Solve a second order nonlinear equation I have a second order nonlinear equation:
$$-u''+ \frac{1}{4}(u')^2+au=x^2.$$
I am only interested in the solutions in $[0, \frac{x^2}{a}+\frac{1}{a^2}]$. One paper claims without proof that the unique solution should be
$$u(x)=(\sqrt{a^2+4}-a)\big(\frac{x^2}{2}+\frac{1}{a}\big).$$
Does anyone have some ideas how to calculate the solution (without guessing the form of quadratic polynomial) and how to prove the uniqueness?
Thank you!
 A: we can trivially see that a solution could be of the form 
$$
u(x) = \alpha x^2 + \beta x + \gamma \tag{1}
$$
trialing wih Eq. (1)
we find
$$
-2\alpha + \frac{1}{4}\left(2\alpha x + \beta\right)^2 + a\left(\alpha x^2 + \beta x + \gamma\right) = x^2
$$
comparing coefficents we find
$$
-2\alpha + \frac{\beta^2}{4} + a\gamma = 0\;\;\text{$x^0$ terms}\\
\alpha\beta + a\beta = \beta(\alpha + a) = 0\;\;\text{$x$ terms}\\
\alpha^2 + a\alpha = \alpha(\alpha+a) = 1\;\;\text{$x^2$ terms}
$$
From the coefficients for $x$ and $x^2$ we require $\alpha + a\neq 0$ thus $\beta = 0$
therefore
$$
-2\alpha +a\gamma = 0 \implies \gamma = \frac{2\alpha}{a}
$$
the solution Eq.(1) is of the form with a constant
$$
u(x) = 2\alpha\left(\frac{x^2}{2} + \frac{1}{a}\right)\tag{2}
$$
now to find $\alpha$
we know that
$$
\alpha^2 + a\alpha = 1 \implies \alpha^2 + a\alpha -1 = 0
$$
solutions of the above quadratic is
$$
\alpha = \frac{-a\pm\sqrt{a^2 +4}}{2} 
$$
or
$$
2\alpha = -a\pm\sqrt{a^2 +4}
$$
now we can put into Eq.(2)
$$
u(x) = \left(\pm\sqrt{a^2 +4}-a\right)\left(\frac{x^2}{2} + \frac{1}{a}\right)
$$
since we are looking for solutions $>0$ at least, we can eliminate (-) from the $\pm$ above.
