Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: DE question?

  1. Jan 25, 2010 #1
    A function y(t) satises the differential equation

    [tex]\frac{dy}{dx}[/tex] = y4 - 6y3 + 5y2

    What are the constant solutions of the equation?

    I have totally no idea what the question is asking can someone explain why substituting y with a constant solves it?
  2. jcsd
  3. Jan 25, 2010 #2
    When you are differentiating y(t) in terms of x, the variable t is not employed. Thus, all the terms y are constant.

    dy(t)/dx=y⁴ - 6y³ + 5y²
    is solved as
    y(t)=(y⁴ - 6y³ + 5y²)x+k

    where k is constant.

    On the other hand, if the question was
    dy/dt=y⁴ - 6y³ + 5y²

    This would be a 5th degree differential equation... maybe solvable.
  4. Jan 25, 2010 #3
    y = constant
    \frac{dy}{dx} = 0
    y⁴ - 6y³ + 5y² = 0
  5. Jan 25, 2010 #4
    solve the polynomial in the above post to get the constant solutions
  6. Jan 27, 2010 #5
    Actually I don't really understand the question. I was hoping someone can explain to me what it actually means. If they want to make y a constant then why did they differentiate y with respect to x and get this equation?
  7. Jan 27, 2010 #6
    I think payumooli interpretation is probably the most accurate. It is true that the question is poorly formed.
    If by constants they mean the zeros, than it is simple.
    If you assume that
    [tex]\ \frac{dy(t)}{dx} = 0[/tex]
    then the equation is
    [tex]y^4 - 6y^3 + 5y^2=0[/tex]

    which is solved by [tex]y^2(5/6 - y)=0[/tex] where [tex]y=0[/tex] or [tex]y=5/6[/tex]

    If you assume that there was a two-variable function y(x,t), where
    [tex]\ \frac{dy(x,t)}{dx} = y^4 - 6y^3 + 5y^2[/tex]
    the case is a bit different. Here, the constants of y(x,t) in relation to x is the right hand side of the equation, since:
    [tex]\ \frac{d(kx+c)}{dx}=k[/tex]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook