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Find the general solution of the differential equation

  1. Feb 12, 2014 #1
    1. The problem statement, all variables and given/known data

    The equation:

    [tex]\frac{dx}{dt}=\frac{t^2+1}{x+2}.[/tex]

    Where the initial value is: x(0) = -2.

    2. Relevant equations

    I believe you have to use the method of seperations of variables.

    3. The attempt at a solution

    So I multiplied both sides with x+2. Then I integrated both sides with respect to t.

    [tex](x+2)\frac{dx}{dt}=t^2+1[/tex]

    [tex]\int(x+2)\frac{dx}{dt}\,dt=\int (t^2+1)\,dt[/tex]

    [tex]\int(x+2)\,dx=\int (t^2+1)\,dt[/tex]

    [tex]\frac{x^2}{2}+2x=\frac{t^3}{3}+t+C[/tex]

    (Note: I've added the two constants of both sides into one constant.)

    Then, I multiplied everything with 2.

    [tex]x^2+4x=\frac{2}{3}t^3+2t+C'[/tex]

    Where C'=2C

    Now I'm not so sure how I should go further then this. Any help would be nice.
     
  2. jcsd
  3. Feb 12, 2014 #2
    It's correct so far -- if you have to solve for x, I think completing the square would work nicely.
     
  4. Feb 12, 2014 #3

    ehild

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    Use the initial condition to find the constant C'.

    ehild
     
  5. Feb 12, 2014 #4

    pasmith

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    [itex]x^2 + 4x = (x + 2)^2 - 4[/itex].

    You might ask "what sign goes before the square root?", and normally the answer would be "use the ODE to work out whether [itex]x'(0)[/itex] is positive or negative", but in this case the ODE tells you that [itex]x'(0)[/itex] is undefined ...
     
  6. Feb 12, 2014 #5
    Thanks allot guys I've found the answer.
     
  7. Feb 12, 2014 #6

    Ray Vickson

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    Just as a matter of interest: what is your solution? (This IVP is a bit tricky!)
     
  8. Feb 13, 2014 #7
    [tex]x = \pm\sqrt{\frac{2}{3}t^3+2t} -2[/tex]

    The constant is zero. It is tricky indeed.
     
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