# Can anyone solve this ODE?

Is the following equation solvable analytically: $$uu''+au^3=b$$ Where $a,b$ are positive real numbers? As you can see, I used a u-sub to get to this equation, but I can't see any tricks. Also, DSolve in Mathematica said the problem was equivalent to the following: $$\int_1^{u}\frac{dK_1}{\sqrt{C_1+2\left(4b\ln K_1-a\frac{K_1^3}{3}\right)}}$$ Which it couldn't solve.

Multiply by $u'/u$: $$u'' u' + au^2 u' = bu'/u$$ Integrate: $$\frac{1}{2} (u')^2 + \frac{a}{3} u^3 = b \log u+C.$$ Rearrange: $$\frac{u'}{\sqrt{2 b \log u + 2C - \frac{2a}{3} u^3}} = 1$$
Integrate each side: $$\int \frac{1}{\sqrt{2 b \log u + 2C - \frac{2a}{3} u^3}} du = t+D$$