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Math Help - How to complete this first order linear diff eq.

  1. #1
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    How to complete this first order linear diff eq.

    Hi, I have to solve the this differential equation:

    y' - y = cos x

    i've found the integrating factor and gotten to the part:
    int: d[ye^-x] = int: [(e^-x) * (cosx) dx]

    and I'm not sure how to integrate [(e^-x) * (cosx) dx].

    can someone help me? Thanks.
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  2. #2
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     \int e^{-x} \cos(x) ~dx= \frac{1}{2}e^{-x}(\sin(x)-\cos(x))+c
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  3. #3
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    Quote Originally Posted by PandaNomium View Post
    Hi, I have to solve the this differential equation:

    y' - y = cos x

    i've found the integrating factor and gotten to the part:
    int: d[ye^-x] = int: [(e^-x) * (cosx) dx]

    and I'm not sure how to integrate [(e^-x) * (cosx) dx].

    can someone help me? Thanks.
    \begin{gathered}<br />
  y' - y = \cos x \hfill \\<br />
  y'{e^{ - x}} - y{e^{ - x}} = \cos x{e^{ - x}} \hfill \\<br />
  y'{e^{ - x}} + y{\left( {{e^{ - x}}} \right)^\prime } = \cos x{e^{ - x}} \hfill \\ <br />
\end{gathered}

    \begin{gathered}<br />
  {\left( {y{e^{ - x}}} \right)^\prime } = \cos x{e^{ - x}} \hfill \\<br />
  y{e^{ - x}} = \int {\cos x\,{e^{ - x}}\,dx}  \hfill \\ <br />
\end{gathered}
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  4. #4
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    thanks, i guess i should rephrase my question. I got that answer using the online integrator, but i'm not sure why it is the answer.

    What properties of integration are being used?
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  5. #5
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    Quote Originally Posted by DeMath View Post
    Try to multiply both sides of the equation by e^{-x}, ie
    The poster has already done this.
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  6. #6
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    Quote Originally Posted by PandaNomium View Post
    thanks, i guess i should rephrase my question. I got that answer using the online integrator, but i'm not sure why it is the answer.

    What properties of integration are being used?
    By parts (two times)

    I = \int {\cos x{e^{ - x}}dx}  =  - {e^{ - x}}\cos x - \int {{e^{ - x}}\sin xdx}  =

    =  - {e^{ - x}}\cos x + {e^{ - x}}\sin x - \underbrace {\int {{e^{ - x}}} \cos xdx}_I \Leftrightarrow

    \Leftrightarrow 2I = {e^{ - x}}\left( {\sin x - \cos x} \right) + C \Leftrightarrow I = \frac{{{e^{ - x}}}}<br />
{2}\left( {\sin x - \cos x} \right) + C.
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  7. #7
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    i appreciate what you've done, thanks. I've never learned integration by parts, so please bear with me.

    I understand every point until the equation is multiplied by 1/2.
    Where does the 2I come in?
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  8. #8
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    Quote Originally Posted by PandaNomium View Post
    i appreciate what you've done, thanks. I've never learned integration by parts, so please bear with me.

    I understand every point until the equation is multiplied by 1/2.
    Where does the 2I come in?
    Look at the last term of the second last line. I is the unknown (integral) that you're trying to find by making it the subject.
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