Unique solution for IVP with $y''(x) + y(x)^2 = 0$? I'm supposed to show that the following initial value problem has a unique non-trivial solution.

*

*$y''(x) + y(x)^2 = 0$

*$y(0) = y(1) = 0$
My approach so far is:
$$y''(x) = -y(x)^2 \Rightarrow y^{-2}dy^2 = -dx^2 \Rightarrow (-y^{-1} + c_1)dy = -c_2 dx \Rightarrow \ln{|y|} + c_1 y = c_2 x + c_3$$
Here is the problem, if we say $y(0) = 0$ but I try to plug that in, I need to calculate $\ln{(0)}$ which is not defined. Did I mess this up?
 A: The argument roughly goes as follows: Choose a unique solution $y$ such that $y(0)=0$ and $y'(0)=1$. This is possible because of Picard-Lindelöf theorem. Prove by differentiating the LHS (so it is constant) that the energy equation
$$
\frac{1}{2}(y')^2 + \frac{1}{3}y^3 = \frac{1}{2}
$$
holds. Because of $y$ increasing near zero, it must hold because of the above equation that $y'$ decreases and $y$ increases until $y'(\bar x) = 0$ for some $\bar x >0$ with positive value (according to the above equation).
Here, $y$ has a local maximum: Since a piecewise constant solution is impossible according to $y''+y^2=0$, $y$ can only decrease there: If it were to increase, then the energy equation is clearly violated. So in $\bar x$, $y'$ decreases again, gets negative and the energy equation states that $y'$ has to decrease further if $y$ decreases as well. This happens until $y(\hat{x})=0$ for some $0 < \bar x < \hat{x}$ (and even further). Then prove that the mapping
$$
v(x) := \hat{x}^2 y\left(\hat{x}\cdot x \right)
$$
is still a solution to the BVP. $v$ is of course non-trivial.
