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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.

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I can explain Mathematica's answer. An explicit solution is highly unlikely. If it exists, the integral we get at the end should be (a) computable by some method and (b) the antiderivative obtained should be invertible. Both are unlikely.

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: $$\int \frac{1}{\sqrt{2 b \log u + 2C - \frac{2a}{3} u^3}} du = t+D$$

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