From bbfe8f4123d7bdbfaf306a6a16c03a996220f9ed Mon Sep 17 00:00:00 2001 From: Sasserisop Date: Sun, 8 Oct 2023 00:26:07 -0600 Subject: [PATCH] more fixes for voparam --- content/Variation of parameters (lec 9-10).md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/content/Variation of parameters (lec 9-10).md b/content/Variation of parameters (lec 9-10).md index 756cde6..8f05756 100644 --- a/content/Variation of parameters (lec 9-10).md +++ b/content/Variation of parameters (lec 9-10).md @@ -11,8 +11,8 @@ $y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$ <- btw $y_{1}$ and $y_{2}$ are ofte Impose the following: 1) $v_{1}'y_{1}+v_{2}'y_{2}=0$ Compute the derivatives and simplify: -$y'_{p}=v_{1}y_{1}'+v_{2}y_{2}'$ -$y_{p}''=v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}''$ +$y_{p}''=v_{1}y_{1}'+v_{2}y_{2}'$ +$y_{p}'=v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}''$ Now we plug those into the second order equation and simplify: 2) $v_{1}'y_{1}'+v_{2}'y_{2}'=\frac{f(t)}{a}$ We now have a system of two equations (1 and 2). Now we can solve for $v_{1}$ and $v_{2}$: