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Math Help - [SOLVED] Differentiation

  1. #1
    Member ronaldo_07's Avatar
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    [SOLVED] Differentiation

    Find the general solution of the second-order inhomogeneous differential equation

    y''+3y'+2y= e^{2x}cos(x)


    So far I have:

     y(x)=Ae^{2x}sin(3x) +Be^{2x}cos(3x)

    \frac{dy}{dx}= \frac{1}{5}(e^{2x}cos(x) +3e^{2x} sin(x))

    \frac{d^{2}y}{dx^{2}}= \frac{1}{25}(-e^{2x}cos(x) +7e^{2x} sin(x))

    After this I substituted \frac{dy}{dx}   and    \frac{d^{2}y}{dx^{2}} in to the origional equation. I'm not sure if what im doing is correct. And after this step I have no Idea what to do. Please Help, Thanks
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  2. #2
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    Quote Originally Posted by ronaldo_07 View Post
    Find the general solution of the second-order inhomogeneous differential equation

    y''+3y'+2y= e^{2x}cos(x)


    So far I have:

     y(x)=Ae^{2x}sin(3x) +Be^{2x}cos(3x)

    \frac{dy}{dx}= \frac{1}{5}(e^{2x}cos(x) +3e^{2x} sin(x))

    \frac{d^{2}y}{dx^{2}}= \frac{1}{25}(-e^{2x}cos(x) +7e^{2x} sin(x))

    After this I substituted \frac{dy}{dx}   and    \frac{d^{2}y}{dx^{2}} in to the origional equation. I'm not sure if what im doing is correct. And after this step I have no Idea what to do. Please Help, Thanks
    First of all use the Auxiliary equation to get the complimentary function:

     \lambda^2 + 3\lambda + 2 = 0

     (\lambda + 2)(\lambda+1) = 0

    Hence  \lambda = -2 \text{ or } \lambda = -1

    So your CF is :

     y_c(x) = Ae^{-2x} + Be^{-x}

    Then consider the homogenous part of the equation which you have done.

    But why did your constants A and B disappear when you differentiated them?

    Also, you have to use the product rule:

     y(x)=Ce^{2x}sin(3x) +De^{2x}cos(3x)

    \frac{dy}{dx}= \frac{d}{dx}\bigg(Ce^{2x}sin(3x) +De^{2x}cos(3x)\bigg)

    = C\frac{d}{dx}\bigg(e^{2x}sin(3x)\bigg) +D\frac{d}{dx}\bigg(e^{2x}cos(3x)\bigg)

    = C\bigg(e^{2x}\frac{d}{dx}\big(sin(3x)\big) + \sin(3x)\frac{d}{dx}\big(e^{2x}\big)\bigg) +D\bigg(e^{2x}\frac{d}{dx}\big(cos(3x)) + \cos(3x)\frac{d}{dx}\big(e^{2x}\big)\bigg)
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  3. #3
    Member ronaldo_07's Avatar
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    Quote Originally Posted by ronaldo_07 View Post
    Find the general solution of the second-order inhomogeneous differential equation

    y''+3y'+2y= e^{2x}cos(x)


    So far I have:

     y(x)=Ae^{2x}sin(3x) +Be^{2x}cos(3x) typo should be x not 3x

    \frac{dy}{dx}= \frac{1}{5}(e^{2x}cos(x) +3e^{2x} sin(x))

    \frac{d^{2}y}{dx^{2}}= \frac{1}{25}(-e^{2x}cos(x) +7e^{2x} sin(x))

    After this I substituted \frac{dy}{dx}   and    \frac{d^{2}y}{dx^{2}} in to the origional equation. I'm not sure if what im doing is correct. And after this step I have no Idea what to do. Please Help, Thanks
    typo, and i missed out the constants by mistake
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  4. #4
    Member ronaldo_07's Avatar
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    After differentiating and substitution in to the origional equation what is the next step?
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