# Analytic solution of: ${u}''+\frac{1}{x}{u}'=-\delta e^{u}$

I am trying to find the analytic solution of $${u}''+\frac{1}{x}{u}'=-\delta e^{u}$$ given the homogeneous mixed boundary conditions $${u'(0)}=0$$ $$u(1)=0$$

How would one attack such a problem? I have been given that the analytic solution is $$u=ln\left ( \frac{8a/\delta }{(ax^2+1)^2} \right )$$ where $a$ solves $8a=\delta (a+1)^2$.

My approach (EDITED):

I was given a hint: use the relation $x=e^{-y}$. I proceeded with that and used the chain-rule as follows $$(u\circ y)'(x)=\frac{d}{dx}u(y(x))=u'(y(x))y'(x)=u'x=-u'e^{y}$$ $$(u\circ y)''(x)=\frac{d}{dx}[u'(y(x))y'(x)]=u''(y(x))[y'(x)]^2+u'(y(x))y''(x)=u''e^{2y}+u'e^{2y}$$ substituting these relations into the ODE yields $$e^{2y}u''=-\delta e^{u}$$ or in an alternative form $$u''=-\delta e^{-2y} e^{u}$$

which is separable (thanks for the help).

• Isn't $y'(x)=-x^{-1}$? – user2520938 Mar 2 '15 at 20:06
• @user2520938 Oh... yes! This way the term involving $u'$ cancels. I will edit my approach above. Thanks man! – Mike Mar 2 '15 at 20:09
• You are welcome! – science Mar 2 '15 at 20:14

$$e^{-u} \frac{d^2u}{dy^2}= \delta e^{-2y}$$