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Math Help - urgent, calculus equation

  1. #1
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    urgent, calculus equation

    Show that the equation f''(x) + 4f'(x) + 4f(x) = 0

    is satisfied if f(x) = (3x-5)e^-2x (thats e to the power of -2x, not e to the power of -2 times x)

    any help would be greatly appreciated. can't seem to get it to equal 0
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  2. #2
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    Quote Originally Posted by s0urgrapes View Post
    Show that the equation f''(x) + 4f'(x) + 4f(x) = 0

    is satisfied if f(x) = (3x-5)e^-2x (thats e to the power of -2x, not e to the power of -2 times x)

    any help would be greatly appreciated. can't seem to get it to equal 0
    The best thing would be for you to post some working and someone will spot where you're going wrong.

    However, here's a tip. A least it's what I would do...

    Don't do all the differentiation.

    f(x)=(3x-5)e^{-2x}

    f'(x)=3e^{-2x}-2f(x)

    etc..
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  3. #3
    Super Member angel.white's Avatar
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    Quote Originally Posted by s0urgrapes View Post
    Show that the equation f''(x) + 4f'(x) + 4f(x) = 0

    is satisfied if f(x) = (3x-5)e^-2x (thats e to the power of -2x, not e to the power of -2 times x)

    any help would be greatly appreciated. can't seem to get it to equal 0
    Initial equation
    f(x)=(3x-5)e^{-2x}

    f(x)=3xe^{-2x}-5e^{-2x}



    First derivative, remember to use the chain rule and the multiplication rule. The multiplication rule says that the derivative of f(x)g(x) is f'(x)g(x)+g'(x)f(x). And the chain rule in this case says the derivative of e^{-2x} is e^{-2x}\left(\frac{dy}{dx}-2x\right) which becomes -2e^{-2x}

    f\prime(x)=3\left(e^{-2x}+(-2)xe^{-2x}\right)-5(-2)e^{-2x}

    f\prime(x)= 3e^{-2x}-6xe^{-2x}+10e^{-2x}

    f\prime(x) = 13e^{-2x}-6xe^{-2x}



    Second derivative, remember the same rules.
    f\prime\prime(x)=12(-2)e^{-2x}-6\left(e^{-2x} + (-2)xe^{-2x}\right)

    f\prime\prime(x)=-24e^{-2x}-6e^{-2x}+12xe^{-2x}

    f\prime\prime(x)=12xe^{-2x}-32e^{-2x}



    Now just plug them into the equation
    f\prime\prime(x)+4f\prime(x)+4f(x)=(12xe^{-2x}-32e^{-2x})+4(13e^{-2x}-6xe^{-2x})+4(3xe^{-2x}-5e^{-2x})

    and simplify
    f\prime\prime(x)+4f\prime(x)+4f(x)=12xe^{-2x}-32e^{-2x}+52e^{-2x}-24xe^{-2x}+12xe^{-2x}-20e^{-2x}

    f\prime\prime(x)+4f\prime(x)+4f(x)=(12+12-24)xe^{-2x}+(52-32-20)e^{-2x}

    f\prime\prime(x)+4f\prime(x)+4f(x)=0
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