# Solving y’ + a(x) y = 0

Let’s consider the 1st-Order linear differential equation

$y'(x)+ a(x) y(x) =0\quad\quad\quad(1)$

We see $y(x) \equiv 0$ is a solution.

However, if (1) has a solution whose value at $x$ is not zero, it must be true that

${1 \over y(x)} y(x)' = -a(x)$

i.e.,

$log(|y(x)|) = -\int a(x) dx + c_1\quad\quad\quad(2)$

where $c_1$ is a constant.

From (2), we obtain

$y(x) = c\;e^{-\int a(x) dx}\quad\quad\quad(3)$

where $c$ is a constant. It is either $+e^{c_1}$ or $-e^{c_1}$.

We assert and prove that for any constant $c$, a function defined by (3)  is a solution of (1):

$(c\;e^{-\int a(x)dx)})' +a(x)c\;e^{-\int a(x) dx}=-c\;a(x)e^{-\int a(x) dx} +a(x) c\;e^{-\int a(x) dx} =0$.

Notice when $c=0$, (3) yields $y(x) \equiv 0$, the zero solution of (1).

Moreover, to see there are no other solutions, let  $u(x)$ be any solution of (1), we have

$(u(x) e^{\int a(x)dx})' = u'(x) e^{\int a(x) dx} + u(x) a(x) e^{\int a(x) dx}$

$= e^{\int a(x)dx } (u'(x) + a(x) u(x))$

$= 0$.

Therefore,

$(u(x) e^{\int a(x) dx})' = 0$.

That is

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

or,

$u(x) = c\;e^{-\int a(x) dx}$

where $c$ is a constant. Hence, any solution of (1) belongs to the family of functions defined by (3)