I have the following functional:
$$ I\left[ y(x) \right] = \int_{0}^{\frac{\pi}{2}} {\left((y')^2 - y^2 + 2xy\right)dy} $$ subject to boundary conditions:
$$ \begin{align} y(0) &= 0 \\ y\left(\dfrac{\pi}{2}\right) &= 0 \end{align} $$
Applying the Euler-Lagnrange's condition $ \left( f_y = \dfrac{\text{d}f_{y'}}{\text{dx}} \right) $; I proceed as follows after replacing $ dy $ with $ y' dx $:
$$ \begin{align} f \equiv f(x, y, y') &= (y')^3 - y' y^2 + 2 x y y' \\ f_y &= -2 y y' + 2 x y' \\ f_{y'} &= 3(y')^2 - y^2 + 2xy \\ \text{and } \dfrac{\text{d}f_{y'}}{\text{dx}} &= 6 y' y'' - 2 y y' + 2 y + 2 x y' \tag{1} \\ \therefore 3 y' y'' + y &= 0 \tag{ Take $ v \equiv v(y) = \frac{dy}{dx} $} \\ 3 v (v v') + y &= 0 \tag{ $ v' = \frac{dv}{dy} $ } \\ \implies 3 v^2 dv + y dy &= 0 \\ \left(\frac{dy}{dx}\right)^3 + \frac{y^2}{2} &= \frac{c^2}{2} \\ \sqrt[3]{2} \frac{dy}{dx} &= \left( c^2 - y^2 \right) ^{\frac{1}{3}} \end{align} $$
- Are the above steps correct?
- How do I proceed now?
Wolfram shows the answer to be some hypergeometric form; which I have no idea about.