# Solving y’ + a(x) y = b(x), Part 2

In my previous post “Solving y’ + a(x) y = b(x), Part 1“, it was revealed that every solution of

$y' + a(x) y = b(x)\quad\quad\quad(1)$

is the sum of a particular solution,  $e^{-\int a(x) dx}\int b(x) e^{\int a(x) dx}dx$, with $c e^{-\int a(x) dx}$, a solution of the homogeneous equation $y '+ a(x) y =0$.

This structure suggests that (1) can be solved in two steps: solve $y' +a(x) y= 0$ first to get $c e^{-\int a(x) dx}$, then change the constant $c$ in it to a function $c(x)$ and solve for $c(x)$ after submitting the variation into (1).

There is an alternative:

Let

$y = u v$

We have

$(u v)' + a(x) (u v) = v u' + u (v' + a(x) v)=b(x)$

From $v' + a(x) v = 0$, we obtain

$v = c_1 e^{-\int a(x) dx}$.

Solving

$v u' = c_1 e^{-\int a(x) dx} u'= b(x)$

subsequently for $u$ yields

$u = \int b(x) e^{\int a(x) dx} dx + c$.

Hence,

$y = u v = e^{-\int a(x)dx} (\int b(x) e^{\int a(x) dx} dx + c)$.