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Consider the ODE $$(x^{2}-1)y''(x)-2xy'(x)+2y(x)=x^{2}-1$$ (i) Show that $y(x)=x$ is a solution of the associated homogeneous equation

(ii) Prove that the function $z(x)=xy(x)$ satisfies the ODE $$u'(x)-\frac{2}{x(x^{2}-1)}u(x)=\frac{1}{x}$$

Part (i) is easy enough and I'm sure I have to somehow use it for (ii), but I can't put the two together. So far I've noticed that seldom the second order term, the two ODES are very similar when you divide the first one by $x(x^2-1)$ and I've tried writing $y$ and its derivatives as a function of $z$ but I'm super stuck.

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  • $\begingroup$ Nope. I did copy it from my homework sheet though, so I guess it could be a mistake from my professor. What makes you think it should be $z(x)=xu(x)$? guessing you meant $u$ as opposed to $v$. $\endgroup$ – gr8astard Mar 17 '20 at 14:58
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You are looking for Reduction of Order, that is, if you know one solution is $y_1 = x$, then a second linearly independent solution is given by $y_2 = y_1 v(x) = x v(x)$.

Let's say we have $y_2 = xv$, then $y_2' = v + x v'$ and $y_2'' = x v'' + 2 v'$.

We have

$$(x^{2}-1)y''(x)-2xy'(x)+2y(x)=x^{2}-1 $$

Substituting

$$(x^2-1) (x v'' + 2 v') - 2 x (v + x v') + 2 (x v) = x^2-1$$

Simplifying

$$x(x^2-1)v'' + 2v'(x^2-1) - 2x(v + xv')+ 2 xv = x(x^2-1)v'' - 2 v' = x^2-1$$

Let $v' = u$, then $v'' = u'$ and substitute

$$x(x^2-1)u' - 2 u = x^2-1$$

Hence

$$u'(x)-\frac{2}{x(x^{2}-1)}u(x)=\frac{1}{x}$$

You can solve this using Integrating Factor.

Note: there is a typo in the problem because $z = x y = x^2$ is not a solution to the ODE.

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