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
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
$\displaystyle \dfrac{dx}{2y} = dy = \dfrac{du}{1-5u}$
Which you pick pairs. You chose
$\displaystyle \dfrac{dx}{2y} = dy$
and
$\displaystyle \dfrac{dx}{2y} = \dfrac{du}{1-5u}$.
Why would you not choose the second pair as
$\displaystyle dy = \dfrac{du}{1-5u}$? It so much easier and the y is not needed.
For your second PDE you have in one of your lines
$\displaystyle u = x + \sqrt{-x^2-2f(-x^2/2-2y^2)}$
Where did this come from?
I guess I overlooked this simpler option. If we use this option, I calculate
$\displaystyle \displaystyle y=-\frac{1}{5}ln(1-5u)+f(x-y^2)$. rearranging for u I get
$\displaystyle \displaystyle u(x,y)=\frac{1}{5}(-e^{-5(y-f(x-y^2))}+1)$ which i believe is the same as attachment when the missing 1/2 is factored in.
Ok, I believe I have it. Silly mistakes as usual. It should just be $\displaystyle -x^2-2(-x^2/2-2y^2)$ 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
$\displaystyle u(x,y) = x+2y+x^2+4y^2$, this satisfies the original DE and IC's
Thanks Danny!