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Thread: Semi Linear PDE Sample Question part 2

  1. #1
    Senior Member bugatti79's Avatar
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    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.

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

    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
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  2. #2
    MHF Contributor
    Jester's Avatar
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    Re: Semi Linear PDE Sample Question part 2

    Quote Originally Posted by bugatti79 View Post
    $\displaystyle \displaystyle ln u = -k sin y -x+f(k)= -(-x-cos y)sin y-x+f(-x-cosy)$
    I believe the $\displaystyle -x$ should be a $\displaystyle - y$. I might also have $\displaystyle f(x+\cos y)$ instead of what you have. Because of this, your final answer will change.
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