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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:

$$x^3y_2'''+7x^2y_2''+xy_2'-16y_2 =3(v'''y_1+v''(3y_1'+y_1'')+v'(2y_1''+y_1'''))+7x^2(v''y_1+2v'y_1'+vy_1'')+x(v'y_1+vy_1')-16(vy_1)$$

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    Should I use a $3\times 3$ Wronskian? I am really confused – 2017-02-05
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    Now sort the last equations by derivatives of $v$ to get the ODE for $y_1$ as the coefficient of $v$. – 2017-02-05

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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.

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    I tried that, but all the terms are positive apart from the ones coming from $-16y$ and so at the end there are just loads of terms! – 2017-02-05
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    should it not be $35x^3v'$ ? – 2017-02-05
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    if it is $35x^3v'$ then we have an euler equation right? By solving it I find that the solutions are $z = Ax^{-5}+Bx^{-7}$. So now does it mean that $v'= Ax^{-5}+Bx^{-7}$. Then I integrate and find $v = -\frac{A}{4}x^{-4}-\frac{B}{6}x^{-6}$, which means that $y_2 = vy_1=-\frac{A}{4}x^{-2}-\frac{B}{6}x^{-4}$ ? – 2017-02-05
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    Yes of course, the coefficient $2$ was on the other term. And yes, and even more also the original equation is an Euler equation that can be directly solved with the corresponding methods. You can redefine the undetermined constants to make the expression more simple. – 2017-02-05
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    oh okay. So could I just solve it by posing $y = x^k$, differentiate it three times and then substituting back in and solving the resulting auxiliary equation? – 2017-02-05
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    Yes. And use that $k=2$ is a known root of the characteristic equation. But that would not meet the task description of order reduction first. – 2017-02-05