# Math Help - General Solution for the characteristics of 2 quasi linear PDE's

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

I calculate the characteristics of 1) and 2) as
1) $\frac{dx}{dt}=y, \frac{dy}{dt}=-x, \frac{du}{dt}=e^u$
hence $\frac{dy}{dx}=\frac{-x}{y}$ however I dont know how this arrived at $x=acos(t) + bsin(t), y=bcos(t)-asin(t), -e^{-u}=t+c$
How are there 2 constants a and b when it is only a first order ODE?

2) $\frac{dx}{dt}=x, \frac{dy}{dt}=y, \frac{du}{dt}=sec(u)$
hence $\frac{dy}{dx}=\frac{y}{x}$ which arrives at
$x=Ae^t, y=Be^t, sin(u)=t+c$

I dont know how x,y and t+c are determined?
Any tips will be appreciated.
Thanks
bugatti79

2. On the first since

$\dfrac{dx}{dt} = y,\;\;\; \dfrac{dy}{dt} = -x$ then $\dfrac{d^2 x}{dt^2} = \dfrac{dy}{dt} = -x$ so

$\dfrac{d^2 x}{dt^2} + x = 0$ which has the solution

$x = a \cos t + b \sin t$ and

$y = \frac{dx}{dt} = - a \sin t + b \cos t$.

On the second, integrate

$\dfrac{dx}{dt} = x,\;\;\; \dfrac{dy}{dt} = y,\;\;\;\dfrac{du}{dt} = \sec u$ separately

3. Originally Posted by Danny
On the first since

$\dfrac{dx}{dt} = y,\;\;\; \dfrac{dy}{dt} = -x$ then $\dfrac{d^2 x}{dt^2} = \dfrac{dy}{dt} = -x$ so

$\dfrac{d^2 x}{dt^2} + x = 0$ which has the solution

$x = a \cos t + b \sin t$ and

$y = \frac{dx}{dt} = - a \sin t + b \cos t$.

On the second, integrate

$\dfrac{dx}{dt} = x,\;\;\; \dfrac{dy}{dt} = y,\;\;\;\dfrac{du}{dt} = \sec u$ separately
It was right in front of my eyes!!

so $\dfrac{d^2 x}{dt^2} + x = 0$ is a standard 2nd order ODE

$\int e^{-u}du=\intdt$ yields $-e^{-u} = t +c$

$\frac{dx}{dt}=x$ yields $ln x = t+c$
$x=e^{t+c}=e^{t} e^{c}=Ae^{t}$ likewise for dy/dt = y

Thanks Danny