For a second order linear ODE of the form $y''+p(t)y'+q(t)y=r(t)$, if $y_1(t)$ is a solution to the homogeneous equation, and let $y(t)=v(t)y_1(t)$. Can we find a $v(t)$ such that $y(t)$ is also the general solution of the ode?
I think the answer is yes. As $y_1''+p(t)y_1'+q(t)y_1=0$, $y(t)=v(t)y_1(t)$, we can plug $y(t)$ into $y''+p(t)y'+q(t)y=r(t)$.
$$v''(t)y_1(t)+v'(t)[2y_1'(t)+p(t)y_1(t)]+v(t)[y_1''(t)+p(t)y_1'(t)+q(t)y_1(t)]=r(t)$$
Since $y_1''+p(t)y_1'+q(t)y_1=0$, then $v''(t)y_1(t)+v'(t)[2y_1'(t)+p(t)y_1(t)]=r(t)$
$$v''(t)+v'(t)\left[2\frac{y_1'(t)}{y_1(t)}+p(t)\right]=\frac{r(t)}{y_1(t)}$$
How can I solve this ODE from here? Should I use variation of parameters? How?