Results 1 to 11 of 11

Math Help - Secondorder Differential equation

  1. #1
    Junior Member
    Joined
    Oct 2008
    Posts
    74
    Awards
    1

    Secondorder Differential equation

    Aloha!

    I have given an inhomogen DE as:
    <br />
n>=1

    X_{n+2}-X_{n+1}+\frac{1}{4}X_{n} = n+4

    Trying to find the general X^h_{n} solution to the associated homogen DE and the special solution X^s_{n} which satisfies the inhomogen DE

    The general X^h_{n}:
    Characteristic eqation is
    r^2-r+\frac{1}{4}


    Which gives x = 0.5

    X_{n} = C \frac{1}{2}^n+Dn \frac{1}{2}^2 = \frac{1}{2}C+ \frac{1}{2}nD where C,D \mathbb{R}

    Correct so far?

    Any help is greatly appriciated!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,387
    Thanks
    52
    Quote Originally Posted by jokke22 View Post
    Aloha!

    I have given an inhomogen DE as:
    <br />
n>=1

    X_{n+2}-X_{n+1}+\frac{1}{4}X_{n} = n+4

    Trying to find the general X^h_{n} solution to the associated homogen DE and the special solution X^s_{n} which satisfies the inhomogen DE

    The general X^h_{n}:
    Characteristic eqation is
    r^2-r+\frac{1}{4}


    Which gives x = 0.5

    X_{n} = C \frac{1}{2}^n+Dn \frac{1}{2}^2 = \color{red}{\frac{1}{2}C+ \frac{1}{2}nD} where C,D \mathbb{R}

    Correct so far?

    Any help is greatly appriciated!
    A couple of things - first off, these are difference equations not differential equations. Second, your particluar solution is correct, i.e.

    X_{n} = C \left(\frac{1}{2}\right)^n+Dn \left(\frac{1}{2}\right)^n (I'm assuming a typo on the second power) but not the second part, in red.

    As for a particular solution, try X_n = an + b and see if you can find an a and b to recover the RHS.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2008
    Posts
    74
    Awards
    1
    Quote Originally Posted by danny arrigo View Post
    A couple of things - first off, these are difference equations not differential equations. Second, your particluar solution is correct, i.e.

    X_{n} = C \left(\frac{1}{2}\right)^n+Dn \left(\frac{1}{2}\right)^n (I'm assuming a typo on the second power) but not the second part, in red.

    As for a particular solution, try X_n = an + b and see if you can find an a and b to recover the RHS.
    Yeah, I see it now! Good of you to clear that up!
    Will do and see what I come up with!

    Yeah, I sometimes mess up the right word since its not in english and use similar but not similar in english! Thanks for the correction though!

    Appriciate the help!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Oct 2008
    Posts
    74
    Awards
    1
    A bit late answer here but If you Danny or someone else could help me out it would be great...

    This is what I've got:
    X_{n} = C \left(\frac{1}{2}\right)^n+Dn \left(\frac{1}{2}\right)^n (I'm assuming a typo on the second power) but not the second part, in red.

    X_n = an+b

    = A(n+2)+B-\frac{1}{2}(A(n+1)+B)+\frac{1}{2}(An+B)

    Which I get to be
    <br />
An+ 1.5A + B

    Solving this equation would give me? Confused since I think I've done it wrong...

    Appreciate all the help I can get!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2008
    Posts
    74
    Awards
    1
    As for a difference equation given at

    Y''-Y'+\frac{1}{4}Y = x-4

    Gives the Characteristic equation as

    r^2-r+\frac{1}{4}

    Which gives x = 0.5

    Which gives

    Y_{n} = C \left(\frac{1}{2}\right)^n+Dn \left(\frac{1}{2}\right)^n




    Which again gives by X_n = an + b

    = A(n+2)+B-\frac{1}{2}(A(n+1)+B)+\frac{1}{2}(An+B)

    = <br />
An+ 1.5A + B

    Just the same except when I'm going to find the solution which combines the general and special solution?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,387
    Thanks
    52
    Quote Originally Posted by jokke22 View Post
    As for a difference equation given at

