Reduction of Order On Functions

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?

• It's separable. $\dfrac {v''}{v'}=(\ln v')'$ Commented Jan 30, 2021 at 22:27
• It's separable, but I'm not sure how to deal with the multiple functions of $x$. Commented Jan 30, 2021 at 22:30
• You try to deduce a general formula for $y_2$ so you dont need to calculate the integral to get an explicit function. Commented Jan 30, 2021 at 23:06

$$y_2=vy_1$$ $$y'_2=v'y_1+vy_1'$$ $$y''_2=v''y_1+2v'y'_1+vy''_1$$ 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: $$y_2= {y_1}\int \dfrac {e^{-\int { p(x) }dx}}{ y_1^2}dx$$