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Math Help - Find the general solution!

  1. #1
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    Find the general solution!

    I have got

    y'+y = sin x.

    How do I start this question?

    Help please!.

    THank you.
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  2. #2
    Forum Admin topsquark's Avatar
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    Re: Find the general solution!

    Quote Originally Posted by tjsdndnjs View Post
    I have got

    y'+y = sin x.

    How do I start this question?

    Help please!.

    THank you.
    The solution will be of the form y = y_0 + y_p.

    To get y_h (the homogeneous solution) set y' + y = 0. To get the particular solution I'd use the method of undetermined coefficients.

    -Dan
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  3. #3
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    Re: Find the general solution!

    Quote Originally Posted by topsquark View Post
    The solution will be of the form y = y_0 + y_p.

    To get y_h (the homogeneous solution) set y' + y = 0. To get the particular solution I'd use the method of undetermined coefficients.

    -Dan
    Really? It's not a second order linear DE, it's first order linear... The integrating factor is \displaystyle \begin{align*} e^{\int{1\,dx}} = e^x \end{align*}, so we have

    \displaystyle \begin{align*} \frac{dy}{dx} + y &= \sin{x} \\ e^{x}\,\frac{dy}{dx} + e^{x}\,y &= e^x\,\sin{x} \\ \frac{d}{dx}\left( e^x\, y \right) &= e^x\,\sin{x} \\ e^x\,y &= \int{ e^x\, \sin{x} \, dx} \\ y &= e^{-x} \int{ e^x \, \sin{x} \, dx} \end{align*}

    Now to solve that ugly integral, use integration by parts...

    \displaystyle \begin{align*} I &= \int{e^x \, \sin{x}\, dx} \\ I &= e^x\,\sin{x} - \int{ e^x \, \cos{x} \, dx } \\ I &= e^x\, \sin{x} - \left( e^x\,\cos{x} - \int{ -e^x\, \sin{x} \, dx }  \right) \\ I &= e^x \, \sin{x} - e^x\, \cos{x} - \int{e^x\, \sin{x} \, dx} \\ I &= e^x\,\sin{x} - e^x\,\cos{x} - I \\ 2I &= e^x\,\sin{x} - e^x\,\cos{x} \\ I &= \frac{1}{2} \, e^x\left( \sin{x} - \cos{x} \right) \end{align*}

    So the solution to your DE is \displaystyle \begin{align*} y = e^{-x} \left[ \frac{1}{2} \, e^x \left( \sin{x} - \cos{x} \right) \right] + C = \frac{1}{2} \left( \sin{x} - \cos{x} \right) + C \end{align*}
    Thanks from topsquark
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  4. #4
    Forum Admin topsquark's Avatar
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    Re: Find the general solution!

    Quote Originally Posted by Prove It View Post
    Really? It's not a second order linear DE, it's first order linear...
    (shrugs) It's my favorite method. I've always found the integrating factor method to be somewhat clumsy. But that's my own opinion.

    -Dan
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