PDE with $\partial u/\partial y$ squared, $\partial u/ \partial x$ not squared. I've seen several problems with $u_x^2 + u_y^2$, such as here Characteristics method applied to the PDE $u_x^2 + u_y^2=u$.

But what about this equation:
  $$u_x + \frac{1}{2}u_y^2+\frac{1}{2}x^2=0$$
  with initial conditions $u(0,y)=f(y)$.

I am told that that fully eliminating parameters is not possible with $f$ unspecified. Apparently the best you can do it is with two equations with one extra variable.
Anyone able to help me out?
 A: First, we reduce the equation as Gregory pointed out:
$$u=v-\frac{x^3}{6}$$
Now we have:
$$v_x+\frac{1}{2}v_y^2=0$$

Edit. The general solution.
It seems to me this equation can be reduced to the Burgers' equation. Let us take another partial derivative w.r.t. $y$:
$$v_{xy}+v_y v_{yy}=0$$
Introducing a new function:
$$w=v_y$$
We can write:

$$w_x+w w_y=0$$

Which is precisely the Burgers' equation. Using the method of characteristics as stated here, we have the implicit general solution as:

$$w(x,y+w(0,y) x)=w(0,y)$$

Where $w(0,y)$ is the initial condition (which can probably be obtained from $v(0,y)$ as $w(0,y)=\frac{d}{dy} v(0,y)$).
If the initial condition is nice enough, we can find the explicit solution, by setting:
$$z=y+w(0,y) x$$
Then:
$$w(x,z)=w(0,y(x,z))$$
Then we find $v(x,z)$ as:

$$v(x,y)=\int w(x,y) dy+C$$

Where $C$ is a constant. You can try this method with various initial conditions.

Some easier ways to get particular solutions in case of simple initial condition:
$1)$ Let's assume the solution to have the form:
$$v(x,y)=g(x)+h(y)$$
Then:
$$g'+\frac{1}{2} h'^2=0$$
But the functions depend on different variables, so for the equation to hold, they have to be linear:
$$g(x)=ax+b \\ h(y)=cy+d$$
The equation gives us:
$$a+\frac{1}{2} c^2=0$$
So:

$$v(x,y)=-\frac{1}{2} c^2x+cy+B$$

Where $B=b+d$.
I do not see how any kind of initial conditions except linear will work here.

$2)$ But this is not the only form we can assume, let's try another:
$$v(x,y)=g(x)h(y)$$
Now we have:
$$hg'+\frac{1}{2}g^2 h'^2=0$$
Dividing by $hg^2$:
$$\frac{g'}{g^2}+\frac{1}{2}\frac{h'^2}{h}=0$$
Both parts of the equation have to be constant:
$$\frac{g'}{g^2}=-\mu$$
$$\frac{1}{2}\frac{h'^2}{h}=\mu$$
Solving, we obtain:
$$g(x)=\frac{1}{\mu x +a}\\ h(y)= \left(\sqrt{\frac{\mu}{2}} y+b \right)^2$$
So we have a solution:

$$v(x,y)=\frac{1}{\mu x +a} \left(\sqrt{\frac{\mu}{2}} y+b \right)^2$$

We can get rid of one of the constants, for example:
$$v(x,y)=\frac{1}{x +A} \left(\sqrt{\frac{1}{2}} y+B \right)^2$$
This is a different solution than $1)$ and can also satisfy the initial conditions for a particular $f(y)$.
A: Assume $p=\frac{\partial v}{\partial x}$, $q=\frac{\partial v}{\partial y}$, and applying the transformation $v=u+x^3/6$ as mentioned  Gregory Sir, the above PDE reduce to $2p+q^2 =0$. Then by using Charpit auxiliary equation we get $p=a$ and $q=b$ where $a$ and $b$ are arbitrary constant satisfy the relation  $2a+b^2=0$. Now using $dv=pdx+qdy$,  we get $v=ax+by+c$ and so $u=-\frac {x^3} 6+ax+by+c$ is the general solution. Using the relation $2a+b^2 =0$ and given initial condition you got the particular solution. 
A: Hint:
$u_x+\dfrac{u_y^2}{2}+\dfrac{x^2}{2}=0$
$u_{xy}+u_yu_{yy}=0$
Let $v=u_y$ ,
Then $v_x+vv_y=0$ with $v(0,y)=f_y(y)$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$
$\dfrac{dv}{dt}=0$ , letting $v(0)=v_0$ , we have $v=v_0$
$\dfrac{dy}{dt}=v=v_0$ , letting $y(0)=f(v_0)$ , we have $y=v_0t+f(v_0)=xv+f(v)$ i.e. $v=F(y-xv)$
