Question

Find a second solution y2 for (x^2-4)y"+4xy'+2y=0; y1=1/(x-2) that isn't a constant multiple of the solution y1.

Answer 1

@SithsAndGiggles

Answer 2

Set \(y_2=\dfrac{u}{x-2}\), then
\[{y_2}'=\frac{u'}{x-2}-\frac{u}{(x-2)^2}\quad\text{and}\quad{y_2}''=\frac{u''}{x-2}-\frac{2u'}{(x-2)^2}+\frac{2u}{(x-2)^3}\]
So you have the ODE
\[\begin{align*}
0&=(x^2-4)y''+4xy'+2y\\\\
0&=(x^2-4)\left(\frac{u''}{x-2}-\frac{2u'}{(x-2)^2}+\frac{2u}{(x-2)^3}\right)+4x\left(\frac{u'}{x-2}-\frac{u}{(x-2)^2}\right)\\
&\quad\quad+2\frac{u}{x-2}\\\\
0&=(x+2)u''-\frac{2(x+2)}{x-2}u'+\frac{2(x+2)}{(x-2)^2}u+\frac{4x}{x-2}u'-\frac{4x}{(x-2)^2}u+\frac{2}{x-2}u\\\\
0&=(x+2)u''+\frac{2x-4}{x-2}u'\end{align*}\]

Answer 3

Missed some simplification, the ODE is equivalent to
\[(x+2)u''+2u'=0\]
From here you can replace \(u'=z\) and you have an ODE linear in \(z\):
\[(x+2)z'+2z=0\]

Answer 4

I got the right answer! Thanks!