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Thread: Differential Equation

  1. #1
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    Differential Equation

    Could someone just point me in the right direction for a general solution to the following question?

    $\displaystyle \frac{d^2y}{dx^2} - 2\frac{d^2y}{dx^2} + 10y = x^2e^{2x}$

    Many thanks.
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  2. #2
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    Quote Originally Posted by pikminman View Post
    Could someone just point me in the right direction for a general solution to the following question?

    $\displaystyle \frac{d^2y}{dx^2} - 2\frac{d^2y}{dx^2} + 10y = x^2e^{2x}$

    Many thanks.
    The general solution of this is of the form $\displaystyle y(x)=g(x)+h(x)$, where $\displaystyle g(x)$ is the general solution of the homogeneous equation:

    $\displaystyle \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 10y = 0$

    and $\displaystyle h(x)$ is a particular solution of the original equation. I would try a solution of the form $\displaystyle h(x)=P(x)e^{2x}$ where $\displaystyle P(x)$ is a quadratic in $\displaystyle x$

    CB
    Last edited by CaptainBlack; Mar 17th 2010 at 11:49 PM.
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    The general solution of this is of the form $\displaystyle y(x)=g(x)+h(x)$, where $\displaystyle g(x)$ is the general solution of the homogeneous equation:

    $\displaystyle \frac{d^2y}{dx^2} - 2\frac{d^2y}{dx^2} + 10y = 0$

    and $\displaystyle h(x)$ is a particular solution of the original equation. I would try a solution of the form $\displaystyle h(x)=P(x)e^{2x}$ where $\displaystyle P(x)$ is a quadratic in $\displaystyle x$

    CB
    I am not that good at maths. Could you explain in a slightly more babyish fashion please?
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  4. #4
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    Quote Originally Posted by pikminman View Post
    I am not that good at maths. Could you explain in a slightly more babyish fashion please?
    I suggest you read this tutorial - http://www.mathhelpforum.com/math-he...-tutorial.html

    In particular, the part about "Method of Undetermined Coefficients".

    Do you know how to solve for the general solution for this?

    $\displaystyle \frac{d^2y}{dx^2} - 2\frac{d^2y}{dx^2} + 10y = 0$
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  5. #5
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    Quote Originally Posted by pikminman View Post
    I am not that good at maths. Could you explain in a slightly more babyish fashion please?
    Your aux equation would be

    $\displaystyle r^2 -2r^2 + 10 = 0 $

    $\displaystyle -r^2 = -10$

    $\displaystyle r^2 = 10$

    $\displaystyle r = ^+_- \sqrt{10}$

    giving you the solution of

    $\displaystyle c_{1}e^{-\sqrt{10}x} + c_{2}e^{\sqrt{10}x}$

    do you know how to do the next step?
    Last edited by 11rdc11; Mar 17th 2010 at 11:07 PM.
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  6. #6
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    Quote Originally Posted by 11rdc11 View Post
    Your aux equation would be

    $\displaystyle r^2 -2r^2 + 10 = 0 $

    $\displaystyle -r^2 = -10$

    $\displaystyle r^2 = 10$

    $\displaystyle r = ^+_- \sqrt{10}$

    giving you the solution of

    $\displaystyle c_{1}e^{-\sqrt{10}t} + c_{2}e^{\sqrt{10}t}$

    do you know how to do the next step?
    Not really...

    Oh! Yes, I meant to say $\displaystyle \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 10y = 0$ not $\displaystyle \frac{d^2y}{dx^2} - 2\frac{d^2y}{dx^2} + 10y = 0$ Sorry...
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  7. #7
    Super Member 11rdc11's Avatar
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    $\displaystyle y_{p} = (ax^2+bx + c)e^{2x}$



    $\displaystyle y'_{p} =(2ax+b)(e^{2x}) +2e^{2x}(ax^2+bx+c)= e^{2x}(2ax+b +2(ax^2 +bx+c)) $

    $\displaystyle = e^{2x}(2ax + b + 2ax^2 +2bx +2c) = e^{2x}(2ax^2 +(2a+2b)x +b +2c)$



    $\displaystyle y''_{p} = 2e^{2x}(2ax^2 +(2a+2b)x +b +2c) +e^{2x}(4ax +2a +2b)$

    $\displaystyle = e^{2x}(4ax^2 +(4a+4b)x +2b +4c +4ax +2a +2b) = e^{2x}(4ax^2 +(8a+4b)x +4b +4c +2a)$

    plugging it all back into the original equation gives

    $\displaystyle e^{2x}(4ax^2 +(8a+4b)x +4b +4c +2a) - 2(e^{2x}(2ax^2 +(2a+2b)x +b +2c))$+ $\displaystyle 10(e^{2x}(ax^2+bx + c)) = x^2e^{2x}$

    which gives you

    $\displaystyle 10a = 1$

    $\displaystyle 4a+10b = 0$

    $\displaystyle 2a+2b +10c =0$

    now solve for a, b, and c and you will have your general solution. I amy have made an alegbra mistake but to sleepy to check right now
    Last edited by 11rdc11; Mar 17th 2010 at 11:43 PM.
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  8. #8
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by pikminman View Post
    Not really...

    Oh! Yes, I meant to say $\displaystyle \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 10y = 0$ not $\displaystyle \frac{d^2y}{dx^2} - 2\frac{d^2y}{dx^2} + 10y = 0$ Sorry...
    Same approach

    $\displaystyle r^2 - 2r + 10 = 0$

    solve the quadratic equation.

    $\displaystyle r^2 -2r = -10$

    $\displaystyle (r-1)^2 = -9$

    $\displaystyle r-1 = \pm 3i$

    $\displaystyle r = 1 \pm 3i$

    so you have completed the homo part of the solution

    $\displaystyle c_{1}e^x\cos{3x} +c_{2}e^x\sin{3x}$
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  9. #9
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    Quote Originally Posted by 11rdc11 View Post
    Same approach

    $\displaystyle r^2 - 2r + 10 = 0$

    solve the quadratic equation.

    $\displaystyle r^2 -2r = -10$

    $\displaystyle (r-1)^2 = -9$

    $\displaystyle r-1 = \pm 3i$

    $\displaystyle r = 1 \pm 3i$

    so you have completed the homo part of the solution

    $\displaystyle c_{1}e^x\cos{3x} +c_{2}e^x\sin{3x}$
    Hey, could you tell me if the following is even nearly correct?

    It's question 1

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