# Thread: Determine the general solution of this QL PDE

1. ## Determine the general solution of this QL PDE

1)Determine the general solution of the following QL PDE

2) Use implicit differentiation to verify the solution.

$u u_x-y u_y=y$

My attempt at 1)

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

Take the second two

$\int-dy=\int du \implies u=-y+A$

Taking the first two

$\frac{dx}{(-y+A)}=\frac{dy}{-y} \implies dx=\frac{(-y+A)dy}{-y}$

Integrating gives

$x=y-A \ln(y) + f(A)$ but $f(A)=u+y$

therefore the general solution implicitly is

$x=y-A \ln(y) + f(u+y)$

How am I doing?

Assuming the baove is right, how do I attempt 2)?

Please note I have this posted in the physics help forum at

Determine the general solution of QL PDE

I will keep each forum informed of the other to ensure no one's time is wasted.

Thanks

2. ## Re: Determine the general solution of this QL PDE

Now replace $A$ with $u + y$ so

$x = y - (u+y) \ln y + f(u + y)$.

Now differentiate

$1 = -u_x \ln y + f'(u+y) u_x$

$0 = 1 - (u_y + 1) \ln y - \frac{u+y}{y} + f'(u+y)(u_y+1)$

Solve for $u_x$ and $u_y$ and sub. into $u u_x - yu_y$ and simplify.

3. ## Re: Determine the general solution of this QL PDE

Nice one Danny!! Thanks :-)