Folks,

Could anyone check my work, thanks.

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

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

Given IC is

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

Answer:

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

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

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

The general solution is

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

My particular solution is

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

How am I doing so far?

thanks