I need to solve $yy'' +y^2 = C$ (constant) Is there an analytic solution? - if so please show me how to derive it.
If not, then boundary conditions for a numerical solution would be $y(0) = 35$, $y(400) = 8$.
I am pretty new at this . . . so apologies if further information/conditions/explanation required. I have access to Mathematica (though I am far from expert) but could not obtain either an analytic or a numerical solution.
 A: I don't think there is a closed form solution that is not a constant, but you can still find a solution.
It is an ODE of the form $f\left( y\left( x \right),\, \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x},\, \frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}} \right) = 0$ where you can reduce the order from $1$ to $2$ using the substitution $z\left( y\left( x \right) \right) := \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x}$:
$$
\begin{align*}
z\left( y\left( x \right) \right) &= \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x}\\
z\left( y\left( x \right) \right) \cdot \frac{\operatorname{d}z\left( y\left( x \right) \right)}{\operatorname{d}y\left( x \right)} &= \frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}^{2}x}\\
\end{align*}
$$
$$
\begin{align*}
y\left( x \right) \cdot \frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}} + \left( y\left( x \right) \right)^{2} - c &= 0\\
y\left( x \right) \cdot z\left( y\left( x \right) \right) \cdot \frac{\operatorname{d}z\left( y\left( x \right) \right)}{\operatorname{d}y\left( x \right)}  + \left( y\left( x \right) \right)^{2} - c &= 0\\
z\left( y\left( x \right) \right) \cdot \frac{\operatorname{d}z\left( y\left( x \right) \right)}{\operatorname{d}y\left( x \right)}  &= \frac{c}{y\left( x \right)} - y\left( x \right)\\
\left( z\left( y\left( x \right) \right) \cdot \frac{\operatorname{d}z\left( y\left( x \right) \right)}{\operatorname{d}y\left( x \right)} \right)\, \operatorname{d}y\left( x \right) &= \left( \frac{c}{y\left( x \right)} - y\left( x \right) \right)\, \operatorname{d}y\left( x \right)\\
z\left( y\left( x \right) \right) \cdot \operatorname{d}z\left( y\left( x \right) \right) &= \left( \frac{c}{y\left( x \right)} - y\left( x \right) \right)\, \operatorname{d}y\left( x \right)\\
\int z\left( y\left( x \right) \right)\, \operatorname{d}z\left( y\left( x \right) \right) &= \int \frac{c}{y\left( x \right)} - y\left( x \right)\, \operatorname{d}y\left( x \right)\\
\frac{1}{2} \cdot \left( z\left( y\left( x \right) \right) \right)^{2} &= c \cdot \ln\left( y\left( x \right) \right) - \frac{1}{2} \cdot \left( y\left( x \right) \right)^{2} + k_{1}\\
\frac{1}{2} \cdot \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} &= c \cdot \ln\left( y\left( x \right) \right) - \frac{1}{2} \cdot \left( y\left( x \right) \right)^{2} + k_{1}\\
\end{align*}
$$
You can further reduce the order of the ODE from $1$ to $0$ by solving for $\operatorname{d}x$:
$$
\begin{align*}
\frac{1}{2} \cdot \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} &= c \cdot \ln\left( y\left( x \right) \right) - \frac{1}{2} \cdot \left( y\left( x \right) \right)^{2} + k_{1}\\
\frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} &= \pm\sqrt{2 \cdot c \cdot \ln\left( y\left( x \right) \right) - \left( y\left( x \right) \right)^{2} + 2 \cdot k_{1}}\\
1\, \operatorname{d}x &= \pm\frac{1}{\sqrt{2 \cdot c \cdot \ln\left( y\left( x \right) \right) - \left( y\left( x \right) \right)^{2} + 2 \cdot k_{1}}}\, \operatorname{d}y\left( x \right)\\
\int 1\, \operatorname{d}x &= \int \pm\frac{1}{\sqrt{2 \cdot c \cdot \ln\left( y\left( x \right) \right) - \left( y\left( x \right) \right)^{2} + 2 \cdot k_{1}}}\, \operatorname{d}y\left( x \right)\\
x + k_{2} &= \pm\int_{a}^{y\left( x \right)} \frac{1}{\sqrt{2 \cdot c \cdot \ln\left( x \right) - x^{2} + 2 \cdot k_{1}}}\, \operatorname{d}x\\
\end{align*}
$$
If we now define this integral as a special function $I_{c,\, k_{1},\, a}\left( y\left( x \right) \right) := \pm\int_{a}^{y\left( x \right)} \frac{1}{\sqrt{2 \cdot c \cdot \ln\left( x \right) - x^{2} + 2 \cdot k_{1}}}\, \operatorname{d}x$, then we can also use the equation for $y(x)$:
$$
\begin{align*}
x + k_{2} &= \pm\int_{a}^{y\left( x \right)} \frac{1}{\sqrt{2 \cdot c \cdot \ln\left( x \right) - x^{2} + 2 \cdot k_{1}}}\, \operatorname{d}x\\
x + k_{2} &= I_{c,\, k_{1},\, a}\left( y\left( x \right) \right)\\
y\left( x \right) &= I_{c,\, k_{1},\, a}^{-1}\left( x + k_{2} \right)\\
\end{align*}
$$
where $I_{c,\, k_{1},\, a}^{-1}\left( z \right)$ is the inverse if $I_{c,\, k_{1},\, a}\left( z \right)$ aka $I_{c,\, k_{1},\, a}^{-1}\left( I_{c,\, k_{1},\, a}\left( z \right) \right) = z$ or
$$
y\left( x \right) = \sqrt{c}\\
$$
Wolfram|Alpha finds the same solution as you can see here.
