Variation of parameters
$ay''+by'+cy=f(t)$
- $y_{h}=c_{1}y_{1}(t)+c_{2}y_{2}(t)$ 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:
- $v_{1}'y_{1}+v_{2}'y_{2}=0$ 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.
- $v_{1}'y_{1}'+v_{2}'y_{2}'=\frac{f(t)}{a}$
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.