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Thread: Secondorder Differential equation

  1. #1
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    Secondorder Differential equation

    Aloha!

    I have given an inhomogen DE as:
    $\displaystyle
    n>=1$

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

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

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


    Which gives $\displaystyle x = 0.5$

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

    Correct so far?

    Any help is greatly appriciated!
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  2. #2
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    Quote Originally Posted by jokke22 View Post
    Aloha!

    I have given an inhomogen DE as:
    $\displaystyle
    n>=1$

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

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

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


    Which gives $\displaystyle x = 0.5$

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

    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.

    $\displaystyle 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 $\displaystyle X_n = an + b$ and see if you can find an $\displaystyle a$ and $\displaystyle b$ to recover the RHS.
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  3. #3
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    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.

    $\displaystyle 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 $\displaystyle X_n = an + b$ and see if you can find an $\displaystyle a$ and $\displaystyle b$ 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!
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    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:
    $\displaystyle 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.

    $\displaystyle X_n = an+b$

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

    Which I get to be
    $\displaystyle
    An+ 1.5A + B$

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

    Appreciate all the help I can get!
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  5. #5
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    As for a difference equation given at

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

    Gives the Characteristic equation as

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

    Which gives $\displaystyle x = 0.5$

    Which gives

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




    Which again gives by $\displaystyle X_n = an + b$

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

    = $\displaystyle
    An+ 1.5A + B$

    Just the same except when I'm going to find the solution which combines the general and special solution?
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  6. #6
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    Quote Originally Posted by jokke22 View Post
    As for a difference equation given at

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

    Gives the Characteristic equation as

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

    Which gives $\displaystyle x = 0.5$

    Which gives

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




    Which again gives by $\displaystyle X_n = an + b$

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

    = $\displaystyle
    An+ 1.5A + B$

    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

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

    expanding gives

    $\displaystyle
    \frac{a n}{4} + a - \frac{3b}{4} = n+4
    $

    so comparing the n terms gives that $\displaystyle a = 4$ and the remaining terms that $\displaystyle a - \frac{3b}{4} = 4$ so $\displaystyle b = 0$ so $\displaystyle Xp_n = 4n$.
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  7. #7
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    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

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

    expanding gives

    $\displaystyle
    \frac{a n}{4} + a - \frac{3b}{4} = n+4
    $

    so comparing the n terms gives that $\displaystyle a = 4$ and the remaining terms that $\displaystyle a - \frac{3b}{4} = 4$ so $\displaystyle b = 0$ so $\displaystyle Xp_n = 4n$.
    Thanks for clearing things up, as you mention the $\displaystyle -4$ 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:
    $\displaystyle 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.

    $\displaystyle X_n = an+b$

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

    Which I get to be
    $\displaystyle
    An+ 1.5A + B$


    This is then wrong and should be:

    $\displaystyle
    \frac{a n}{4} + a - \frac{3b}{4} = n+4
    $

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

    Correct?
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  8. #8
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    Quote Originally Posted by jokke22 View Post
    Thanks for clearing things up, as you mention the $\displaystyle -4$ 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:
    $\displaystyle 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.

    $\displaystyle X_n = an+b$

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

    Which I get to be
    $\displaystyle
    An+ 1.5A + B$

    This is then wrong and should be:

    $\displaystyle
    \frac{a n}{4} + a - \frac{3b}{4} = n+4
    $

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

    Correct?
    Yep and if the problem was

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

    $\displaystyle
    \frac{a n}{4} + a - \frac{3b}{4} = n-4
    $

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

    Hope that helps.
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  9. #9
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    Thanks a bunch!

    It helps a LOT!

    To verify.

    The general solution would then be:

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

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

    Just replace $\displaystyle X_0$ and $\displaystyle X_1$ with $\displaystyle _n$ and calculate?
    Last edited by jokke22; May 5th 2009 at 08:08 AM.
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  10. #10
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    Quote Originally Posted by danny arrigo View Post
    Yep and if the problem was

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

    $\displaystyle
    \frac{a n}{4} + a - \frac{3b}{4} = n-4
    $

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

    Hope that helps.
    What would the $\displaystyle b$ actually be in this case?
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  11. #11
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    Quote Originally Posted by jokke22 View Post
    What would the $\displaystyle b$ actually be in this case?
    Just substitute $\displaystyle a = 4$ into $\displaystyle a - \frac{3b}{4} = -4$ and solve for b
    Spoiler:
    $\displaystyle b = \frac{32}{3}$
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