more voparam fixes

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Sasserisop 2023-10-08 11:43:02 -06:00
parent 1761284b70
commit 8808993cd7
3 changed files with 6 additions and 6 deletions

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@ -195,6 +195,7 @@ let it be $y_{p}(t)$
then the sum of the solutions then the sum of the solutions
$y(t)=c_{1}y_{1}(t)+c_{2}y_{2}(t)+y_{p}(t)$ must solve $ay''+by'+cy=f(t)$ $y(t)=c_{1}y_{1}(t)+c_{2}y_{2}(t)+y_{p}(t)$ must solve $ay''+by'+cy=f(t)$
Theorem: If $a(t),\ b(t),\ c(t)$ are continuous on $I$ , then IVP: $a(t)y''+b(t)y'+c(t)y=f(t)$ ; $y(t_{o})=y_{o}$ \ , $y'(t_{o})=y_{1}$ has a unique solution. Theorem: If $a(t),\ b(t),\ c(t)$ are continuous on $I$ , then IVP: $a(t)y''+b(t)y'+c(t)y=f(t)$ ;
where $y(t_{o})=y_{o}$ , $y'(t_{o})=y_{1}$ has a unique solution.
we will do the proofs next class. we will do the proofs next class.
#end of lecture 7 #end of lecture 7

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@ -11,7 +11,7 @@ $y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$ <- btw $y_{1}$ and $y_{2}$ are ofte
Impose the following: Impose the following:
1) $v_{1}'y_{1}+v_{2}'y_{2}=0$ 1) $v_{1}'y_{1}+v_{2}'y_{2}=0$
Compute the derivatives and simplify: Compute the derivatives and simplify:
$y'_{p}=v_{1}y_{1}'+v_{2}y_{2}'$ $y_{p}'=v_{1}y_{1}'+v_{2}y_{2}'$
$y_{p}''=v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}''$ $y_{p}''=v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}''$
Now we plug those into the second order equation and simplify: Now we plug those into the second order equation and simplify:
2) $v_{1}'y_{1}'+v_{2}'y_{2}'=\frac{f(t)}{a}$ 2) $v_{1}'y_{1}'+v_{2}'y_{2}'=\frac{f(t)}{a}$
@ -22,6 +22,7 @@ also, $W[y_1,y_{2}]$ is the Wrońskian, and it equals to: $\det \begin{pmatrix}y
Finally, the general solution is: Finally, the general solution is:
$$y(t)=y_{h}+y_{p}\qquad \text{where}\qquad y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$$ $$y(t)=y_{h}+y_{p}\qquad \text{where}\qquad y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$$
## What you need to remember ## What you need to remember
#remember
So, what do you need to commit to memory? I believe memorizing these three is a good tradeoff between memory allocated and speed for when you're solving a #voparam problem: So, what do you need to commit to memory? I believe memorizing these three is a good tradeoff between memory allocated and speed for when you're solving a #voparam problem:
1) $y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$ 1) $y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$
2) $v_{1}'=-\frac{{f(t)y_{2}(t)}}{aW[y_{1},y_{2}]}$ 2) $v_{1}'=-\frac{{f(t)y_{2}(t)}}{aW[y_{1},y_{2}]}$
@ -74,9 +75,7 @@ $y(0)=0=c_{1}-y_{p}(0)=c_{1}-\frac{1}{5} \implies c_{1}=\frac{1}{5}$
$y'(0)=\frac{4}{5}=2c_{2}+v_{1}'(0)+2v_{2}(0)-\frac{1}{5} \implies c_{2}=1$ $y'(0)=\frac{4}{5}=2c_{2}+v_{1}'(0)+2v_{2}(0)-\frac{1}{5} \implies c_{2}=1$
$$y(t)=\frac{1}{5}\cos(2t)+\sin(2t)-\frac{1}{5}e^t+v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)$$ $$y(t)=\frac{1}{5}\cos(2t)+\sin(2t)-\frac{1}{5}e^t+v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)$$
#end of lecture 9 #end of lecture 9 #start of lecture 10
#start of lecture 10
#ex #second_order #voparam #mouc #ex #second_order #voparam #mouc
$$y''-2y'+y=e^t\ln(t)+2\cos(t)$$ $$y''-2y'+y=e^t\ln(t)+2\cos(t)$$
Find homogenous solution first: Find homogenous solution first:

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@ -12,7 +12,7 @@ I have written these notes for myself, I thought it would be cool to share them.
[Variation of parameters (lec 9-10)](variation-of-parameters-lec-9-10.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) (raw notes, not reviewed or revised yet.)
[Free vibrations (lec 11-12)](free-vibrations-lec-11-12.html) (raw notes, not reviewed or revised yet.) [Free vibrations (lec 11-12)](free-vibrations-lec-11-12.html) (raw notes, not reviewed or revised yet.)
[Resonance in free vibrations (lec 13-14)](resonance-in-free-vibrations-lec-13.html) [Resonance & AM (lec 13-14)](resonance-am-13-14.html)
[Laplace transform (lec 14)](laplace-transform-lec-14.html) [Laplace transform (lec 14)](laplace-transform-lec-14.html)
</br> </br>