Results 1 to 5 of 5

Math Help - Differential Equations

  1. #1
    Junior Member
    Joined
    Oct 2007
    Posts
    55

    Differential Equations

    The question states to find the solution of the following -

    <br />
\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 2e^x<br />

    Ive got quite far, but when trying to find the value of 'C' i keep getting this:

    C-2C+C=2

    Which gives no value for C, any help will be appreciated, as i really want to finish this question..
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,751
    Thanks
    652
    Hello, brd_7!

    I don't know what "C" you're referring to . . .


    Find the solution of: . <br />
\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y \:= \;2e^x

    The characteristic equation is: . m^2-2m + 1\:=\:0\quad\Rightarrow\quad (m-1)^2\:=\:0

    . . which has roots: . m \;=\;1,\,1

    The homogeneous solution is: . y \;=\;C_1e^x + C_2xe^x


    Then the particular solution is of the form: . y_p \;=\;Ax^2e^x

    . . and (by your favorite method) we find that: . y_p \:=\:x^2e^x


    Therefore, the solution is: . \boxed{y \;=\;C_1e^x + C_2xe^x + x^2e^x}


    Without further information, we cannot evaluate the C's.

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2007
    Posts
    55
    Why do you use:
     y_p \;=\;Ax^2e^x
    And not:
    y_p \;=\;Ae^x
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by brd_7 View Post
    Why do you use:
     y_p \;=\;Ax^2e^x
    And not:
    y_p \;=\;Ae^x
    Because Ae^x is part of the homogenous solution (try using it and solving for A. Uh oh ....). So you might try the next simplest, which is Axe^x but uh oh ...... So try the next simplest, which is .....
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2007
    Posts
    55
    Ah right, ok, because i used Ae^x and it didnt work, which is where i got suck originally, i never knew you could do that, now i do, im very thankful!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. 'Differential' in differential equations
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: October 5th 2010, 10:20 AM
  2. Replies: 2
    Last Post: May 18th 2009, 03:49 AM
  3. Differential Equations
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 12th 2009, 05:44 PM
  4. Replies: 5
    Last Post: July 16th 2007, 04:55 AM
  5. Replies: 3
    Last Post: July 9th 2007, 05:30 PM

Search Tags


/mathhelpforum @mathhelpforum