ODE: $u'' +\frac {(1-4z)}{z^2(1-2z)}u' =0$

Question

I have got the following ODE: $$u'' +\frac {(1-4z)}{z^2(1-2z)}u' =0$$ The answer book rearranges as follows: $$\frac{u''}{u'} = -\frac{1-4z}{z^2(1-2z)}$$

I know I have to split the fraction on the RHS. But what is happening on the LHS:$\frac{u''}{u'}$ Can this be integrated at once?

Answer 1

Note that $\frac{u''}{u'}$ is the logarithmic derivative of $u'$ $$\mathcal{L (u')}=\frac{u''}{u'}$$ You can also susbtitute $p=u' \implies u''=p'$

Then $$\frac{u''}{u'} = -\frac{1-4z}{z^2(1-2z)}$$ $$\frac{dp}{pdz} = -\frac{1-4z}{z^2(1-2z)}$$ $$\int \frac{dp}{p} = -\int \frac{1-4z}{z^2(1-2z)} dz$$ $$\ln ( p) = -\int \frac{1-4z}{z^2(1-2z)} dz+K$$ $$\ln ( u') = -\int \frac{1-4z}{z^2(1-2z)} dz+K$$

or write $$\frac{u''}{u'}=\frac{du'}{u'dz}$$ and integrate the separable equation.....

Answer 2

Hint

$$[\ln u']'=\frac{u''}{u'}$$