    Y''-Y'+\frac{1}{4}Y = x\color{red}{-4}

    Gives the Characteristic equation as

    r^2-r+\frac{1}{4}

    Which gives x = 0.5

    Which gives

    Y_{n} = C \left(\frac{1}{2}\right)^n+Dn \left(\frac{1}{2}\right)^n




    Which again gives by X_n = an + b

    = A(n+2)+B - \frac{1}{2} (A(n+1)+B)+\frac{1}{2}(An+B) \;\color{blue}{*}

    = <br />
An+ 1.5A + B

    Just the same except when I'm going to find the solution which combines the general and special solution?
    Not sure of the red term - the - 4. Your earlier post had a +4. Not sure why you have the 1/2's *. Picking it up here

    a(n+2)+b -\left(a(n+1)+b\right) - \frac{1}{4}(an+b) = n+4

    expanding gives

     <br />
\frac{a n}{4} + a - \frac{3b}{4} = n+4<br />

    so comparing the n terms gives that a = 4 and the remaining terms that a - \frac{3b}{4} = 4 so b = 0 so Xp_n = 4n.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Oct 2008
    Posts
    74
    Awards
    1
    Quote Originally Posted by danny arrigo View Post
    Not sure of the red term - the - 4. Your earlier post had a +4. Not sure why you have the 1/2's *. Picking it up here

    a(n+2)+b -\left(a(n+1)+b\right) - \frac{1}{4}(an+b) = n+4

    expanding gives

     <br />
\frac{a n}{4} + a - \frac{3b}{4} = n+4<br />

    so comparing the n terms gives that a = 4 and the remaining terms that a - \frac{3b}{4} = 4 so b = 0 so Xp_n = 4n.
    Thanks for clearing things up, as you mention the -4 is there because its a different equation than the original one... Sorry for mixing things up, will post new If I can't get it right.

    As for the original
    This is what I've got:
    X_{n} = C \left(\frac{1}{2}\right)^n+Dn \left(\frac{1}{2}\right)^n (I'm assuming a typo on the second power) but not the second part, in red.

    X_n = an+b

    = A(n+2)+B-\frac{1}{2}(A(n+1)+B)+\frac{1}{2}(An+B)

    Which I get to be
    <br />
An+ 1.5A + B


    This is then wrong and should be:

     <br />
\frac{a n}{4} + a - \frac{3b}{4} = n+4<br />

    a = 4 and the remaining terms that a - \frac{3b}{4} = 4 so b = 0 so Xp_n = 4n

    Correct?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,387
    Thanks
    52
    Quote Originally Posted by jokke22 View Post
    Thanks for clearing things up, as you mention the -4 is there because its a different equation than the original one... Sorry for mixing things up, will post new If I can't get it right.

    As for the original
    This is what I've got:
    X_{n} = C \left(\frac{1}{2}\right)^n+Dn \left(\frac{1}{2}\right)^n (I'm assuming a typo on the second power) but not the second part, in red.

    X_n = an+b

    = A(n+2)+B-\frac{1}{2}(A(n+1)+B)+\frac{1}{2}(An+B)

    Which I get to be
    <br /> <i>An+ 1.5A + B</i>

    This is then wrong and should be:

     <br />
\frac{a n}{4} + a - \frac{3b}{4} = n+4<br />

    a = 4 and the remaining terms that a - \frac{3b}{4} = 4 so b = 0 so Xp_n = 4n

    Correct?
    Yep and if the problem was

    x_{n+2} + x_{n+1} - \frac{1}{4} x_n = n - 4 as you said then

     <br />
\frac{a n}{4} + a - \frac{3b}{4} = n-4<br />

    so a = 4 still but a - \frac{3b}{4} = - 4 which would give a diffferent b.

    Hope that helps.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Oct 2008
    Posts
    74
    Awards
    1
    Thanks a bunch!

    It helps a LOT!

    To verify.

    The general solution would then be:

    X_{n} = C \left(\frac{1}{2}\right)^n+Dn \left(\frac{1}{2}\right)^n +4_n

    And if you were to give these initial condition as
    x_o = 0, x_1 = 5

    Just replace X_0 and X_1 with _n and calculate?
    Last edited by jokke22; May 5th 2009 at 09:08 AM.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Oct 2008
    Posts
    74
    Awards
    1
    Quote Originally Posted by danny arrigo View Post
    Yep and if the problem was

    x_{n+2} + x_{n+1} - \frac{1}{4} x_n = n - 4 as you said then

     <br />
\frac{a n}{4} + a - \frac{3b}{4} = n-4<br />

    so a = 4 still but a - \frac{3b}{4} = - 4 which would give a diffferent b.

    Hope that helps.
    What would the b actually be in this case?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,387
    Thanks
    52
    Quote Originally Posted by jokke22 View Post
    What would the b actually be in this case?
    Just substitute a = 4 into a - \frac{3b}{4} = -4 and solve for b
    Spoiler:
    b = \frac{32}{3}
    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(2)
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: September 6th 2009, 09:54 AM
  5. Partial differential equation-wave equation - dimensional analysis
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: August 28th 2009, 12:39 PM

Search Tags


/mathhelpforum @mathhelpforum