Question

P(x)y"+a(x)y'=0

Answer 1

Given $P(x)y''+a(x)y'=0$ _________________________(1).

To solve this, put $y' = v$ so $y'' = v'$, we get

$P(x)v' + a(x)v = 0$

$\implies P(x)\frac{dv}{dx} = -a(x) v \\ \implies \frac{dv}{v} = -\frac{a(x)}{P(x)} dx$

So by integration on both sides, we get

$\int \frac{dv}{v} = -\int(\frac{a(x)}{P(x)} )dx$ .

Hence, finally we get $v(x)$.

Then, again by integrating $v$ with respect to $x$, we get $y$.