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Math Help - [SOLVED] Please find the mistake

  1. #1
    Member Altair's Avatar
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    [SOLVED] Please find the mistake

    I have to find the particular solution of the following DE.

    D^2 - 3D + 2 = sinx

    Here is how I did it.

    \frac{sinx}{D^2-3D+2}

    \frac{Im(e^{ix})}{D^2-3D+2}

    \frac{cosx + isinx}{(i)^2 -3(i) +2}

    \frac{cosx + isinx}{1-3i}

    Then after rationalizing,

    \frac{cosx-3sinx}{10}

    Where as the answer given by the book is,

    \frac{sinx + 3cosx}{10}

    Can you please tell where is my mistake ?
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  2. #2
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    Quote Originally Posted by Altair View Post
    I have to find the particular solution of the following DE.

    D^2 - 3D + 2 = sinx

    Here is how I did it.

    \frac{sinx}{D^2-3D+2}

    \frac{Im(e^{ix})}{D^2-3D+2}

    \frac{cosx + isinx}{(i)^2 -3(i) +2}

    \frac{cosx + isinx}{1-3i}

    Then after rationalizing,

    \frac{cosx-3sinx}{10}

    Where as the answer given by the book is,

    \frac{sinx + 3cosx}{10}

    Can you please tell where is my mistake ?
    You've found the real part of \frac{(\cos x + i \sin x)(1 + 3i)}{10}. The particular solution is the imaginary part, since \sin x is the imaginary part of e^{ix} .....
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  3. #3
    Member Altair's Avatar
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    Quote Originally Posted by mr fantastic View Post
    You've found the real part of \frac{(\cos x + i \sin x)(1 + 3i)}{10}. The particular solution is the imaginary part, since \sin x is the imaginary part of e^{ix} .....
    Got it. But how do I get the imaginary part?
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  4. #4
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    Quote Originally Posted by Altair View Post
    Got it. But how do I get the imaginary part?
    You're kidding me? How did you get the imaginary part if you don't know how to get the real part.

    \frac{(\cos x + i \sin x)(1 + 3i)}{10}.

    Expand the numerator. Throw away the bits that don't have an i in them. What's left (divided by 10) gives the imaginary part!!
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  5. #5
    Member Altair's Avatar
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    Quote Originally Posted by mr fantastic View Post
    You're kidding me? How did you get the imaginary part if you don't know how to get the real part.

    Expand the numerator. Throw away the bits that don't have an i in them. What's left (divided by 10) gives the imaginary part!!
    You know what our professor simply told us to "multiply the i term with the i term and the real one with real term". And failed to give the answer why. You gave it. Thanks. And I surely didn't know.
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  6. #6
    Member Altair's Avatar
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    \frac{3icosx + isinx}{10}

    How do I get rid of the i ? Just by telling that as Sin is the imaginary part so this i is just the indicator ?
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  7. #7
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    Quote Originally Posted by Altair View Post
    \frac{3icosx + isinx}{10}

    How do I get rid of the i ? Just by telling that as Sin is the imaginary part so this i is just the indicator ?
    The real part of a + ib is a.

    The imaginary part of a + ib is b.

    -Dan
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  8. #8
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    No ... you multiply the numerator as you would with normal binomials (FOIL if that's what you call it)

    \frac{(\cos x + i \sin x)(1 + 3i)}{10}

    =\frac{\cos x + 3i\cos x + i\sin x + 3i^{2}\sin x}{10}

    Simplify, combinining like terms with like terms and eventually you'll get the numerator in the form of a + bi. As pointed out, a will be your real part and bi will be the imaginary.
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