Consider the 2nd order linear ODE: $y^{\prime\prime}+p(x)y^{\prime}+q(x)y=0$
This is usually solve by performing the substitution: $y(x)=u(x)\exp(\int\frac{-p(x)}{2}dx)$.
We get $y^{\prime}(x)=u^{\prime}(x)\exp(\int\frac{-p(x)}{2}dx)-u(x)\frac{p(x)}{2}\exp(\int\frac{-p(x)}{2}dx)$ and $y^{\prime\prime}(x)=u^{\prime\prime}(x)\exp(\int\frac{-p(x)}{2}dx)-u^{\prime}(x)p(x)\exp(\int\frac{-p(x)}{2}dx)-u(x)\frac{p^{\prime}(x)}{2}\exp(\int\frac{-p(x)}{2}dx)+u(x)\frac{p^{2}(x)}{4}\exp(\int\frac{-p(x)}{2}dx)$
So the original equation become $u^{\prime\prime}(x)\exp(\int\frac{-p(x)}{2}dx)-u^{\prime}(x)p(x)\exp(\int\frac{-p(x)}{2}dx)-u(x)\frac{p^{\prime}(x)}{2}\exp(\int\frac{-p(x)}{2}dx)+u(x)\frac{p^{2}(x)}{4}\exp(\int\frac{-p(x)}{2}dx)+p(x)\left(u^{\prime}(x)\exp(\int\frac{-p(x)}{2}dx)-u(x)\frac{p(x)}{2}\exp(\int\frac{-p(x)}{2}dx)\right)+q(x)u(x)\exp(\int\frac{-p(x)}{2}dx)=0$ which simplify to $u^{\prime\prime}(x)+\left(q(x)-\frac{p^{\prime}(x)}{2}-\frac{p^{2}(x)}{4}\right)u(x)=0$
Which is supposedly easier to solve because we no longer has the first derivative term. So here are my questions:
Why is the equation easier to solve without the first derivative term? Aren't we just shoving the problem from the first derivative term into the zeroth derivative term? We sort of do that when it comes to solving quadratic equation, but when you solve quadratic equation we are given a magic tool (square root) to instantly solve any quadratic equation without middle term, but here we don't have that tool.
What are the good method to solve once we remove the first derivative term? Let's say assuming $p(x)$ and $q(x)$ are very nice functions, but not so nice that $\left(q(x)-\frac{p^{\prime}(x)}{2}-\frac{p^{2}(x)}{4}\right)$ is just a constant.
What motivate this choice of substitution? Why $\exp$ of the antiderivative of $-\frac{p(x)}{2}$. I mean, it's justified afterward by doing the algebra and somehow the terms cancel, but is there a deep reason why one would even think of doing that substitution and expect that it has a good chance of working, over many other possible choices?