first order linear PDE, general solution Given the PDE, find the general solution.
$$
-4yu_x + u_y-yu=0
$$
So,
$$
\frac{dy}{dx} = \frac{1}{-4y}
$$
$ x+2y^2=k $ is a characteristic curve. If I do the parametrization
$$
y=s \\
x=k-2y^2=k-2s^2
$$
The PDE becomes the ODE
$$
\frac{du}{ds} -yu = 0 \rightarrow \frac{du}{ds}-su = 0 \\
u=f(k)e^{\frac{s^2}{2}}
$$
Which gives me the answer
$$
u(x,y)=f(x+2y^2)e^{\frac{1}{2}y^2}
$$
However, if I divide the original PDE by $-4y$, then I get:
$$
u_x + \frac{1}{-4y}u_y+\frac{1}{4}u=0
$$
with the same characteristic curve. But now I can do the following parametrization:
$$
\frac{dx}{ds}=1 \rightarrow x=s\\
\frac{dy}{ds}=-\frac{1}{4y}\rightarrow -2y^2=s+k
$$
The PDE then becomes:
$$
\frac{du}{ds} + \frac{1}{4}u = 0 
\Rightarrow u=f(k)e^{-\frac{s}{4}}
$$
The final answer this time is:
$$
u(x,y) = f(x+2y^2)e^{-\frac{x}{4}}
$$
My question is: are both answers the same? WolframAlpha gives as the answer the second option.
If I say $x+2y^2=0 \rightarrow y^2=-x/2$, I can change the $exp$ term from one to another, but is this right to do? And why should I get $x+2y^2=0$, and not any other value of the characteristic $k$? Is it somewhat preferably to write the solution in terms of $x$ instead of $y$?
 A: The solutions are indeed the same:
$$u(x,y)=f(x+2y^2)e^{y^2/2}=g(x+y^2)e^{-(1/4)(x+2y^2)}e^{y^2/2}=g(x+2y^2)e^{-x/4}$$
You still have a function with $x-y^2$ as argument, but not the same as in the first case.
$$f(x+2y^2)=g(x+2y^2)h(x+2y^2)\;\mathbb{with}\;h(x+2y^2)=e^{-(1/4)(x+2y^2)}$$
Added
You can get now any of the particular solutions you are requested to give. I solve here the second one in your comment. The others can be solved in the same fashion.
It can be write "$u(x,y)=−y$ along $y^2=x$" this way:
$$u(y^2,y)=-y$$
Substituting into the general solution,
$$u(x,y)=f(x+2y^2)e^{y^2/2}$$
gives:
$$u(y^2,y)=f(y^2+2y^2)e^{y^2/2}=-y$$
Rearranging:
$$f(3y^2)=-ye^{-y^2/2}$$
Now, the variable change $z=3y^2$ drives to the function $f$ corresponding to these initial conditions.
$$f(z)=\pm\sqrt{z/3}e^{-z/6}$$
Now, the argument of $f$ have to be $x+2y^2$, and substituted into $u$ is
$$u(x,y)=\pm\sqrt{\frac{x+2y^2}{3}}e^{-(x+2y^2)/6}e^{y^2/2}=\pm\sqrt{\frac{x+2y^2}{3}}e^{(y^2-x)/6}$$
As surface in $\mathbb R^3$ the two halves fit smoothly together as to form a unique surface. Nevertheless, due to the introduction of the square in the varible change, it's necessary to "tune" them to fit to the intial conditions. We use both them in a suitable domain to get:
$$u(x,y)=\begin{cases}
   -\sqrt{\frac{x+2y^2}{3}}e^{(y^2-x)/6}\;\mathbb{if}\;y>0\\
   \sqrt{\frac{x+2y^2}{3}}e^{(y^2-x)/6}\;\mathbb{if}\;y\le 0
\end{cases}$$
