Solving a Coupled Second-Order Differential Equation I'm trying to solve the real-valued differential equation
$$xy\ddot{y}+2y\dot{x}\dot{y}+\alpha\ddot{x}+x\dot{y}^2=0$$
where $\alpha\in\mathbb{R}$ and $\alpha>0$ for both $x(t)$ and $y(t)$.  How would I even approach this problem?
EDIT: Sorry, I forgot a conservation law:
$$\alpha^2\dot{x}^2+x^4\dot{y}^2 = P^2$$
where $P$ is a conserved quantity.
EDIT 2: The original Lagrangian this comes from is 
$$\mathcal{L}=\frac{1}{2}\left(-\alpha\dot{x}^2+x^2\dot{y}^2+x^2y^2\dot{\theta}^2+x^2y^2\sin^2\theta\dot{\phi}^2\right)$$
 A: As I said before, knowing more detail of the problem may lead to its solution. So you are studying the motion of a free point-mass in a mildly curved space-time with Lorenzian metric
$$ d\sigma^2 = - \alpha^2 \, dx^2 + x^2\Big(dy^2 + y^2 \big(d\theta^2 + \sin(\theta) \, d\phi^2\big)\Big)$$ Let me rename the variables $x=t$ and $y = r$. Then 
$$ d\sigma^2 = - \alpha^2 \, dt^2 + t^2\Big(dr^2 + r^2 \big(d\theta^2 + \sin(\theta) \, d\phi^2\big)\Big)$$
If you consider $(t,r,\theta, \phi)$ as spherical coordinates plus time, the Cartesian coordinates are
\begin{align}
x &= r \, \cos(\phi) \, \sin(\theta)\\
y &= r \, \sin(\phi) \, \sin(\theta)\\
z &= r \, \cos(\theta)\\
\end{align}
so the space part of the metric becomes
$$g_E^2 = dr^2 + r^2 \big(d\theta^2 + \sin(\theta) \, d\phi^2\big) = dx^2 + dy^2 + dz^2$$ so your space-time metric is basically
$$d\sigma^2 = -\alpha^2 \, dt^2 + t^2 \, g^2_E =  -\alpha^2 \, dt^2 + t^2 \, \big( dx^2 + dy^2 + dz^2\big)$$
So your Lagrangian is technically the Lorenzian archlength
$$\mathcal{L} = \sqrt{  -\alpha^2 \, \dot{t}^2 + t^2 \, \big( \dot{x}^2 + \dot{y}^2 + \dot{z}^2\big)}$$ i.e.
$$\mathcal{L}^2 = { -\alpha^2 \, \dot{t}^2 + t^2 \, \big( \dot{x}^2 + \dot{y}^2 + \dot{z}^2\big)}$$ where $\dot{t} = \frac{dt}{ds}, \,\, \dot{x} = \frac{dx}{ds}, \, \,\dot{y} = \frac{dy}{ds}, \,\, \dot{x} = \frac{dz}{ds}$ with respect to some parameter $s$ which will be determined later.
The Euler-Lagrange equations look like this
\begin{align}
\frac{d}{ds} \left(\frac{\partial \mathcal{L}}{\partial \dot{t}}\right) & = \frac{\partial \mathcal{L}}{\partial t}\\
\frac{d}{ds} \left(\frac{\partial \mathcal{L}}{\partial \dot{x}}\right) & = \frac{\partial \mathcal{L}}{\partial x}\\
\frac{d}{ds} \left(\frac{\partial \mathcal{L}}{\partial \dot{y}}\right) & = \frac{\partial \mathcal{L}}{\partial y}\\
\frac{d}{ds} \left(\frac{\partial \mathcal{L}}{\partial \dot{z}}\right) & = \frac{\partial \mathcal{L}}{\partial z}
\end{align}
Now 
\begin{align}
\frac{1}{2} \, \frac{\partial}{\partial \dot{t}} \big( \mathcal{L}^2\big) & = \mathcal{L}  \, \frac{\partial \mathcal{L}}{\partial \dot{t}} = - \alpha^2 \, \dot{t} \\
\frac{1}{2} \,\frac{\partial}{\partial \dot{x}} \big( \mathcal{L}^2\big) & = \mathcal{L}  \, \frac{\partial \mathcal{L}}{\partial \dot{x}} = t^2 \, \dot{x}\\
