I just started reading about second order differential equations and I have issue with exact equations. In the book I'm reading from, says,
The equation $p(x)y''+q(x)y'+r(x)y=s(x)$ is exact if we have
$p''-q'+r=0$. These type of equations can be solve with the following
method,$$\frac d{dx}(py'+(q-p)y)=s(x)\xrightarrow{\int} \text{ first order
linear differential equation} \Rightarrow \ldots$$
I tried to find elementary proofs for why exact equations can be solve as mentioned in the book but some of them was too advanced for me like Wikipedia and I found a similar question on this site here which states that
If an equation $P(x)y''+Q(x)y'+R(x)y=0$
can be written in the form:
$$[P(x)y']'+[f(x)y]'=0$$
then the equation is said to be exact.
And it cause more confusion for me because in the book says RHS is a function $s(x)$ but here RHS is zero.
Can you please prove the statement in the book?
Best Answer
$$p(x)y''+q(x)y'+r(x)y=s(x)$$ $$p(x)y''+q(x)y'+(q'-p'')y=s(x)$$ $$p(x)y''+(q(x)y)'-p''y=s(x)$$ $$p(x)y''\color {red}{+p'y'}+(q(x)y)'\color{red}{-p'y'}-p''y=s(x)$$ $$(p(x)y')'+(q(x)y)'-(p'y)'=s(x)$$ It's easy to integrate now.