PX153 - E4 - method of variation of parameters

a more general method for finding the particular integral
consider a 2nd order inhomogeneous ODE:

a y^{″} + b y^{'} + c y = f (x)

we first find the complementary function:

y_{c} = C u_{1} (x) + D u_{2} (x)

we assume that we can write the particular integral in the following form:

y_{p} (x) = P (x) u_{1} (x) + Q (x) u_{2} (x)

for some arbitrary functions $P (x)$ and $Q (x)$

we want to find $P (x)$ and $Q (x)$ in terms of $u_{1} (x), u_{2} (x), f (x)$
first, calculate the derivative:

y_{p}^{'} (x) = P^{'} (x) u_{1} (x) + P (x) u_{1}^{'} (x) + Q^{'} (x) u_{2} (x) + Q (x) u_{2}^{'} (x)

we put the following constraint: $P^{'} u_{1} + Q^{'} u_{2} = 0$
the second derivative is then:

y_{p}^{″} (x) = P^{'} (x) u_{1}^{'} (x) + P (x) u_{1}^{″} (x) + Q^{'} (x) u_{2}^{'} (x) + Q (x) u_{2}^{″} (x)

in the ODE, this gives:

a (P^{'} u_{1}^{'} + Q^{'} u_{2}^{'}) + a (P u_{1}^{″} + Q u_{2}^{″}) + b (P u_{1}^{'} + Q u_{2}^{'}) + c (P u_{1} + Q u_{2}) = f (x)

a (P^{'} u_{1}^{'} + Q^{'} u_{2}^{'}) + P (a u_{1}^{″} + b u_{1}^{'} + c u_{1}) + Q (a u_{2}^{″} + b u_{2}^{'} + c u_{2}) = f (x)

a (P^{'} u_{1}^{'} + Q^{'} u_{2}^{'}) = f (x) . . . [2]

P^{'} u_{1} + Q^{'} u_{2} = 0 . . . [1]

from $[1]$ :

Q^{'} = - \frac{u_{1}}{u_{2}} P^{'}

subbing in $[2]$ :

P^{'} (\frac{u_{1}^{'} u_{2} - u_{1} u_{2}^{'}}{u_{2}}) = \frac{f (x)}{a}

⟹ P^{'} = - \frac{1}{a} \frac{u_{2} f (x)}{u_{1} u_{2}^{'} - u_{1}^{'} u_{2}}

⟹ Q^{'} = \frac{1}{a} \frac{u_{1} f (x)}{u_{1} u_{2}^{'} - u_{1}^{'} u_{2}}

problem class notes

$y = u y_{1} + v y_{2}$ is a solution
we can use two equations: $u^{'} y_{1} + v^{'} y_{2} = 0$ $u^{'} y_{1}^{'} + v^{'} y_{2}^{'} = f (x)$
you should end up with 4 equations