Results 1 to 9 of 9

Math Help - Differential Equation

  1. #1
    Newbie
    Joined
    Nov 2009
    Posts
    23

    Differential Equation

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

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

    Many thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by pikminman View Post
    Could someone just point me in the right direction for a general solution to the following question?

    \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 y(x)=g(x)+h(x), where g(x) is the general solution of the homogeneous equation:

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

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

    CB
    Last edited by CaptainBlack; March 18th 2010 at 12:49 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2009
    Posts
    23
    Quote Originally Posted by CaptainBlack View Post
    The general solution of this is of the form y(x)=g(x)+h(x), where g(x) is the general solution of the homogeneous equation:

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

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

    CB
    I am not that good at maths. Could you explain in a slightly more babyish fashion please?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Jan 2010
    Posts
    354
    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?

    \frac{d^2y}{dx^2} - 2\frac{d^2y}{dx^2} + 10y = 0
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    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

    r^2 -2r^2 + 10 = 0

    -r^2 = -10

    r^2 = 10

    r = ^+_- \sqrt{10}

    giving you the solution of

    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; March 18th 2010 at 12:07 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Nov 2009
    Posts
    23
    Quote Originally Posted by 11rdc11 View Post
    Your aux equation would be

    r^2 -2r^2 + 10 = 0

    -r^2 = -10

    r^2 = 10

    r = ^+_- \sqrt{10}

    giving you the solution of

    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 \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 10y = 0 not \frac{d^2y}{dx^2} - 2\frac{d^2y}{dx^2} + 10y = 0 Sorry...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    y_{p} = (ax^2+bx + c)e^{2x}



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

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



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

    = 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

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

    which gives you

    10a = 1

    4a+10b = 0

    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; March 18th 2010 at 12:43 AM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    Quote Originally Posted by pikminman View Post
    Not really...

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

    r^2 - 2r + 10 = 0

    solve the quadratic equation.

    r^2 -2r = -10

    (r-1)^2 = -9

    r-1 = \pm 3i

    r = 1 \pm 3i

    so you have completed the homo part of the solution

    c_{1}e^x\cos{3x} +c_{2}e^x\sin{3x}
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Nov 2009
    Posts
    23
    Quote Originally Posted by 11rdc11 View Post
    Same approach

    r^2 - 2r + 10 = 0

    solve the quadratic equation.

    r^2 -2r = -10

    (r-1)^2 = -9

    r-1 = \pm 3i

    r = 1 \pm 3i

    so you have completed the homo part of the solution

    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

    Google Docs
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: May 8th 2011, 01:27 PM
  2. Replies: 1
    Last Post: April 11th 2011, 02:17 AM
  3. [SOLVED] Solve Differential equation for the original equation
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: February 21st 2011, 02:24 PM
  4. Partial differential equation-wave equation - dimensional analysis
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: August 28th 2009, 12:39 PM
  5. Replies: 13
    Last Post: May 19th 2008, 09:56 AM

Search Tags


/mathhelpforum @mathhelpforum