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Math Help - I lied, one more (first-order DE). :)

  1. #1
    Junior Member cinder's Avatar
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    I lied, one more (first-order DE). :)

    x\frac{dy}{dx}+y=\sin(x), x>0, initial condition: y(\frac{\pi}{2})=1

    What I have so far...

    \frac{dy}{dx}+\frac{1}{x}y=\frac{\sin(x)}{x}<br />
\Rightarrow \int \frac{1}{x}dx=\ln|x| \Rightarrow e^{\ln|x|}=x<br />

    I'm not sure if I did the previous part right. And after that I'm unsure how to proceed.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by cinder View Post
    x\frac{dy}{dx}+y=\sin(x), x>0, initial condition: y(\frac{\pi}{2})=1

    What I have so far...

    \frac{dy}{dx}+\frac{1}{x}y=\frac{\sin(x)}{x}<br />
\Rightarrow \int \frac{1}{x}dx=\ln|x| \Rightarrow e^{\ln|x|}=x<br />

    I'm not sure if I did the previous part right. And after that I'm unsure how to proceed.
    x\frac{dy}{dx}+y=\sin(x)

    may be written as:

     \frac{d}{dx} (x~y) =\sin(x)

    so:

    x~y = -\cos(x) +C

    and the rest should be simple.

    RonL
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  3. #3
    Eater of Worlds
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    You're OK so far. You have your integrating factor.

    \frac{d}{dx}(xy)=sin(x)

    Integrate:

    \int\frac{d}{dx}(xy)=\int{sin(x)}dx

    xy=-cos(x)+C

    y=\frac{-cos(x)}{x}+\frac{C}{x}

    IC is y(\frac{\pi}{2})=1

    Using this and solving for C=\frac{\pi}{2}

    Therefore, you have:

    y=\frac{\pi}{2x}-\frac{cos(x)}{x}
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  4. #4
    Junior Member cinder's Avatar
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    I guess I'm not seeing how it can be rewritten as \frac{d}{dx}(xy)=sin(x).
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by cinder View Post
    I guess I'm not seeing how it can be rewritten as \frac{d}{dx}(xy)=sin(x).
    Think of it this way:
    \frac{d}{dx}(xy) = 1 \cdot y + x \frac{dy}{dx}
    by the product rule.

    So
    \frac{d}{dx}(xy) = y + x \frac{dy}{dx} = sin(x)
    which is your initial equation.

    -Dan
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