General Solution for the characteristics of 2 quasi linear PDE's

Hi all,

I am stuck on how the general solutions where obtained for the following quasi linear PDE's. I understand this to be quasi linear because the highest derivative of u is linear even though e^u is nonlinear.

Anyway,

1) $\displaystyle yu_x-xu_y=e^u$ and 2) $\displaystyle xu_x+yu_y=sec(u)$

I calculate the characteristics of 1) and 2) as

1) $\displaystyle \frac{dx}{dt}=y, \frac{dy}{dt}=-x, \frac{du}{dt}=e^u$

hence $\displaystyle \frac{dy}{dx}=\frac{-x}{y}$ however I dont know how this arrived at $\displaystyle x=acos(t) + bsin(t), y=bcos(t)-asin(t), -e^{-u}=t+c$

How are there 2 constants a and b when it is only a first order ODE?

2) $\displaystyle \frac{dx}{dt}=x, \frac{dy}{dt}=y, \frac{du}{dt}=sec(u)$

hence $\displaystyle \frac{dy}{dx}=\frac{y}{x}$ which arrives at

$\displaystyle x=Ae^t, y=Be^t, sin(u)=t+c$

I dont know how x,y and t+c are determined?

Any tips will be appreciated.

Thanks

bugatti79