Finding the solution of $u_x + y u_y = 0$ using $u(0, y) = y^3$ I am currently studying the textbook Partial Differential Equations – An introduction, second edition, by Walter A. Strauss. The section The Variable Coefficient Equation of chapter 1 says the following:

The equation
$$u_x + y u_y = 0 \label{4}\tag{4}$$
is linear and homogeneous but has a variable coefficient ($y$). We shall illustrate for equation \eqref{4} how to use the geometric method somewhat like Example 1. The PDE \eqref{4}  itself asserts that the directional derivative in the direction of the vector $(1, y)$ is zero. The curves in the $xy$ plane with $(1, y)$ as tangent vectors have slopes $y$ (see Figure 3). Their equations are
$$\dfrac{dy}{dx} = \dfrac{y}{1} \label{5}\tag{5}$$
This ODE has the solutions
$$y = Ce^x \label{6}\tag{6}$$
These curves are called the characteristic curves of the PDE \eqref{4} . As $C$ is changed, the curves fill out the $xy$ plane perfectly without intersecting. On each of the curves $u(x, y)$ is a constant because
$$\dfrac{d}{dx}u(x, Ce^x) = \dfrac{\partial{u}}{\partial{x}} + Ce^x \dfrac{\partial{u}}{\partial{y}} = u_x + yu_y = 0.$$
Thus $u(x, Ce^x) = u(0, Ce^0) = u(0, C)$ is independent of $x$. Putting $y = Ce^x$ and $C = e^{−x}y$, we have
$$u(x, y) = u(0, e^{-x}y).$$
It follows that
$$u(x, y) = f(e^{-x}y) \label{7}\tag{7}$$
is the general solution of this PDE, where again $f$ is an arbitrary function of only a single variable. This is easily checked by differentiation using the chain rule (see Exercise 4). Geometrically, the “picture” of the solution $u(x, y)$ is that it is constant on each characteristic curve in Figure 3.


I am now trying to do the following exercise:

Find the solution of \eqref{4}  that satisfies the auxiliary equation $u(0, y) = y^3$.

I used \eqref{6}  to get that $u(0, y) = y = C$, which means that $y = C = y^3$. However, I think that this is an incorrect conclusion. But is this not the reasoning that the author applied above to get \eqref{7}? So I don't understand why this reasoning was incorrect in this case.
I would greatly appreciate it if people would please take the time to carefully explain this.
 A: You seem to make some confusion due to the use of the variables $x,y$ for multiple (and different) purposes. The auxiliary (initial value) equation you write as $u(0,y)=y^3$ but you could equally well write it as  $u(0,t)=t^3$ for any real value of $t$ (better not use $y$ here).
For a given point $(x,y)$ in the plane you have from the result just before (7):  $u(x,y) = u(0,ye^{-x})$. Setting $t=y e^{-x}$ and using the previous we get:
$$ u(x,y)=u(0,ye^{-x}) = u(0,t) = t^3 = (y e^{-x})^3 = y^3 e^{-3x}.$$
Hope this brings you some peace of mind.
A: Let us verify the solution, using the Cauchy method.
Trying to find the solution in the form of
$$u(x,y) = X(x)Y(y),\tag1$$
one can get
$$X'(x)Y(y) + yX(x)Y'(y) = 0,$$
$$\frac{X'(x)}{X(x)} + y\frac{Y'(y)}{Y(y)} = 0.\tag2$$
Since the first term of $LHS(2)$ does not depend of $y$ and the second term does not depend of $x,$ then solutions of $(2)$ exists only if these terms are the opposite constants.
Assume the second term as $\lambda,$ then
$$\frac{X'(x)}{X(x)} = -\lambda,\quad \frac{Y'(y)}{Y(y)} = \frac\lambda y,\tag3$$
$$
\begin{cases}
\ln C_1^{-1}X = - \lambda x\\[4pt]
\ln C_2^{-1}Y = \lambda \ln y
\end{cases}\Rightarrow
\begin{cases}
X = C_1e^{-\lambda x}\\[4pt]
Y = C_2y^\lambda,
\end{cases}
$$
$$u(x,y,\lambda) = C_\lambda(y e^{-x})^\lambda.\tag4$$
The general solution can be defined as the arbitrary linear combination of such solutions  over the domain $D(\lambda)$ in the form of
$$u(x,y) = \int\limits_{D(\lambda)} w(\lambda)\ y^\lambda\ e^{-\lambda x}\text{ d}\lambda
= f(e^{-x}y).\tag5$$
If $u(0,y) = y^3,$ then

*

*from $(4)$ should $\lambda = 3, C_\lambda = 1,u(x,y,3) = y^3e^{-3x};$

*from $(5)$ should $w(\lambda) = \delta(\lambda-3),\quad u(x,y)= y^3e^{-3x}.$
Also, some details by the paper.

*

*The directional derivative is
$$u^\,_{\{1,y\}}= \text{ grad }u\cdot \{1,y\}= \{u_x,u_y\}\cdot \{1,y\} = u_x+yu_y;$$


*Since characteristic lines are defined by the equation $u(x,y) = C,$ then the full differential
$$\text{ d}u = u_x \text{ d}x + u_y \text{ d}y = 0$$
on this lines is zero, and
$$\frac{\text{ d}y}{\text{ d}x} = -\dfrac{u_x}{u_y}.$$
Taking in account PDE $(OP.4),$ this leads to the ODE $(OP.5).$


*To get the solution from $(OP.7),$ Maclaurin series can be used. However, the kernel of the representation $(5)$ continuously depends of $\lambda$ and looks more suitable if $\lambda$ is not integer.
A: $u_x+yu_y=0\\ u_{xx}+yu_{xy}=0\\ a=\ln{y}+x\\ b=y\\ u_{ab}=0\\ u=f(a)+g(b)=f(\ln{y}-x)+g(y)\\ u_x+yu_y=0\Longrightarrow g(y)=C=\text{cost.}\\ f(\ln{y})=y^3\Longrightarrow f(t)=e^{3t}-C\Longrightarrow u=y^3e^{-3x}$
