# Thread: Semi Linear PDE's:2y u_x+u_y=1-5u; 4yu_x-xu_y+2x-4y=0

1. ## Semi Linear PDE's:2y u_x+u_y=1-5u; 4yu_x-xu_y+2x-4y=0

Folks,

Could anyone check my work as attached?

pdf 250 part ii. Find the explicit general solution
pdf 252. Find the particular solution for the given IC.

Thanks

2. A couple of comments. For your first PDE, you have a 1/2 missing. But on a note, you want to solve the characteristic equations

$\dfrac{dx}{2y} = dy = \dfrac{du}{1-5u}$

Which you pick pairs. You chose

$\dfrac{dx}{2y} = dy$

and

$\dfrac{dx}{2y} = \dfrac{du}{1-5u}$.

Why would you not choose the second pair as

$dy = \dfrac{du}{1-5u}$? It so much easier and the y is not needed.

$u = x + \sqrt{-x^2-2f(-x^2/2-2y^2)}$

Where did this come from?

3. Originally Posted by Danny
A couple of comments. For your first PDE, you have a 1/2 missing. But on a note, you want to solve the characteristic equations

$\dfrac{dx}{2y} = dy = \dfrac{du}{1-5u}$

Which you pick pairs. You chose

$\dfrac{dx}{2y} = dy$

$\dfrac{dx}{2y} = \dfrac{du}{1-5u}$.

Why would you not choose the second pair as

$dy = \dfrac{du}{1-5u}$? It so much easier and the y is not needed.
I guess I overlooked this simpler option. If we use this option, I calculate

$\displaystyle y=-\frac{1}{5}ln(1-5u)+f(x-y^2)$. rearranging for u I get

$\displaystyle u(x,y)=\frac{1}{5}(-e^{-5(y-f(x-y^2))}+1)$ which i believe is the same as attachment when the missing 1/2 is factored in.

Originally Posted by Danny
$u = x + \sqrt{-x^2-2f(-x^2/2-2y^2)}$
Ok, I believe I have it. Silly mistakes as usual. It should just be $-x^2-2(-x^2/2-2y^2)$ under the square root and I also forgot a constant of integration which would be a function of k. Hence I get the particular solution to be
$u(x,y) = x+2y+x^2+4y^2$, this satisfies the original DE and IC's