# Semi Linear PDE Sample Question part 2

• August 4th 2011, 01:42 PM
bugatti79
Semi Linear PDE Sample Question part 2
Folks,

Could anyone check my work, thanks.

Find the general and particular solution and show u(0,0)=e.

$sin y u_x+u_y=(x cos y- sin^2 y) u$

Given IC is

$\displaystyle ln u(x, \frac{\pi}{2})=x^2+x-\frac{\pi}{2}$ for $-1 \le x \le 3$

$\displaystyle \frac{dy}{dx}=\frac{1}{sin y} \implies k = -x - cos y$

$\displaystyle \frac{du}{dy}=\frac{(x cos y- sin^2 y)u}{1}$ $\implies ln u = \int ((-k -cos y)cos y - sin^2 y)dy$

$\displaystyle ln u = -k sin y -x+f(k)= -(-x-cos y)sin y-x+f(-x-cosy)$

The general solution is

$\displaystyle ln u=x sin y+siny cos y-x+f(-x-cos y)$

My particular solution is

$ln u(x,y) =x sin y+siny cosy+cos y+x^2+cos^2 y- \frac{\pi}{2}$

How am I doing so far?

thanks
• August 5th 2011, 06:20 AM
Jester
Re: Semi Linear PDE Sample Question part 2
Quote:

Originally Posted by bugatti79
$\displaystyle ln u = -k sin y -x+f(k)= -(-x-cos y)sin y-x+f(-x-cosy)$

I believe the $-x$ should be a $- y$. I might also have $f(x+\cos y)$ instead of what you have. Because of this, your final answer will change.