Characteristic curves of pde $u_x + xy^3 u_y = 0$

Question

I am new to these sorts of questions, and the method of characteristics. I've been asked to consider the equation:

$ \frac{∂u}{∂x} + xy^{3}\frac{∂u}{∂y} = 0$

I need to find the characteristic curves in the explicit form, and obtain the general solution of the PDE and check that this solution satisfies the PDE by differentiation. I ended up with the general solution being:

$ \frac{x^{2}}{2} + \frac{1}{2y^{2}} $

Would appreciate any help.

Answer 1

$$ \frac{∂u}{∂x} + xy^{3}\frac{∂u}{∂y} = 0$$

$u= \frac{x^{2}}{2} + \frac{1}{2y^{2}} \quad$ is not the general solution.

This is only a particular solution.

You apparently got the characteristic equation $$x^2+\frac{1}{y^2}=c$$ So, the general solution is : $$u(x,y)=F\left(x^2+\frac{1}{y^2}\right)$$ where $F$ is an arbitrary function, to be determined according to some boundary condition.