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Homework Help: Cant solve this differential equation

  1. Feb 26, 2007 #1
    1. The problem statement, all variables and given/known data
    a. find the general solution to this differential equation
    dy/dx = x(y-1)^2
    b. find the particular solution to the given initial condition f(0) = 1
    c. use the solution found in b to find the range of f

    2. Relevant equations
    none really

    3. The attempt at a solution
    this question seemed simple, but i cant really get the right answer.
    here is my attempt

    *i use different letters after manipulating constants, because they are still constants after adding/subtracting/dividing/multiplying

    dy/dx = x(y-1)^2

    *separation of variables*

    x dx = (y-1)^-2 dy

    *integrate both sides*

    x^2/2 + C1= -(y-1)^-1 + C2

    x^2/2 = -(y-1)^-1 + K

    solve for y

    (x^2/2 = -1/(y-1) + K) * 2

    x^2 = -2/(y-1) + L

    x^2 = -2(y-1)^-1 + L

    (x^2 + H = -2(y-1)^-1 ) * -1/2
    x^2/2 + G = (y-1)^-1
    raise everything to the negative 1

    (x^2/2 +G)^-1 = y-1

    x^2/2 +G = (x^2 + 2G)/2

    ((x^2 + 2G)/2)^-1 = 2/(x^2 + 2G)

    2/(x^2 + 2G) + 1 = y

    but i can tell already this is the wrong answer, and with a wrong general solution, i cant do the other parts of the problem.
  2. jcsd
  3. Feb 26, 2007 #2


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    Homework Helper

    There's a sign error on the first term in your solution.
  4. Feb 26, 2007 #3


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    You solved the given differential equation [itex]\frac{dy}{dx}=x(y-1)^2[/itex] and obtained the correct general solution (except you dropped a - sign on the [itex]x^2[/itex] term after you multiplied by -1/2) [tex]y_g=\frac{2}{2G-x^2}+1[/tex] for which [tex]\frac{dy_g}{dx}=\frac{4x}{(2G-x^2)^2}=x(y_g-1)^2[/tex] and is then indeed a valid solution to the given equation, however the initial condition that f(0)=1 is something of a problem since when we solved the equation we had divided both sides by (y-1)^2 for which we need to assume that [itex]y\neq 1[/itex] since that would involve division by zero. But maybe we can ignore that and solve it anyways... If we set x=0 and y=1 in the general solution [tex]y_g=\frac{2}{2G-x^2}+1[/tex] we get [tex]1=\frac{1}{G}+1\Rightarrow \frac{1}{G}=0[/tex] which only holds if we let [itex]G\to\pm\infty[/itex] and this gives the particular solution [tex]y_p=\lim_{G\to\pm\infty}\left(\frac{2}{2G-x^2}+1\right)=0+1=1[/tex], which is the equilibrium solution.
    Last edited: Feb 26, 2007
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