\ln t}$ eq2) $e^tv_{1}'+(te^t+e^t){v_{2}'}=e^t{\ln t}$
subtract eq1 from eq2: $v_{2}'=\ln(t)$ subtract eq1 from eq2: $v_{2}'=\ln(t)$
@ -107,7 +107,7 @@ $=t\ln(t)-t$ no constant of integration.
compute $v_{1}$ now: compute $v_{1}$ now:
$v_{1}'=-tv_{2}'$ $v_{1}'=-tv_{2}'$
$=-t\ln t$ $=-t\ln t$
integrate to get v_1: integrate to get $v_1$:
$v_{1}=-\int t\ln t \, dt$ $v_{1}=-\int t\ln t \, dt$
integrate by parts (btw integration by parts will be the most important integration technique in this course): integrate by parts (btw integration by parts will be the most important integration technique in this course):
$v_{1}=-\frac{1}{2}(t^2\ln t)-\int t^2\frac{1}{t} \, dt$ $v_{1}=-\frac{1}{2}(t^2\ln t)-\int t^2\frac{1}{t} \, dt$
@ -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
$y_{p}''(t)=(\frac{1}{2}t^2\ln t+\frac{1}{4}t^2)e^t+(t\ln t-t)te^t$ $y_{p}''(t)=(\frac{1}{2}t^2\ln t+\frac{1}{4}t^2)e^t+(t\ln t-t)te^t$
$y_{p}(t)=-\sin(t)+\frac{1}{2}t^2\ln(t)e^t-\frac{3}{4}t^2e^t$ $y_{p}(t)=-\sin(t)+\frac{1}{2}t^2\ln(t)e^t-\frac{3}{4}t^2e^t$
general solution is produced by adding the homogenous eq with $y_{p}(t)$ general solution is produced by adding the homogenous eq with $y_{p}(t)$
general: general solution:
$$y(t)=c_{1}e^t+c_{2}te^t+y_{p}(t)$$ $$y(t)=c_{1}e^t+c_{2}te^t-\sin(t)+\frac{1}{2}t^2\ln(t)e^t-\frac{3}{4}t^2e^t$$
We are done.