# problem with non linear pde

I have the following pde which i cannot solve. any suggestions, tips on how to approach a solution?

$$\left(1-x^3 \frac{\partial y}{\partial x} \partial_{y}\right) f(y(x))-1/4 \left(1-\frac{1}{x^3 (\frac{\partial y}{\partial x})} \partial_y\right) (\partial_y f(y(x)))^2 =b^2$$ or in a cleaner way: $$\left(1-w(y)\partial_y\right)f(y)-1/4\left(1-\frac{1}{w(y)}\partial_y \right) \left(\partial_y f(y)\right)^2=b^2$$ with $$w(y)=x^3\frac{\partial y}{\partial x}, \ \ y=y(x), \ \ b=const.$$

--Edited from comments--- This question comes from a problem in classical general relativity, involving an Ad$S_5$ background metric. The differential equation comes from trying to embed a string in this background. The function $f$ is related to the shape of the string, whereas $y$ and $x$ are related to the worldsheet coordinates. The unknowns are both $f$, and $y$.

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You're solving for both $f$ and $y$? – Mike Hall Mar 23 '11 at 1:17
What is the unknown function here, f(y) or y(x)? Either way, this looks like an ODE, not a PDE. – Michael Renardy Mar 23 '11 at 1:18
Where in blazes did you get such a monster? – drbobmeister Mar 23 '11 at 2:10
Dear Regina, I think your problem is quite interesting and so I feel motivated to do something I normally wouldn't be so concerned with, and that is to ask you to add a few words addressing the origins of this problem. I'm pretty sure many would agree doing so would help people help you in your search for an answer. – drbobmeister Mar 23 '11 at 5:12
To tell you about the origins of the problem: Im a physicist and doing a calculation in classical GR; a problem involving an AdS_5 background metric. The differential equation comes from trying to embed a string in this background. The function $f$ is related to the shape of the string, $y, x$ are related to the worldsheet coordinates. I could say more, but that would go beyond explaining where that equation comes from. As said: Id be thankful about any help. I have a feeling that there should be a way to solve this but am not mighty enough to do so. – Regina Mar 23 '11 at 13:06