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The following PDE: $(x-y)\;\frac{\partial u}{\partial x} + (y-x-u)\;\frac{\partial u}{\partial y} = u$ with $u(x,0) = 1$

satisfies:

a) $u^2(x - y +u) + (y-x-u) = 0$

b) $u^2(x + y +u) + (y-x-u) = 0$

c) $u^2(x - y +u) + (y+x+u) = 0$

d) $u^2(x - y +u) + (y+x-u) = 0$

My attempt:

I tried to solve the following ODE

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

First I got $d(x+y +u) = 0$ from where I deduced that $x + y + u = c_1$.

Secondly I substituted $u$ in $y-x-u$ to get $y-x+x+y-c_1 = 2y - c_1$ and the solved

$\frac{dy}{2y - c_1} = \frac{du}{u}$ to get $\frac{\sqrt{y - (c_1/2)}}{u} = c_2$. Hence resubstituting $c_1$, we have $$2c_2^2u^2 = y - x -u.$$

The intial value is not helping me because it gives $c_2^2 = -(x+1)/2$ which is not a constant. Also my answer is nowhere close to the options provided?

henceproved
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1 Answers1

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In fact neither $c_1$ nor $c_2$ are constants. Better said, choosing values for them defines a particular characteristic curve. The boundary conditions (in this case $u(x,0)=1$) give the initial conditions for each characteristic curve. What you are calculated is the equations that along the curve $(x,0)$ (the $y$ axis) the constants $c_1$ and $c_2$ must satisfy (the second one is obtained by the same procedure as th first):

$c_2^2 = -(x+1)/2$

$x+1=c_1$

For both equations to be true they must to be related this way: $2c_2^2=-c_1$ (substituting $x+1$ in the first by the value from the second)

But the constants have to be related in general, so is,

$\dfrac{y - x -u}{u^2}=2c_2^2=-c_1=-(x+y+u)$ or

$u^2(x + y +u) + (y-x-u) = 0$

Rafa Budría
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