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Homework Help: Differential equations - exact equations

  1. Feb 24, 2007 #1
    Find the solution to the initial value problem.
    2ty^3 + 3t^2y^2 dy/dt = 0 , y(1) = 1

    I found out whether the equation was exact and it was, and i continued as follows. NOTE: the answer is y(t) = t^(-4/3) i just don't know how to get it. or maybe my solution is just in another form.

    im going to use f instead of that greek symbol i do not how to type out here.

    M(t,y) = 2ty^3
    N(t,y) = 3t^2y^2

    don't know why, just copied from textbook.
    M(t,y) = df(t,y)/dt = 2ty^3
    N(t,y) = df(t,y)/dy = 3t^2y^2

    f(t,y) = integ(M(t,y)dt) + h(y)
    f(t,y) = integ(2ty^3 dt) + h(y)
    f(t,y) = 2y^3(1/2t^2) + h(y)
    f(t,y) = y^3t^2 + h(y)

    df(t,y)/dy = t^2(3y^2) + dh(y)/dy
    df(t,y)/dy = 3t^2y^2 + dh(y)/dy
    3t^2y^2 = 3t^2y^2 + dh(y)/dy
    dh(y)/dy = 1
    h(y) = integ(1 dy) + c
    h(y) = y + c

    f(t,y) = y^3t^2 + y + c
    y^3t^2 + y = C

    sub y(1) = 1

    (1)^3(1)^2 + 1 = C
    C= 2
    .'. y(t)^3t^2 + y(t) = 2

    and i can't isolate y(t) so i left that as final solution. any thoughts on whether this is right or wrong.
  2. jcsd
  3. Feb 24, 2007 #2


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    Science Advisor

    You've made one silly little error- and it isn't "differential equations", it's basic algebra- maybe an arithmetic error!

    Surely your textbook defines M and N?
    Your differential equation is [itex]2ty^3+ 3t^2y^2dy/dt= 0[/itex] which you can write in "differential form" as [itex](2ty^3)dt+ (3t^3y^2)dy= 0[/itex].

    You want to determine whether the left side is an "exact differential"- that is, if it can be written as df for some function f(t,y). By the chain rule, [tex]df= \frac{\partial f}{\partial t}dt+ \frac{\partial f}{\partial y}dy[/tex]

    So you want to know if there exist f such that
    [tex]\frac{\partial f}{\partial t}= 2ty^3[/tex]
    [tex]\frac{\partial f}{\partial t}= 3t^2y^2[/tex]

    One way of determining whether such an f exists without finding it is to remember that the "mixed" derivatives are equal (as long as they are continuous). Here, if such an f exists,
    [tex]\frac{\partial^2 f}{\partial y\partial t}= \frac{\partial (2ty^3}{\partial y}= 6ty^2[/tex]
    [tex]\frac{\partial^2 f}{\partial t\partial y}= \frac{\partial (3t^2y^2}{\partial t}= 6ty^2[/tex]

    Yes, those are the same so this equation is exact and such a function f(t,y) exists!

    Good. In "reversing" partial integration with respect to t, in which you treat y as a constant, the "constant of integration might be a function of y- that's your "h(y)".

    Strictly speaking, that should be [itex]\frac{\partial f}{\partial y}[/itex] but that's fine. The dh/dy is an "ordinary" derivative since h depends only on y.
    Yes, that partial derivative must be equal to the "N(t,y)" from the equation.

    What? Is that a typo? Surely, subtracting [itex]3t^2y^2[/itex] from both sides, dh/dy= 0!

    Since dh/dy= 0, h(y)= c, a constant. (And since h depends only on y, it really is a constant.)

    First, it should be [itex]f(t,y)= y^3t^2+ c[/itex]. Second you don't really need the "c" here but it doesn't hurt. Since the equation is basically df= 0, f(t,y)= C and you can combine c and C- exactly as you do next.

    Correction: [itex]y^3t^2= C[/itex]

    With the correction [itex](1^3)(1^2)= C[/itex] so C= 1.
    [itex]y^3t^2= 1[/itex]

    Well, with the correction, you can write [itex]y= ^3\sqrt{1/t^2}= t^{-\frac{2}{3}}[/itex] but you should be aware that, with first order differential equations you quite often can't solve for one variable as a function of the other. You could, for example, use "implicit differentiation" to see if the solution satisfies the differential equation: differentiating [itex]y^3t^2= 1[/itex] with respect to t, we get [itex]3y^2t^2 y'+ 2y^3t= 0[/itex]. Yes, that's exactly the original equation.

    If we check you erroneous solution, [itex]y(t)^3t^2 + y(t) = 2[/itex] we get [itex]3y^2t^2y'+ 2y^3t+ y'= 0[/itex] or [itex]2y^3t+ (3y^2t^2+ 1)y'= 0[/itex] which is NOT the original equation.

    By the way, in addition to being exact, this is also a "separable" equation: we can write [itex]2ty^3 + 3t^2y^2 dy/dt = 0[/itex] as [itex]3t^2y^2 dy/dt= -2ty^3[/itex] and then, dividing both sides by [itex]y^3t^2[/itex], get [itex]3dy/y= -2dt/t[/itex]. Integrating both sides, 3 ln(y)= -2ln(t)+ C so
    [itex]ln(y^3)= ln(t^{-2})+ C[/itex], [itex]y^3= Ct^{-2}[/itex] and, finally, [itex]t^2y^3= C[/itex] as before.
  4. Feb 24, 2007 #3
    oh wow, thanks so much
    didnt correct that, so use to crossing things out and writing 1 afterwards..
  5. Feb 24, 2007 #4
    i mean didnt see that mistake.. lol
    i did correct it
    thanks a lot.
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