Solve $(1+x^{2}) \frac{\partial{u}}{\partial{x}} + y\frac{\partial{u}}{\partial{y}} = 0$ I have the following equation:
$$
(1+x^{2}) \frac{\partial{u}}{\partial{x}} + y\frac{\partial{u}}{\partial{y}} = 0.
$$ 
Explain why we cannot deduce a solution by imposing $u(x,0)$.
So far what I have done is:
Separate the variables.
$$
\frac{dy}{y} = \frac{dx}{1+x^{2}}
$$
Take $x = \tan \ \theta$, so that $\frac{dx}{d\theta}=\sec^2\theta, dx=\sec^2\theta \ d\theta$,
$$
1+x^2=(\tan \ \theta)^2+1=\tan^2\theta+1=sec \ \theta .
$$
What I have now is:
LHS: 
$$
\int\frac{dy}{y} = \ln(y)
$$
RHS: 
\begin{align}
\int\frac{1}{x^2+1} \ dx &= \int\frac{1}{\sec \ \theta} \ (\sec^2\theta \ d\theta) \\
&= \int \sec \ \theta \ d\theta \\
&= \ln(\tan \ \theta + \sec \ \theta) \ + \ C \\
&= \ln[x+(x^2+1)]+C \\
&=\ln(x^2+x+1)+C
\end{align}
Therefore, $\ln \ (y) = \ln\ (x^{2}+x+1) + C$
$$
y = (x^{2}+x+1)e^{C}.
$$
Let $e^{C}=A$, $y = A(x^{2}+x+1)$ and $A = y(x^{2}+x+1)^{-1}$. Then, 
$$
u(x,y) = f(A) = f(y(x^{2}+x+1)^{-1}).
$$
I am unsure on how to proceed and wondering if anyone can help me with this. Thank you!
 A: The key question is " why we cannot deduce a solution by imposing $u(x,0)$ ".
In order to clarify this point, we will express the solution of the PDE on a general form. The method of characteristics leads to :
$$\frac{dx}{1+x^2}=\frac{dy}{y}=\frac{du}{0}$$
This Relationship is valid on the characteristics which are : 
\begin{cases}
    du=0 \quad \rightarrow \quad u=c_1\\
    \frac{dy}{y}= \frac{dx}{1+x^2} \quad \rightarrow \quad \ln(y)-\arctan(x)=c_2\\
  \end{cases}
Thus, the general solution on implicit form is :
$$\Phi\left(u\:,\:\ln(y)-\arctan(x)\right)=0$$
where $\Phi$ is any derivable function of two variables. This is equivalent to :
$$u(x,y)=F\left(\ln(y)-\arctan(x)\right)\text{  or }=G\left(y\;e^{-\arctan(x)}\right)$$
where $F$ or $G$ are any derivable functions.
Then $u(x,0)=F\left(\infty\right)\text{  or }=G\left(0\right)$
Any chosen form, $u(x,0)$ can only be a constant or $\infty$ , so cannot be a function of $x$. This is the reason why where is no solution if $u(x,0)$ non constant is imposed.
A: Basically, you are solving the PDE
$$
(1+x^{2}) \frac{\partial{u}}{\partial{x}} + y\frac{\partial{u}}{\partial{y}} = 0
$$ 
by assuming that $x = x(t)$, $y = y(t)$, so
$$
\frac{d}{d t} u\big(x(t),y(t)\big) = x'(t) u_x + y'(t) u_y = 0,
$$
and then
$$
x'(t) = 1 + x^2(t), \qquad y'(t) = y(t), \qquad u'(t) = 0.
$$
A solution given in $y = 0$ would mean that, for $t = 0$, $y(0) = 0$.
Now, what can you say about
$$
\int_{y(0)}^{y(t)} \frac{d}{ds} \log y(s)\,ds = \int_0^t \,ds\,?
$$
