# Having trouble solving a PDE with method of characteristics

I am currently trying to solve a problem I have already solved, but am trying to solve it the way our professor solved it. The PDE is given by:

$$yu_x-2xyu_y=2xu, \>\>\>\>\>\>\>u=y^3 \text{ when } x=0, 1\leq y\leq 2$$

Now, I found the characteristic equations (noting that $u(0,y)=z(y)$): $$\dot x(s)=y, \>\dot y(s)=-2xy, \>\dot z(s)=2xz$$ I multiplied $\dot x(s)$ by $2x$ and added it to $\dot y(s)$ to get: $$2x\dot x+\dot y=0$$ Using Separation of Variables Method I got: $$x^2+y=y_0, \>\>\>\>\Longrightarrow\>\>\>\> y=y_0-x^2$$ Where $y_0$ is based off of the initial data. Then , my professor says: $$\frac{\partial x}{\partial s}=y_0-x^2$$ And claims it is easy to solve the above, and then the ODE: $$\dot z(s)=2xz$$ is trivial to solve. My question is, how is the above easy? I tried Separation of Variables which led to: $$x=-\sqrt{y_0}\frac{1+e^{2\sqrt{y_0} s}}{1-e^{2\sqrt{y_0}s}}$$ I don't see how thats easy. I know the method where you solve for: $$\frac{\dot z(s)}{z(s)}=-\frac{\dot y(s)}{y(s)}$$ and separate the variables to solve for this, but I want to know how to work it out this way. If anyone could help it would be appreciated!

Solving the ODE $x'(s)=y_0-x^2(s)$ is 'easy' in a course of PDEs, because the variables can easily be separated. Of course the integration can sometimes be tricky but here it is a standard integral. We have

$$s+C=\int ds +C = \int \frac{dx}{y_0-x^2}=\frac{1}{\sqrt{y_0}} \tanh^{-1} \left( \frac{x}{\sqrt{y_0}} \right)$$

and thus $$x(s)=\sqrt{y_0} \tanh(\sqrt{y_0} s) \left(=\sqrt{y_0} \frac{e^{2\sqrt{y_0}s}-1}{e^{2\sqrt{y_0}s} +1} \right).$$

Then the ODE $z'(s)=2xz=2\sqrt{y_0} \tanh(\sqrt{y_0}s)z$ is again separable since

$$2 \underbrace{\int \sqrt{y_0} \tanh(\sqrt{y_0}s) ds}_{\ln(\cosh(\sqrt{y_0} s)}+C=\int \frac{dz}{z}=\ln(z)$$

and thus

$$z(s)=C\cosh^2(\sqrt{y_0}s).$$

I wouldn't call it trivial but in a course of PDEs these steps can of course be left out as an exercise.