0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$u= \frac{x^{2}}{2} + \frac{1}{2y^{2}} \quad$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.