I'm trying to solve the following first order PDE:
$$3u_x+y^2u_y=\frac{x}{y}u$$
Applying the method of characteristics we have the following:
$$\frac{dx}{3}=\frac{dy}{y^2}=\frac{ydu}{xu}$$
It can easily be seen that by combining the first equality we can get our first constant : $$\frac{dx}{3}=\frac{dy}{y^2}\\=>\frac{x}{3}=-\frac{1}{y}+c_1\\=>\frac{x}{3}+\frac{1}{y}=c_1$$
This is where I'm stuck. I can't think of a proper combination which will allow me to integrate and find the second constant I'm looking for. Any ideas?
Edit: In case it helps, the solution manual says the second constant is : $$c_2=\ln{u}-\frac{x^2}{6y}-\frac{x^3}{54}$$
use the last equation and the first one $$\frac x 3 dx= \frac yu du$$
Use the first constant of integration $$\frac x3 +\frac 1y=K$$ $$\implies y=-\frac 1{(x/3-K)}$$ $$\frac x 3 dx=\frac {du}{u(K-x/3)}$$ $$\frac x 3(K-\frac x 3) dx=\frac {du}{u}$$ After integration $$K\frac {x^2}6-\frac {x^3}{27}+C_2=\ln |u|$$ Substitute $K=\frac x3+\frac 1y$ You get the final answer you posted $$\frac {x^3}{54}+\frac {x^2}{6y}+C_2=\ln |u|$$