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Homework Help: Differential Equation HW

  1. Jun 22, 2005 #1
    Hello, I have tried to solve the following problem but did not succeed to do so.

    [y(y^3 - x)]dx + [x (y^3 + x)]dy = 0
    I sense that the key factor here is (y^3 - x ) and (y^3 +x), but could not figure out how to lead the equation to

    dy/dx + P(x)y = Q(x) form.

    The general answer for the problem is 2xy^3 - x^2 = Cy^2.

    Once I can change the equation to dy/dx + P(x)y = Q(x) form, I can do the rest (probably anybody can...)

    Thanks for your help in advance.
  2. jcsd
  3. Jun 22, 2005 #2
    This is of the form f(xy)ydx +F(xy)xdy=0
    here the Integrating factor is ::
    [tex] \frac{1}{xy[f(xy)-F(xy)]}[/tex]
    and general Integral equation is ::
    [tex]\int \frac{f(xy)+F(xy)}{f(xy)-F(xy)} \frac{d(xy)}{xy} + log\frac{x}{y} = c[/tex]
  4. Jun 22, 2005 #3
    that looks awfully a lot like exact DE: you'll have to get partial derivatives and the whole nine yards. Try that.
  5. Jun 22, 2005 #4
    Well , there is an alternative if you use differentials
    d(xy)= ydx +xdy
    [tex]d(\frac{y}{x}) = \frac{xdy-ydx}{x^2}[/tex]

    Here u can rearrange ur diff eqn to

    :: y3 d(xy) + x3d(y/x)=0
    divide by y3 u will get the required answer after integrating
  6. Jun 22, 2005 #5
    Exact D.E.

    I believe it is an exact DE since I have just learned that part. But when I did
    \partial M (x, y) / \partial y = 4y^3 - x and
    \partial N (x, y) / \partial x = y^3 + 2x so they are not the same.

    But I could not find integrating facutor to make their answers equal. What should I do now?
  7. Jun 22, 2005 #6
    I will try that

    Thank you "himanshu121". I will try that for now to see if I can understand that formula.

    Well, I couldn't get it.

    I have just started this differential equation class (independent). I thought I understood them well. However, when it comes to solve problems, I am experiencing a hard time. For example, this equation, I could not see why it Have some suggestion to improve my understanding?
    Last edited: Jun 22, 2005
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