Suppose $y_1 (x)$ is a solution to $y'' + p(x)y' + q(x)y = 0$. Use reduction of order to show that a second solution is $y_2 (x) = y_1 (x) \int \frac{e^{-\int p(x)dx}}{(y_1 (x))^2} dx$.
I plugged the guess $v(x)y_1 (x)$ into the ODE and obtained $v''y_1 + 2v'y_1' + pv'y_1 = 0$.
The substitution $w(x) = v'$ yields the first-order ODE $w' = -\frac{2y_1' + py_1}{y_1}w$, but this is hard to solve because $w$, $y_1$ and $p$ are all functions of $x$. Additionally, this exercise appears in the section on The Method of Frobenius, so I may be doing something different than the intended approach.
Is this work correct, and if so, how can I obtain the stated result from this point?
$$y_2=vy_1$$ $$y'_2=v'y_1+vy_1'$$ $$y''_2=v''y_1+2v'y'_1+vy''_1$$ So that: $$y''_2 + p(x)y'_2 + q(x)y_2 = 0$$ Becomes: $$ v''y_1+2v'y'_1+vy''_1+ p(x)(v'y_1+vy_1') + q(x)vy_1 = 0$$ $$ v''y_1+2v'y'_1+ p(x)v'y_1 = 0$$
$$ v''y_1+v'(2y'_1+ p(x)y_1) = 0$$ This is separable. $$ (\ln v')'=-\dfrac {2y'_1+ p(x)y_1 }{y_1}$$ $$ \ln v'=-\int \dfrac {2y'_1+ p(x) y_1}{y_1}dx$$ $$ \ln v'=-2 \ln (y_1)-\int { p(x) }dx$$ $$ v'=\dfrac {e^{-\int { p(x) }dx}}{ y_1^2}$$ $$ v=\int \dfrac {e^{-\int { p(x) }dx}}{ y_1^2}dx$$ Finally: $$ y_2= {y_1}\int \dfrac {e^{-\int { p(x) }dx}}{ y_1^2}dx$$