Method of variation of parameters.

These notes are set so that you get to prove the main results by solving smaller problems that when put together give the big result. The answers to the problems are in the videos. You will get the most out of these notes if you do (or try) the problems before looking at the videos.

Lagrange's method of variation of parameters.

This is a method of finding a solution to the inhomogeneous equation $$ y''+p(x) y' + q(x)y=f(x) $$ assume that we have a linearly independent pair of solutions $y_1$ $y_2$ of the homogeneous equation $$ y'' + p(x) y' + q(x)y =0 $$ As you have verified in the homework if we set $$ y_p = u_1 y_1 + u_2 y_2 $$ where $u_1$ and $u_2$ are solutions to the pair of equation \begin{align*} u_1' y_1 + u_2'y_2&= 0\\ u_1' y_1' + u_2'y_2'&=f(x) \end{align*} (the second of these is the result of plugging $y_p$ into the equation and noticing that many terms cancel because $y_1$ and $y_2$ are solutions to the homogeneous equation) then we can solve for $u_1'$ and $u_2'$, then find $u_1$ and $u_2$ by integration, and then $ y_p = u_1 y_1 + y_2 y_2$ is a solution to $ y''+p(x) y' + q(x)y=f(x)$.

I find the notation a bit easier to deal with without the subscripts. So let $\phi$ and $\psi$ be linearly independent solutions to the homogeneous equations set $$ y=u\phi + v \psi $$ where we choose $u$ and $v$ to be solutions to the equations \begin{align*} \phi u'+\psi v'&=0\\ \phi' u'+ \phi' v'&=f(x) \end{align*} then $y$ is a solution to the inhomogeneous equation.

Problem: As first example let us look at the problem I made such a mess of in class: find a particular solutions to $$ y''-y = e^{2x}. $$

Solution.

Problem: Find a particular solution to $$ y'' + 4y = 16 \sec(2x) $$

Solution.

We can formalize this process some. The equations \begin{align*} \phi u'+\psi v'&=0\\ \phi' u'+ \phi' v'&=f(x) \end{align*} can be solved to get $$ u'= \frac{-f\psi}{W} , \qquad v' = \frac{f\phi}{W} $$ where $$ W = \left| \begin{matrix} \phi&\psi\\ \phi'&\psi' \end{matrix}\right| = \phi\psi'-\phi'\psi $$ is the Wronskian of the functions $\phi$ and $\psi$. Thus \begin{align*} u&= \int \frac{-f\psi}{W}\,dx \\ v&= \int \frac{f\phi}{W} \,dx \end{align*}

Problem: Show that $\phi=x^2$ and $\psi = x^3$ are linearly independent solutions to homogeneous equation $$ x^2 y'' -4xy' + 6y=0 $$ and use them to find a particular solution the inhomogeneous equaton $$ x^2 + y'' -4xy' + 6y = 6+12 x^2. $$

Solution.