Method of Characteristics for $(1+x^2)u_x+2u_y=y\cdot(1+u^2)$ Suppose I had a problem like 
$$(1+x^2)u_x+2u_y=y\cdot(1+u^2),\;\; u(1,y)=1$$
This is of the form 
$$a(x,y)u_x+b(x,y)u_y+c(x,y,u)$$
So I would use the Method of Characteristics:
$$\frac{dx}{dt}=(1+x^2)\implies x=\tan(t+c_1)$$
$$\frac{dy}{dt}=2\implies y=2t+c_2$$
Using the initial condition we get that
$$t=\tan^{-1}x -1,\;\; c_2=y-2\tan^{-1}(x)+2$$
Where I'm having a bit of an issues is with the $du/dt$ term.  Would I set it up like 
$$\frac{du}{dt}=(2t+c_2)(1+u^2)\implies u=\tan(t^2+c_2t+c_3)?$$
Then sub in the $c_2$ to get 
$$u=\tan\left[t^2+(y-2\tan^{-1}x+2)t+c_3)\right]?$$
Then use the initial condition again to get that 
$$u(1,y)=\tan\left[t^2+(y-2\tan^{-1}(1)+2)t+c_3)\right]$$
and solve for c_3?  
 A: $$(1+x^2)u_x+2u_y=y\cdot(1+u^2),\qquad u(1,y)=1$$
Another approach to compare to what was discussed in the comments :
Change of function : $u(x,y)=\tan\left(U(x,y)\right)  \quad\to\quad \begin{cases} \frac{\partial u}{\partial x}=(1+u^2)\frac{\partial U}{\partial x} \\ \frac{\partial u}{\partial y}=(1+u^2)\frac{\partial U}{\partial y} \end{cases}$
$$(1+x^2)U_x+2U_y=y$$
with condition : $u(1,y)=1=\tan\left(U(1,y)\right)  \quad\to\quad U(1,y)=\frac{\pi}{4}+n\pi$
Change of variable : $\begin{cases}x=\tan(X) \\ U(x,y)=V(X,y)\end{cases}$ $\quad\to\quad (1+x^2)\frac{\partial U}{\partial x}=\frac{\partial V}{\partial X}$
$$V_X+2V_y=y$$
Condition : $U(1,y)=\frac{\pi}{4}+n\pi=V\left((\frac{\pi}{4}+m\pi),y\right)$
$\frac{1}{4}y^2$ is a particular solution
Change of function : $V(X,y)=\frac{1}{4}y^2+W(X,y)$
$$W_X+2W_y=0$$
Condition : $\frac{\pi}{4}+n\pi=\frac{1}{4}y^2+W\left((\frac{\pi}{4}+m\pi),y\right)$
With the method of characteristics, the characteristic equations are :
$$\frac{dX}{1}=\frac{dy}{2}=\frac{dW}{0}$$
From $\frac{dX}{1}=\frac{dy}{2}$ the first characteristic curve : $\quad 2X-y=c_1$
From $dW=0$ the second characteristic curve : $W=c_2$
General solution on implicit form : $\Phi(2X-y\:,\: W)=0$ any differentiable function $\Phi$ of two variables.
Solving the implicit equation for the second variable $\quad\to\quad W=F(2X-y)\quad$ any differentiable function $F$.
The condition :  $\frac{\pi}{4}+n\pi= \frac{1}{4}y^2+F\left(2(\frac{\pi}{4}+m\pi)-y\right)$ determines the function $F$ : 
$$F(Y)=\frac{\pi}{4}+n\pi- \frac{1}{4}\left(-Y+\frac{\pi}{2}+2m\pi\right)^2$$
$$W=F(2X-y)=\frac{\pi}{4}+n\pi- \frac{1}{4}\left(-(2X-y)+\frac{\pi}{2}+2m\pi\right)^2$$
$$V(X,y)=\frac{1}{4}y^2+W(X,y)=\frac{1}{4}y^2+\frac{\pi}{4}+n\pi- \frac{1}{4}\left(-(2X-y)+\frac{\pi}{2}+2m\pi\right)^2$$
$$U(x,y)=\frac{1}{4}y^2+\frac{\pi}{4}+n\pi- \frac{1}{4}\left(-(2\tan^{-1}(x)-y)+\frac{\pi}{2}+2m\pi\right)^2$$
$$u(x,y)=\tan\left(\frac{1}{4}y^2+\frac{\pi}{4}+n\pi- \frac{1}{4}\left(-(2\tan^{-1}(x)-y)+\frac{\pi}{2}+2m\pi\right)^2 \right)$$
After simplification :
$$u(x,y)=\tan\left(\frac{y^2+\pi-\left(y-2\tan^{-1}(x)+\frac{\pi}{2}+2m\pi\right)^2}{4} \right)$$
any integer $m$.
