This question is similar to this question, however I do not understand the method I should be using. I need to solve $$x^3y'''+7x^2y''+xy'-16y =0 \,\,\,\,\,\,\, \text{knowing that} \,\,\,\,\,\ y_1 = x^2$$
Okay so normally for second order ODE I simply use the formula $$y_2 =y_1\int \frac{e^{-\int p(x)dx}}{y_1^2}dx$$ where $p(x)$ comes from $$y''+p(x)y'+q(x)y = 0$$ I know the formula comes from the intuition that the second solution has the following relation with the first one: $y_2 = v(x)y_1$ where $v(x)$ is the function to be found. However how do I find a second solution for the equation above?
Edit : Trial
Trying $y_2 = vy_1$ I get $y_2' = v'y_1+vy_1'$ and $y_2''=v''y_1+v'y_1'+v'y_1'+vy_1'' = v''y_1+2v'y_1'+vy_1''$ and also $y_2''' = v'''y_1+v''y_1'+2(v''y_1'+v'y_1'')+v'y_1'''+v''y_1''=v'''y_1+v''(3y_1'+y_1'')+v'(2y_1''+y_1''')$ but they are all positive and if I sub them into the main equation I will just get a very big amount of terms. Indeed we would have:
$$x^3y_2'''+7x^2y_2''+xy_2'-16y_2 =3(v'''y_1+v''(3y_1'+y_1'')+v'(2y_1''+y_1'''))+7x^2(v''y_1+2v'y_1'+vy_1'')+x(v'y_1+vy_1')-16(vy_1)$$
Just use $y=v·y_1$ and plug in the derivatives. The non-derived terms of $v$ should cancel leaving a second order equation for $z=v'$.
\begin{align} -16y&=-16x^2v\\ xy'&=x^3v'+2x^2v\\ 7x^2y''&=7x^4v''+28x^3v'+14x^2v\\ x^3y'''&=x^5v'''+6x^4v''+6x^3v'\\ \hline x^3y'''+7x^2y''+xy'-16y&=x^5v'''+13x^4v''+35x^3v' \end{align} which gives $x^2z''+13xz'+35z=0$ as reduced equation.