Results 1 to 8 of 8

Math Help - 2nd order nonhomo DE

  1. #1
    Member zangestu888's Avatar
    Joined
    Jan 2009
    Posts
    91

    2nd order nonhomo DE

    Am not sure at all how to do this, all i know is the intial guess the DE is

    y''+ 5y'=30x^2

    my intial guess is ax^2 +bx + c

    the deritives are y'=2ax + b and y''=2a

    then i get (2a) + 5(2ax+b)=30x^2 or 2a + 10ax + 5b=30x^2
    10ax + 2a + 5b=30x^2
    if i equate them dont i get a=b=0? or am i missing something any help appreciated thanks in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Sampras's Avatar
    Joined
    May 2009
    Posts
    301
    Quote Originally Posted by zangestu888 View Post
    Am not sure at all how to do this, all i know is the intial guess the DE is

    y''+ 5y'=30x^2

    my intial guess is ax^2 +bx + c

    the deritives are y'=2ax + b and y''=2a

    then i get (2a) + 5(2ax+b)=30x^2 or 2a + 10ax + 5b=30x^2
    10ax + 2a + 5b=30x^2
    if i equate them dont i get a=b=0? or am i missing something any help appreciated thanks in advance!
    Let  \mathcal{L} \{y(x) \} = Y(s) . Take the Laplace transform of both sides. So  \mathcal{L} \{y''+5y' \} = \mathcal{L} \{30x^2 \} . Solve for  Y(s) . Then take inverse Laplace transforms to get  y(x) .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by zangestu888 View Post
    Am not sure at all how to do this, all i know is the intial guess the DE is

    y''+ 5y'=30x^2

    my intial guess is ax^2 +bx + c

    the deritives are y'=2ax + b and y''=2a

    then i get (2a) + 5(2ax+b)=30x^2 or 2a + 10ax + 5b=30x^2
    10ax + 2a + 5b=30x^2
    if i equate them dont i get a=b=0? or am i missing something any help appreciated thanks in advance!
    My post in this thread: http://www.mathhelpforum.com/math-he...-integral.html

    suggests the modification you need to make to your particular solution.

    Quote Originally Posted by Sampras View Post
    Let  \mathcal{L} \{y(x) \} = Y(s) . Take the Laplace transform of both sides. So  \mathcal{L} \{y''+5y' \} = \mathcal{L} \{30x^2 \} . Solve for  Y(s) . Then take inverse Laplace transforms to get  y(x) .
    From the work posted by the OP I think it's clear that this technique is probably beyond what s/he has learned so far.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member zangestu888's Avatar
    Joined
    Jan 2009
    Posts
    91
    thanks for ur help but we havent startd this laplace transformation, were doing undertermined coeefcients now, any other way? are my coeefficients correct.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,402
    Thanks
    1327
    The characteristic equation for this r^2+ 5r= 0 which has roots 0 and -5 so the homogenous equation has solutions C_1e^{-5x} and C_2e^{-x}= C_2, a constant. Since 30x^2= 30x^2e^{0x} is of that "type", you need to multiply by "x". Try y= x(ax^2 +bx + c)= ax^3+ bx^2+ cx
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by zangestu888 View Post
    thanks for ur help but we havent startd this laplace transformation, were doing undertermined coeefcients now, any other way? are my coeefficients correct.
    Did you read my post in the thread whose link I gave you?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member zangestu888's Avatar
    Joined
    Jan 2009
    Posts
    91
    o noo thanks! il look at it now!! thanks!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member zangestu888's Avatar
    Joined
    Jan 2009
    Posts
    91
    okay ithink i got it am gonna redo it since e^0 is actually in the homogenous side of the DE i must do a multiple of x k thanks am gonna work on it now, a question i can use variation of paramters?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 7
    Last Post: December 17th 2010, 10:25 AM
  2. [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
  3. 2nd order nonhomo equation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 3rd 2009, 08:40 AM
  4. Replies: 2
    Last Post: February 23rd 2009, 05:54 AM
  5. Replies: 2
    Last Post: November 25th 2008, 09:29 PM

Search Tags


/mathhelpforum @mathhelpforum