\frac{1}{2} \,\frac{\partial}{\partial \dot{y}} \big( \mathcal{L}^2\big) & = \mathcal{L}  \, \frac{\partial \mathcal{L}}{\partial \dot{y}} = t^2 \, \dot{y}\\
\frac{1}{2} \, \frac{\partial}{\partial \dot{z}} \big( \mathcal{L}^2\big) & = \mathcal{L}  \, \frac{\partial \mathcal{L}}{\partial \dot{z}}= t^2 \, \dot{z}
\end{align}
leading to 
\begin{align}
 \frac{\partial \mathcal{L}}{\partial \dot{t}} &= - \frac{\alpha^2 \, \dot{t}}{ \mathcal{L} } \\
\frac{\partial \mathcal{L}}{\partial \dot{x}} &= \frac{t^2 \, \dot{x}}{ \mathcal{L} }\\
\frac{\partial \mathcal{L}}{\partial \dot{x}} &= \frac{t^2 \, \dot{y}}{ \mathcal{L} }\\
\frac{\partial \mathcal{L}}{\partial \dot{x}} &= \frac{t^2 \, \dot{z}}{ \mathcal{L} }\\
\end{align}
Similarly
\begin{align}
\frac{1}{2} \, \frac{\partial}{\partial {t}} \big( \mathcal{L}^2\big) & = \mathcal{L}  \, \frac{\partial \mathcal{L}}{\partial {t}} = t \big(\dot{x}^2 + \dot{y}^2 + \dot{z}^2\big)\\
\frac{1}{2} \,\frac{\partial}{\partial {x}} \big( \mathcal{L}^2\big) & = \mathcal{L}  \, \frac{\partial \mathcal{L}}{\partial {x}} = 0\\
\frac{1}{2} \,\frac{\partial}{\partial {y}} \big( \mathcal{L}^2\big) & = \mathcal{L}  \, \frac{\partial \mathcal{L}}{\partial {y}} =0\\
\frac{1}{2} \, \frac{\partial}{\partial {z}} \big( \mathcal{L}^2\big) & = \mathcal{L}  \, \frac{\partial \mathcal{L}}{\partial {z}}= 0
\end{align} Finally, the raw Euler-Lagrange equations look as follows
\begin{align}
 - \alpha^2 \, \frac{d}{ds}\left(\frac{\dot{t}}{ \mathcal{L} } \right) &= \frac{t \big(\dot{x}^2 + \dot{y}^2 + \dot{z}^2\big)}{\mathcal{L}}\\
\frac{d}{ds}\left( \frac{t^2 \, \dot{x}}{ \mathcal{L}} \right)  &= 0 \\
\frac{d}{ds}\left( \frac{t^2 \, \dot{y}}{ \mathcal{L}} \right)  &= 0 \\
\frac{d}{ds}\left( \frac{t^2 \, \dot{z}}{ \mathcal{L}} \right)  &= 0 
\end{align} The last three equations can be immediately integrated once
\begin{align}
 \frac{d}{ds}\left(\frac{\dot{t}}{ \mathcal{L} } \right) &=  - \frac{t \big(\dot{x}^2 + \dot{y}^2 + \dot{z}^2\big)}{\alpha^2 \, \mathcal{L}}\\
\frac{t^2 \, \dot{x}}{ \mathcal{L}} &= u_0 \\
\frac{t^2 \, \dot{y}}{ \mathcal{L}} &= v_0 \\
\frac{t^2 \, \dot{z}}{ \mathcal{L}} &= w_0 
\end{align} leading to
\begin{align}
 \frac{d}{ds}\left(\frac{\dot{t}}{ \mathcal{L} } \right) &=  - \frac{t \big(\dot{x}^2 + \dot{y}^2 + \dot{z}^2\big)}{\alpha^2 \, \mathcal{L}}\\
\dot{x} &= \frac{u_0}{t^2} \,\mathcal{L}\\
\dot{y} &= \frac{v_0}{t^2} \,\mathcal{L} \\
\dot{z} &= \frac{w_0}{t^2} \,\mathcal{L}
\end{align} 
Square the last three equations and plug them in the first one
\begin{align}
\frac{d}{ds}\left(\frac{\dot{t}}{ \mathcal{L} } \right) &= - \frac{t}{\alpha^2 \, \mathcal{L}} \, \big(\dot{x}^2 + \dot{y}^2 + \dot{z}^2\big) = - \frac{t}{\alpha^2 \, \mathcal{L}} \, \frac{\big({u_0}^2 + {v_0}^2 + {w_0}^2\big)}{t^4} \, \mathcal{L}^2\\
&= - \, \frac{{u_0}^2 + {v_0}^2 + {w_0}^2}{\alpha^2 \, t^3} \, \mathcal{L}
\end{align} or rewrite it as
