# Show $z(x)=xy(x)$ satisfies this first order ODE given this second order ODE and one of its solutions

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.

• 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 $v$. – gr8astard Mar 17 '20 at 14:58

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.