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Unseparable Differential Equations

  1. Feb 14, 2009 #1
    I have a question about differential equations... The equation for a general differential equation that is not separable is:
    dy/dx + P(x)y = Q(x)

    So my question is can you have a Q(x) that has both x and y variables?

    Example:
    1. The problem statement, all variables and given/known data
    xy' +y = y^2

    3. The attempt at a solution
    dy/dx + (1/x)y = (1/x)y^2

    The integrating factor V(x) = e ^[int. P(x)]
    = e ^[int. 1/x dx]
    = x

    Therefore y = [1/V(x)][int. Q(x)V(x)dx + C]
    = [1/x] [int. ???? + C]
     
  2. jcsd
  3. Feb 14, 2009 #2
    This is separable! (and non-linear, by the way) Proceed as you normally would in that case.
     
    Last edited: Feb 15, 2009
  4. Feb 15, 2009 #3
    Re:Differential Equations

    Yes, sorry as i was looking over i finally remembered its called a first order linear differential equation.. sorry :$... but i tried to solve by separating but dont seem to get the right answer.. here's what i got for a solution (with the initial condition being y(1)=-1)

    -1/y - ln|y| = ln|x| + 1

    but how would i isolate y??
     
  5. Feb 15, 2009 #4
    Re: Differential Equations

    How did you get that?

    We have [tex]\frac{1}{y(y-1)} y' = \frac{1}{x}[/tex]

    and the antiderivative of

    [tex]\frac{1}{y(y-1)} = \frac{1}{y-1} - \frac{1}{y}[/tex]

    is ?
     
  6. Feb 15, 2009 #5
    even still.. i dont see how you can isolate for y without y in the answer??
     
  7. Feb 15, 2009 #6

    rock.freak667

    User Avatar
    Homework Helper

    Why would you need to isolate y? If you were given y(1)=-1, that means that when x=1,y=-1. No need to isolate y.
     
  8. Feb 15, 2009 #7
    You didn't answer my question. Without seeing your response I can't know why you can't isolate y.
     
  9. Feb 15, 2009 #8
    Re: Differential Equations

    oh sorry...
    so since you said
    therefore having taking the antiderivative of both sides,
    the next line would be:
    1/y-1 - 1/y = ln|x| + C
    then subbing in y=-1 and x=1 to solve for C:
    1/1 +1/1 = ln|1| + C
    C = 2
    so
    1/y-1 - 1/y = ln|x| + 2 (sorry i think i might have put 1 before as C)
    so now i need to enter an answer that is y=
    and i dont know how to isolate for y??
     
  10. Feb 15, 2009 #9
    Re: Differential Equations

    You still didn't answer my question, tascja!

    The antiderivative of 1/(y-1) - 1/y is ln(y-1) - ln(y) = ln((y-1)/y) ...
     
  11. Feb 15, 2009 #10
    omg i thought you were implying that 1/(y-1) - 1/y was the antiderivative of 1/y(y-1) that's why i was confused, sorry.. again... okay so
    ln((y-1)/y)) = ln|x| +c
    (y-1)/y = e^[ln|x| +c]
    1- 1/y = e^[ln|x|]e^C *e^C is just arbitrary number
    y = 1/(1-Cx) * where with the initial conditions C=2
    y = 1/(1-2x)

    THANK YOU FOR ALL YOUR HELP!!! :D
     
  12. Feb 15, 2009 #11
    If what I wrote was unclear it is I who should be apologising! Well done!
     
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