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This is the PDE that's been troubling me:

$$3u_x+y^2u_y=\frac{x}{y}u$$

From the characteristic equations we have $$x'(t)=3$$ $$y'(t)=y^2$$ $$u'(t)=\frac{x}{y}u$$ From the first two we can get one relationship: $$\frac{dx}{dy}=\frac{3}{y^2}=>\frac{dx}{3}=\frac{dy}{y^2}$$ $$=>\frac{x}{3}=-\frac{1}{y}+C_1=>x+\frac{3}{y}=C_1 \space\space (1)$$

Now I need one relationship for $u$ which will include a constant $C_2=f(C_1)$ where f arbitrary. I can't think of a good way to get the second equation though. Any ideas?

Edit: This is the best I can think of for now: $$\frac{dy}{y^2}=\frac{du}{\frac{x}{y}}=>\frac{xdy}{y^3}=\frac{du}{u}$$ I'm not sure if I can do this now, but from equation (1) I will write x in terms of y and this gives: $$\frac{C_1}{y^3}-\frac{3}{y^4}=\frac{du}{u}$$ $$=>\frac{1}{y^3}-\frac{C_1}{2y^2}=\ln|u|$$

Now I can replace $C_1$ with $x+\frac{3}{y}$ and have a solution of u. However, I am supposed to find a general solution, not a specific one. I've surely taken a wrong step. Now that I see this again, I even forgot to add the second constant.

A clear solution would help me a lot.

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Your first characteristic is correct : $$x+\frac{3}{y}=C_1$$ There is a typo in : $\quad\frac{dy}{y^2}=\frac{du}{\frac{x}{y}}\quad$ which should be $\quad\frac{dy}{y^2}=\frac{du}{\frac{x}{y}u}\quad$ without consequence since it is corrected in the derived equations.

Your calculus is correct : $\ln|u|=\frac{1}{y^3}-\frac{C_1}{2y^2}+c_2\quad$ (without forgetting the integration constant). $$\ln|u|=\frac{1}{y^3} -\frac{x+\frac{3}{y}}{2y^2}+c_2$$ $$\ln|u|=-\frac{x}{2y^2}-\frac{1}{2y^3}+c_2$$ $$u\exp\left(\frac{x}{2y^2}+\frac{1}{2y^3}\right)=C_2$$ The general solution comes from $C_2=f(C_1)$ $$u=\exp\left(-\frac{x}{2y^2}-\frac{1}{2y^3}\right)f\left(x+\frac{3}{y}\right)$$ $f$ is an arbitrary function, to be determined according to the boundary condition.

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