MATH201/content/Prof variation of parameters.md

Variation of parameters

ay''+by'+cy=f(t)

1. y_{h}=c_{1}y_{1}(t)+c_{2}y_{2}(t) <- h is homogenous, ie: f(t)=0 Lagrange proposed this method to find the particular solution y_{p}: y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t) <- btw y_{1} and y_{2} are often called a fundamental pair. we put y_p into the equation and make it equal to the RHS. To do so, find the derivatives first: y'_{p}=v_{1}'y_{1}+v_{1}y_{1}'+v_{2}'y_{2}+v_{2}y_{2}' to avoid second derivatives in the equation and problems with uniqueness, Lagrange imposed:
2. v_{1}'y_{1}+v_{2}'y_{2}=0 this simplifies our work down the road as well. y'_{p}=\cancel{ v_{1}'y_{1} }+v_{1}y_{1}'+\cancel{ v_{2}'y_{2} }+v_{2}y_{2}' so y_{p}''=v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}'' now we plug into the second order equation: a(v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}'')+b(v_{1}y_{1}'+v_{2}y_{2}')+c(v_{1}y_{1}+v_{2}y_{2})=f(t) v_{1}(\cancel{ ay_{1}''+ by_{1}' +cy_{1} })+v_{2}(\cancel{ ay_{2}''+ by_{2}'+cy_{2} })+a(v_{1}'y_{1}'+v_{2}'y_{2}') Both y_{1} and y_{2} are solutions to the homogenous counterpart. So the first two terms above equal to zero.
3. v_{1}'y_{1}'+v_{2}'y_{2}'=\frac{f(t)}{a} we now have a system of two equations: \det \begin{pmatrix}y_{1} & y_{2} \\y_{1}'& y_{2}'\end{pmatrix} = Wronskian = W[y_{1},y_{2}]\ne 0 by definition y_1 and y_2 are linearly independent solutions so the above can never be 0! v_{1}'=-\frac{{f(t)y_{2}t}}{aW[y_{1},y_{2}]}; v_{2}'=\frac{{f(t)y_{1}(t)}}{aW[y_{1},y_{2}]} <- integrate both sizes to get v_{1} and v_{2}. When integrating, you don't need to add a generic constant. Finally, your solution is: y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t) where you get y_{1},y_{2} from your homogenous solution.