Quasilinear PDE:$(x^2+1)u_x +2xu_y = 0, u(0,y)=\phi(y)$ I am trying to solve the PDE 
$$(x^2+1)u_x +2xu_y = 0, u(0,y)=\phi(y)$$
The solution is given here at the end of the third page and begining of the fourth and the author concluded at a step that $\dot{y} = 2y \tan(t)$. But I dont understand:
Since $\dot{y}=2x$ and $x=\tan(t)$, isnt $\dot{y}=2\tan(t) \Rightarrow y = -\ln|cos(t)| + C$? I dont uderstand what he did and the answer. What is wro g with my idea?
Thanks in advance!
 A: It looks like an error in the notes to me.  If you plug in the solution found in the notes (which is what the author suggests you do in Exercise 0.4), you get
$$
(x^2 + 1) u_x = (x^2 + 1) \left[ 2 y x \phi'(y(1 + x^2))\right]
$$
but
$$
2x u_y = 2x \left[ (1 + x^2) \phi'(y(1 + x^2))\right]
$$
and so
$$
(x^2+1)u_x +2xu_y = \left[ (x^2 + 1) (2 y x + 2x) \right] \phi'(y(1 + x^2)) \neq 0.
$$
The solution found in the notes would be the solution to the equation $(x^2+1)u_x - 2xyu_y = 0$ (note the change in sign in the second term) with the same boundary conditions.
A: As a side note: A different solution may be obtained by solving the equation by finding the characteristic curves. This may be seen by the following.
$$(1+x^2) u_{x} + 2x u_{y} = 0$$ 
leads to
$$ \frac{dx}{1+x^2} = \frac{dy}{2x} = \frac{du}{0}$$.
The first equation is
$$\frac{dx}{1+x^2} = \frac{dy}{x} \to \frac{2x}{1+x^2} \, dx = dy$$ 
which yields $y- \ln(1+x^2) = c_{1}$. The second curve can be found from the equation $du = 0\cdot dy$ which yields $u = c_{2}$. The intersection between the two curves provides
$$u(x,y) = f(y - \ln(1+x^2)).$$
Applying the condition $u(0,y) = \phi(y)$ yields a solution of the form
$$u(x,y) = \phi(y - \ln(1+x^2)).$$