$$\frac{1}{\mathcal{L}} \, \frac{d}{ds} \left(\frac{1}{ \mathcal{L} } \, \frac{dt}{ds}\right) = - \, \frac{{u_0}^2 + {v_0}^2 + {w_0}^2}{\alpha^2 \, t^3} $$ The proper time is $d\tau = \mathcal{L}ds$ so the equations turn into
 $$\frac{d}{d\tau} \left( \frac{dt}{d\tau }\right) = - \, \frac{{u_0}^2 + {v_0}^2 + {w_0}^2}{\alpha^2 \, t^3} $$
$$\frac{d^2 t}{d\tau^2} =  - \, \frac{k_0}{t^3}$$ where $k_0 = \frac{{u_0}^2 + {v_0}^2 + {w_0}^2}{\alpha^2}$ is a constant. Finally, in terms of proper time, the system of Euler-Lagrange equations describing the geodesics become
\begin{align}
\frac{d^2 t}{d\tau^2} &=  - \, \frac{k_0}{t^3}\\
\frac{dx}{d\tau} &= \frac{u_0}{t^2}\\
\frac{dy}{d\tau} &= \frac{v_0}{t^2}  \\
\frac{dz}{d\tau}&= \frac{w_0}{t^2} 
\end{align} 
So you solve the first equation, which is completely decoupled from the rest, and obtain the function $t = t(\tau)$. Then 
\begin{align}
x(\tau) &= x_0 + \int \frac{u_0}{t(\tau)^2} \, d\tau\\
y(\tau) &= y_0 + \int \frac{v_0}{t(\tau)^2} \, d\tau \\
z(\tau)&= z_0 + \int \frac{w_0}{t(\tau)^2} \, d\tau
\end{align} 
A: Is this a homework problem or is it some sort of practice question? If not, any possible symmetries coming from the formulation of the problem? If not, I don't know how far one can go, but let's see. It may help if you tell us how  you derived the equation, weather it is an Euler-Lagrange equation of some sort. Maybe it's coming from a problem in differential geometry related to curves on surfaces?
Invert the function $x=x(t)$ and write it as a function $t=t(x)$. Then
$$\frac{d}{dt} = \frac{dx}{dt} \, \frac{d}{dx}$$
Form the metric $$\alpha^2 \left(\frac{dx}{dt}\right)^2 + x^4 \, \left(\frac{dy}{dt}\right)^2 = P^2$$ one gets 
$$P^2 = \alpha^2 \left(\frac{dx}{dt}\right)^2 + x^4 \, \left(\frac{dx}{dt}\right)^2\left(\frac{dy}{dx}\right)^2 =  \left(\frac{dx}{dt}\right)^2 \, \left(\alpha^2 + x^4 \, \left(\frac{dy}{dx}\right)^2\right)$$ and so
$$ \left(\frac{dx}{dt}\right)^2 = \frac{P^2}{\alpha^2 + x^4 \, \left(\frac{dy}{dx}\right)^2}$$
$$\frac{dx}{dt}= \frac{P}{\sqrt{\alpha^2 + x^4 \, \left(\frac{dy}{dx}\right)^2}}$$ Thus
$$\frac{d}{dt} = \left(\frac{P}{\sqrt{\alpha^2 + x^4 \, \left(\frac{dy}{dx}\right)^2}} \right)\,\, \frac{d}{dx}$$
Now the equation $$x y \, \frac{d^2 y}{dt^2} + 2 \, y \, \frac{dx}{dt}\frac{dy}{dt} + \alpha \frac{d^2x}{dt^2} + x \left(\frac{dy}{dt}\right)^2 = 0$$ can be written as
$$\frac{d}{dt}\left(x y \, \frac{dy}{dt} + \alpha \frac{dx}{dt}\right) +  y \, \frac{dx}{dt}\frac{dy}{dt} = 0$$ Writing the latter equation in terms of $x$ as an independent variable
$$\left(\frac{P^2}{\sqrt{\alpha^2 + x^4 \, \left(\frac{dy}{dx}\right)^2}} \right)\,\, \frac{d}{dx} \left(\frac{x y \frac{dy}{dx} + \alpha}{\sqrt{\alpha^2 + x^4 \, \left(\frac{dy}{dx}\right)^2}}\right) +  \frac{P^2 \, y \, \frac{dy}{dx}}{\alpha^2 + x^4 \, \left(\frac{dy}{dx}\right)^2} = 0$$
