Solution of an initial value problem (MCQ) (CSIR DEC 2015)

Question

The solution of the initial value problem

$ (x-y) u_{x} + (y-x-u) u_{y} = u $ with the initial condition $u(x,0) = 1$ satisfies

1. $ u^2(x-y+u) + (y-x-u) = 0$
2. $ u^2(x+y+u) + (y-x-u) = 0$
3. $ u^2(x-y+u) - (x+y+u) = 0$
4. $ u^2(y-x+u) + (x+y-u) = 0$

This is what I am able to do

The characteristic equations are :

$\frac{dx}{x-y} = \frac{dy}{y-x-u} = \frac{du}{u} $

From this we get

$dx + dy + du = 0$

Integrating we get a characteristic curve

$x+y+u =C_{1}$

I am unable to get a second characteristic curve. Please help.

The correct answer is 2.

Thanks in advance!

Answer 1

Second characteristic curve :

$\frac{dx}{x-y} = \frac{dy}{y-x-u} = \frac{du}{u} =\frac{dx-dy+du}{(x-y)-(y-x-u)+u} = \frac{d(x-y+u)}{2(x-y+u)} = \frac{du}{u} $

$\frac{1}{2}\ln|x-y+u|-\ln|u|=$constant $\quad\to\quad \frac{x-y+u}{u^2}=C_2$

The implicit solution is $\quad x+y+u=F\left(\frac{x-y+u}{u^2}\right) \quad $any differentiable function $F$.

The condition $u(x,0)=1$ determines the function $F(x+1)=x+1 \quad\to\quad F(X)=X$.

$F\left(\frac{x-y+u}{u^2}\right)=\frac{x-y+u}{u^2}$

$$x+y+u=\frac{x-y+u}{u^2}$$