Infinite number of solutions using method of characteristics I'm currently having a go at a question about the method of characteristics (and struggling!). I want to show that the Cauchy problem $u_{x} + 5x^{4}u_{y} = 2$ with $u(s,s^5)=2s+1$ has infinitely many $C^1(\mathbb{R}^2)$ solutions. 
Solving for the characteristic curves gives $x_{1}(t) = t + s_{0}$,  $x_{2}(t)=(t+s_{0})^5$ and $z(t) = 2(t_{0}+s_{0}) + 1$.
Now since the integral curves cover the graph of $u$, there exists some positive $t_{0}$ such that $(x,y) = (x_{1}(t_{0}),x_{2}(t_{0}))$ = $(t_{0} + s_{0},(t_{0}+s_{0})^5)$.
Hence, $u(x,y)=z(t_{0}) = 2(t_{0}+s_{0}) + 1$.
From this, we can clearly see that $u(x,y)=2x+1$ and $u(x,y)=2y^{1/5} + 1$ are solutions of the given PDE problem, but I have no idea how to show that we can find infinitely many solutions.
I did see this post, but I don't particularly understand the top comments' solution. Any help is much appreciated!
 A: Note that your second solution doesn't solve the BVP (always check solutions by substitution, when possible!). Apply the method of characteristics


*

*$\frac{d}{dt} x = 1$, letting $x(0) = s$ we know $x = t + s$,

*$\frac{d}{dt} y = 5x^4$, letting $y(0) = s^5$ we know $y = (t + s)^5$,

*$\frac{d}{dt} u = 2$, letting $u(0) = 2s+1$ we know $u = 2(t + s)+1$.


Solutions $u(x,y)$ are obtained by substitution of $t, s$ in the last equation. This is done by using the two equations of two unknowns given by the expression of $x, y$. However, this system
$$
\left\lbrace
\begin{aligned}
t + s &= x\\
t+s &= \sqrt[5]{y}
\end{aligned}
\right.
$$
has no solution, except along the curve $y - x^5 = 0$ where it has infinitely many solutions.

Let's derive parameter-free solutions.
The Lagrange-Charpit equations
$$
\frac{dx}{1} = \frac{dy}{5x^4} = \frac{du}{2}
$$
give the characteristic families $y - x^5 = c_1$ and $u - 2x = c_2$. The general solution reads
$$
u = 2x + f(y - x^5)
$$
in explicit form, where $c_2 = f(c_1)$ involves an arbitrary function $f$. The boundary condition $u(s,s^5) = 2s+1$ imposes $f(0) = 1$,
which suggests that infinitely many functions $f$ are suitable.
A: $$u_x+5x^4u_y=2$$
Charpit-Lagrange system of characteristic ODEs :
$$\frac{dx}{1}=\frac{dy}{5x^4}=\frac{du}{2}$$
A first characteristic equation comes from solving $\frac{dx}{1}=\frac{dy}{5x^4}$ :
$$y-x^5=c_1$$
A second characteristic equation comes from solving $\frac{dx}{1}=\frac{du}{2}$ :
$$u-2x=c_2$$
The general solution of the PDE on implicit form $c_2=F(c_1)$ is :
$$u-2x=F(y-x^5)$$
$$u(x,y)=2x+F(y-x^5)$$
$F$ is an arbitrary function (to be determined if some boundary conditions are correctly specified).
Condition : $\quad u(s,s^5)=2s+1$
$y-x^5=s^5-x^5=0$
$$u(s,s^5)=2s+1=2s+F(s^5-s^5)=2s+F(0)\quad\implies\quad F(0)=1$$
Of course they are an infinity of functions which are equal to $1$ when the argument is $0$. For exemple : $F(s)=\cos(s)$ or $F(s)= 1+s^4$ or $F(s)=e^{3s}$ etc.
As a consequence they are an infinity of solutions of the PDE which satisfies the condition. For example :
$u(x,y)=2x+\cos(y-x^5)$
$u(x,y)=2x+1+(y-x^5)^4$
$u(x,y)=2x+e^{3(y-x^5)}$
etc.
