# Solving y’ + a(x) y = b(x), Part 3

Yet another way to find the solution of $y' + a(x) y = b(x)\quad\quad\quad(1)$

is to seek a function $f(x) > 0 \; \forall x$ such that the result of multiply (1) by $f(x)$, namely $f(x)y' + f(x) a(x) y =f(x) b(x)\quad\quad\quad(2)$

can be written as $(f(x) y)' = b(x) f(x)$,

i.e., $f(x) y =\int b(x) f(x)\; dx + c_{3}$

where $c_{3}$ is a constant.

Or, $y = {1 \over f(x)} (\int b(x) f(x)\; dx + c_{3})\quad\quad\quad(3)$

since $f(x) > 0$.

Let us proceed to find such $f(x)$.

From (2) we see that if $f'(x)=a(x) f(x)$

then the left side of (2) is certainly $(f(x) y)'$ and consequently  for $f(x) > 0 \;\forall x$, ${1 \over f(x)} f'(x) = a(x)\quad\quad\quad(4)$

or, $log(f(x)) = \int a(x) dx + c_{1}$

where $c_{1}$ is a constant.

Therefore, a solution to (4) is $f(x) = c_{2}e^{\int a(x)\;dx}\quad\quad\quad(5)$

where $c_{2} = exp(c_{1})$. This is a positive function $\forall x$ indeed.

With (5), (3) becomes $y = e^{-\int a(x)\;dx} (\int b(x) e^{ \int b(x) dx} dx + c)\quad\quad\quad(6)$

where $c = {c_{3} \over c_{2}}$.

In fact, for any constant c, $e^{-\int a(x) dx} (\int b(x) e^{\int a(x) dx} dx + c)$

is a solution of (1): $(e^{-\int a(x) dx} (\int b(x) e^{\int a(x) dx} dx+ c))' +$ $a(x)\dot(e^{-\int a(x)\;dx} (\int b(x) e^{\int a(x) dx} dx + c))$ $= -a(x) (e^{-\int a(x) dx} (\int b(x) e^{\int a(x) dx} dx + c))$ $+ e^{-\int a(x) dx} b(x) e^{\int a(x) dx}$ $+ a(x)(e^{-\int a(x) dx} (\int b(x) e^{\int a(x) dx} dx + c))$ $=b(x)$

Therefore, the solution of (1) is $y = e^{-\int a(x) dx} (\int b(x) e^{\int a(x) dx} dx + c)$

where $c$ is any constant.

Exercise: prove that $e^x$  is a positive function.