$$ \frac{d}{dx} \left(\frac{x y \frac{dy}{dx} + \alpha}{\sqrt{\alpha^2 + x^4 \, \left(\frac{dy}{dx}\right)^2}}\right) +  \frac{y \, \frac{dy}{dx}}{\sqrt{\alpha^2 + x^4 \, \left(\frac{dy}{dx}\right)^2}} = 0$$ In addition to that, one can also write $z(x) = \frac{1}{2} y(x)^2$ and so the equation becomes
$$ \frac{d}{dx} \left(\frac{x \frac{dz}{dx} + \alpha}{\sqrt{\alpha^2 + \frac{x^4}{2 \, z} \, \left(\frac{dz}{dx}\right)^2}}\right) +  \frac{ \frac{dz}{dx}}{\sqrt{\alpha^2 + \frac{x^4}{2 \, z} \, \left(\frac{dz}{dx}\right)^2}} = 0$$
Alternatively, you can multiply the initial equation by $x$ and get
$$x^2 y \ddot{y} + 2xy \, \dot{x} \dot{y} + x^2 \dot{y}^2 + \alpha \, x \ddot{x} = 0$$ which becomes
$$0 = \frac{d}{dt} \big(x^2 y \dot{y}\big) + \alpha \, x \ddot{x} = \frac{d}{dt} \big(x^2 y \dot{y}\big) + \alpha \,\big( x \ddot{x} + \dot{x}^2\big) - \alpha \, \dot{x}^2 = \frac{d}{dt} \big(x^2 y \dot{y} +  \alpha \, x \dot{x} \big) - \alpha \, \dot{x}^2$$
Now had your second equation been $$\alpha^2 \, \dot{x}^2 + x^4 y^2 \dot{y}^2 = P^2$$ then a miracle could have happened and since $x^2y\dot{y} = \sqrt{P^2 - \alpha^2 \, \dot{x}^2}$ the equation would have become
$$\frac{d}{dt} \Big(\sqrt{P^2 - \alpha^2 \, \dot{x}^2} + \alpha \, x \dot{x}\Big) - \alpha \, \dot{x}^2 = 0$$ which would have been a whole new story. This equation would have had much more potential. 
Currently, you can end up with $$\frac{d}{dt} \Big( y \sqrt{P^2 - \alpha^2 \, \dot{x}^2} + \alpha \, x \dot{x}\Big) - \alpha \, \dot{x}^2 = 0$$ Maybe now one can write $y=y(t)$ as a function $t=t(y)$ and $x=x(y)$. Then
$$\frac{d}{dt} = \dot{y} \frac{d}{dy}$$ so 
$$\dot{y} \frac{d}{dy}\left( y \sqrt{P^2 - \alpha^2 \, \dot{x}^2} + \alpha \, x \dot{x}\right) - \alpha \, \dot{x}^2 = 0$$
$$\dot{y} \frac{d}{dy}\left( y \sqrt{P^2 - \alpha^2 \, \dot{y}^2 \left(\frac{dx}{dy}\right)^2} + \alpha \, x  \dot{y} \, \frac{dx}{dy}\right) - \alpha \,  \dot{y}^2 \left(\frac{dx}{dy}\right)^2 = 0$$ Express $$\dot{y} = \frac{P}{\sqrt{\alpha^2\left(\frac{dx}{dy}\right)^2 + x^4}} = f\left(x,\frac{dx}{dy}\right)$$ so we get
$$f \frac{d}{dy}\left( y \sqrt{P^2 - \alpha^2 \, f^2 \left(\frac{dx}{dy}\right)^2} + \alpha \, x  f \, \frac{dx}{dy}\right) - \alpha \,  f^2 \left(\frac{dx}{dy}\right)^2 = 0$$
$$\frac{d}{dy}\left( y \sqrt{P^2 - \alpha^2 \, f^2 \left(\frac{dx}{dy}\right)^2} + \alpha \, x  f \, \frac{dx}{dy}\right) - \alpha \,  f \, \left(\frac{dx}{dy}\right)^2 = 0$$
$$\frac{d}{dy}\left( y \sqrt{P^2 - \alpha^2 \, f\left(x,\frac{dx}{dy}\right)^2 \left(\frac{dx}{dy}\right)^2} + \alpha \, x  f\left(x,\frac{dx}{dy}\right) \, \frac{dx}{dy}\right) - \alpha \,  f\left(x,\frac{dx}{dy}\right) \, \left(\frac{dx}{dy}\right)^2 = 0$$ This is again a $y$ inhomogeneous second order ODE of type $F\big(y, x, x'(y), x''(y)\big) = 0$. 
Are you sure that the restriction is not $$\alpha^2 \, \dot{x}^2 + x^4 y^2 \dot{y}^2 = P^2 \, ? :)$$
