I am tasked with solving
$$\cos(y)\frac{\partial u}{\partial x} + x \frac{\partial u}{\partial y} = \frac{x}{u}$$
And can't seem to get past the characteristics, namely:
$$\frac{dx}{\cos(y)} = \frac{dy}{x}$$
and
$$\frac{du}{x/u} = \frac{dx}{\cos(y)}.$$
How might I proceed here?
$$\cos(y)\frac{\partial u}{\partial x} + x \frac{\partial u}{\partial y} = \frac{x}{u}$$ Charpit-Lagrange characteristic ODEs : $$\frac{dx}{\cos(y)}=\frac{dy}{x}=\frac{du}{\left(\frac{x}{u}\right)}$$ A first characteristic equation comes from solving $\frac{dx}{\cos(y)}=\frac{dy}{x}\quad\implies\quad 2xdx-2\cos(y)dy=0$ $$x^2-2\sin(y)=c_1$$ A second characteristic equation comes from solving $\frac{dy}{x}=\frac{du}{\left(\frac{x}{u}\right)}\quad\implies\quad 2udu-2dy=0$ $$u^2-2y=c_2$$ The general solution of the PDE on the form of implicit equation $c_2=F(c_1)$ is : $$u^2-2y=F\big(x^2-2\sin(y)\big)$$ $F$ is an arbitrary function. $$\boxed{u(x,y)=\pm\sqrt{2y+F\big(x^2-2\sin(y)\big)}}$$ The function $F$ has to be determined according to some boundary conditions which are missing in the wording of the question. The sign of the square root also has to be determined according to the boundary conditions.
