# Thread: I need help with this differential equation

1. ## I need help with this differential equation

u''+u'/z=(K*g/q^2*z+T*K/q^2*z^2)*u

2. Originally Posted by Emilly
u''+u'/z=(K*g/q^2*z+T*K/q^2*z^2)*u

$
u'' + \frac{u'}{z} - \left(\frac{K\,g}{q^2 z} + \frac{T\,K}{q^2 z^2} \right) u = 0
$
?

3. yes

4. Originally Posted by Danny

$
u'' + \frac{u'}{z} - \left(\frac{K\,g}{q^2 z} + \frac{T\,K}{q^2 z^2} \right) u = 0
$
?
First let $a = \frac{K\,g}{q^2}$ and $b = \frac{T\,K}{q^2}$ so your ODE is

$
z^2 u'' + z u' -(a z+b)u = 0
$

Under the transformation $z = \frac{x^2}{4a}$ this ODE becomes

$x^2 u'' + x u' - (x^2 + 4b)u= 0$ a modified Bessel equation which has standards solutions.