Partial differential equation (first order)

Question

I don't have ideas to solve the following PDE of the 1st order

$$ (x^2 - y^2 + 1)u_{x} + 2xyu_{y} = 0 $$

Could you give me a hint ?

Thanks, R.

Answer 1

Hint: use the method of characteristics.

Answer 2

$(x^2-y^2+1)u_x+2xyu_y=0$

$\dfrac{x^2-y^2+1}{2xy}u_x+u_y=0$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dy}{dt}=1$ , letting $y(0)=0$ , we have $y=t$

$\dfrac{dx}{dt}=\dfrac{x^2-y^2+1}{2xy}=\dfrac{x^2-t^2+1}{2xt}$

$2x\dfrac{dx}{dt}=\dfrac{x^2}{t}-\dfrac{t^2-1}{t}$

$\dfrac{d(x^2)}{dt}-\dfrac{x^2}{t}=-t+\dfrac{1}{t}$

$\dfrac{1}{t}\dfrac{d(x^2)}{dt}-\dfrac{x^2}{t^2}=-1+\dfrac{1}{t^2}$

$\dfrac{d}{dt}\left(\dfrac{x^2}{t}\right)=-1+\dfrac{1}{t^2}$

$\dfrac{x^2}{t}=x_0-t-\dfrac{1}{t}$

$x^2=x_0t-t^2-1=x_0y-y^2-1$

$\dfrac{du}{dt}=0$ , letting $u(0)=f(x_0)$ , we have $u(x,y)=f(x_0)=f\left(\dfrac{x^2+y^2+1}{y}\right)$