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Math Help - Solve the Differential Equation

  1. #1
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    Solve the Differential Equation

    I have been having some trouble with the following question - i'm not sure how the substitution works and how to change back at the end.

    Any help greatly appreciated!

    Show that posing x = t +1/z changes the equation x' = x^2 - tx +1 to a linear equation. Then Solve.
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  2. #2
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    Quote Originally Posted by rmangan View Post
    I have been having some trouble with the following question - i'm not sure how the substitution works and how to change back at the end.

    Any help greatly appreciated!

    Show that posing x = t +1/z changes the equation x' = x^2 - tx +1 to a linear equation. Then Solve.
    Have you made the change of variable? What did you get for the ODE in z?

    You should have started from:

    \frac{d}{dt} \left(t+\frac{1}{z}\right) =\left(t+\frac{1}{z}\right)^2 -t\left(t+\frac{1}{z}\right)+1

    and simplified.

    CB
    Last edited by CaptainBlack; March 20th 2009 at 10:28 AM.
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  3. #3
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    The substitution is in fact x = t + 1/z. I have substituted in but came across a few difficulties. Firstly, i was unsure as to how to calculate (t + 1/z)'. I came up with t'. Is this correct?

    Assuming it is, the ODE yielded t' - (1/z)t = 1 + 1/z^2.

    I tried to solve this and ended up with t(z) = zln|z| - 1/2z.

    Does this seem correct. My major problem was how to change it back to get a solution x(t).

    THanks so much for your help.
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  4. #4
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    Quote Originally Posted by rmangan View Post
    The substitution is in fact x = t + 1/z. I have substituted in but came across a few difficulties. Firstly, i was unsure as to how to calculate (t + 1/z)'. I came up with t'. Is this correct?
    \frac{d}{dt}\left(t+\frac{1}{z}\right)=1+(-1)\frac{1}{z^2}z'

    CB
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  5. #5
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    Ok, then I got z' +tz = -1.


    I solved z' + tz = 0 first and got Ce^(-t^2/2)

    I couldn't seem to find a particular solution so I varied the constant and ended up with the general solution z(t) = g(t)/g'(t) + c, where g(t) is the integral of e^(t^2/2)

    Does this seem correct? The question asks me to express this in terms of the initial conditions (t(0), x(0) ). How do I translate it back to a solution for x(t)?

    Thanks again for all your help.
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