Why do we do this step in linear differential equation? Assume, given a linear equation:
$y' + a(x)y + b(x) = 0$
I know that using Bernoulli's method, we substitute $y = u(x)v(x) \Rightarrow y'=u'(x)v(x)+v'(x)u(x)=u'v+v'u$.
Then we get view: $u'v+v'u + a(x)uv + b(x) = 0$
But the crucial part is when we somewhy factorise $u$ and set equation $v' + a(x)v = 0$
Are there any explanations why we do this step? This might be a stupid question, so I apologize in advance. Thanks
 A: $$v' + a(x)v = 0$$
Is easy to solve since it's separable:
$$\dfrac {v'}{v}=-a(x)$$
$$(\ln v)'=-a(x)$$
Integrate you find $v(x)$ then the original equation becomes:
$$u'v+v'u + a(x)uv + b(x) = 0$$
$$\implies u'v + b(x) = 0$$
$$u'=-\dfrac {b(x)}{v(x)}$$
You integrate and you get $u$.
$$u(x)=-\int \dfrac {b(x)}{v(x)}dx$$
In conclusion in order to solve a first order linear inhomogeneous differential equation on the form:
$$y' + a(x)y + b(x) = 0$$
You solve two easier differential equations:
$$v'+av=0$$
And:
$$u'=-\dfrac {b(x)}{v(x)}$$
Then:
$$y(x)=uv$$
A: For Linear equations like this:
$$y' + a(x)y + b(x) = 0$$
Best and easiest approach I know is:
$$Integrating\ Factor, c(x)=e^{\int{a(x)\ dx}}  $$
Multiply this both sides. You get:
$$c(x)\ y' + c(x)\ a(x)\ y + c(x)\ b(x) = 0$$

clearly:  $ c'(x)=c(x)\ a(x)$

Above equation reduced to:
$$c(x)\ y' + c'(x)\ y + c(x)\ b(x) = 0$$
If you see the steps of the above mentioned method carefully. You'll know why exactly you're solving the linear differential problem in the way you are.
