Results 1 to 8 of 8

Math Help - 2nd Order ODE

  1. #1
    Newbie
    Joined
    Jul 2011
    Posts
    8

    2nd Order ODE

    Can anyone take me through solving this 2nd Order ODE?


    y'' - 4y' + 3y = e^3^x




    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45

    Re: 2nd Order ODE

    Quote Originally Posted by remember View Post
    y'' - 4y' + 3y = e^3^x
    Just routine knowing the corresponding theory. Have you studied it?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: 2nd Order ODE

    First calculate the homogenous DE:
    y''-4y'+3y=0 and determine the general solution, afterwards you've to find a particular solution.

    Can you do that?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jul 2011
    Posts
    8

    Re: 2nd Order ODE

    I understand that i must first find a complementary function, which i worked out as:

    y(x)=Ae^x + Be^3^x - where A and B are constants.


    My problem is with the next step. I've previously done examples where the homogeneous expression is equal to a quadratic expression... what do i do in the case of an exponential?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: 2nd Order ODE

    That looks good for the general solution, now the particular solution will look like: a\cdot x \cdot e^{3x} (because of e^{3x}), you have to find a. Calculate the first derivative and second derivative of a\cdot x \cdot e^{3x} and substitute in the DE to find a.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jul 2011
    Posts
    8

    Re: 2nd Order ODE

    I have worked out that:

    y = a e^3^x x
    y' = e^3^x (3x + 1)
    y'' = 3e^3^x (3x + 2)


    When you say substitute into the DE, do you mean like this:

    3e^3^x (3x + 2) - 4(e^3^x (3x + 1) + 3(a e^3^x x) = e^3^x?

    Then expand the brackets and work out a?

    Is my final solution in the form:

    y(x) = a x e^3^x + Ae^x + Be^3^x ?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5

    Re: 2nd Order ODE

    Quote Originally Posted by remember View Post
    I have worked out that:

    y = a e^3^x x
    y' = e^3^x (3x + 1)
    y'' = 3e^3^x (3x + 2)


    When you say substitute into the DE, do you mean like this:

    3e^3^x (3x + 2) - 4(e^3^x (3x + 1) + 3(a e^3^x x) = e^3^x?

    Then expand the brackets and work out a?

    Is my final solution in the form:

    y(x) = a x e^3^x + Ae^x + Be^3^x ?
    Yes. Review the theory (method of undetermined coefficients) in your class notes or textbook.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: 2nd Order ODE

    If you calculate a:
    y=a\cdot x\cdot e^{3x}
    y'=a\left(e^{3x}+3x\cdot e^{3x}\right)=a\cdot e^{3x}+3ax \cdot e^{3x}
    y''=3a\cdot e^{3x}+3a \left(e^{3x}+3x\cdot e^{3x}\right)
    =6a \cdot e^{3x}+9ax\cdot e^{3x}

    If you substitute this in the equation you get:
    6a\cdot e^{3x}+9ax\cdot e^{3x}-4\left(a\cdot e^{3x}+3ax\cdot e^{3x}\right)+3\left(ax\cdot e^{3x}\right)=e^{3x}
    \Leftrightarrow 2a\cdot e^{3x}=e^{3x}
    \Leftrightarrow a=\frac{1}{2}

    So the solution of the ODE:
    y=A\cdot e^{x}+B\cdot e^{3x}+\frac{1}{2}\cdot x\cdot e^{3x}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Re-writing higher order spatial derivatives as lower order system
    Posted in the Differential Equations Forum
    Replies: 11
    Last Post: July 27th 2010, 08:56 AM
  2. Replies: 1
    Last Post: October 27th 2009, 04:03 AM
  3. Replies: 2
    Last Post: February 23rd 2009, 05:54 AM
  4. Replies: 2
    Last Post: November 25th 2008, 09:29 PM
  5. Replies: 4
    Last Post: August 12th 2008, 04:46 AM

Search Tags


/mathhelpforum @mathhelpforum