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Exact Eq.

  1. Jul 25, 2005 #1
    The equation P(x)y"+Q(x)y'+R(x)y=0 is said to be exact if it can be written in the form [P(x)y']'+[f(x)y]'=0, where f(x) is to be determined in terms of P(x), Q(x), and R(x). The latter equation can be integrated once immediately, resultingin a first order linear equation for y that can be solved. By equating the coefficients of the preceding equations and then eliminating f(x), show that a necessary condition for exactness is P"(x)-Q'(x)+R(x)=0.

    I don't understand how to equate the coefficients and I was stuck after taking the integral of [P(x)y']'+[f(x)y]'=0 and got [P(x)y']+[f(x)y]=0
  2. jcsd
  3. Jul 25, 2005 #2
    all right, let's start with the basics:
    1. Py'' + Qy' + Ry = 0

    is said to be exact if it can be written in the form
    2. (My')' + (Ny)' = 0

    Where P, Q, R, M, N, are functions only of x.
    prime is derivative wrt x. and we have existence
    of M'', P'', N', Q'. and untill I know better assume
    none of the above is 0.

    if we take the derivatives indicated in 2.
    M'y' + My'' + N'y + Ny' = 0

    My'' + (M' + N)y' + N'y = 0.
    Py'' + Qy' + Ry = 0

    now try equating the coefficients and generating
    the necessary condition. [in this case, you need
    to eliminate both M, and N.]

    to solve (My')' +(Ny)' = 0
    we integrate once to get My' + Ny = C
    where C is an integration constant

    we can put this in standard form
    y' + (N/M)y = C/M

    this is solvable by picking a suitable integrating factor....

    [now, which conditions on P, Q, R, N, M were necessary?]
    Last edited: Jul 25, 2005
  4. Jul 25, 2005 #3
    So [P(x)y']'+[f(x)y]'=0 -> P(x)y"+(P'(x)+f(x))y'+f '(x)y=0, and equating them I got:
    Q(x)=P'(x)+f(x) -> f '(x)= -P"(x)+Q'(x)
    R(x)= f '(x)

    Then setting -P"(x)+Q'(x) = R(x) -> P"(x)-Q'(x)+R(x)=0

    Will that suffice for showing the condition?
    Last edited: Jul 25, 2005
  5. Jul 25, 2005 #4
    Another question: how do I find an integrating factor for the Diff. Eq y"+xy'+y=0?

    I got it into this form: (y')'+(xy)'=0, integrated and got y'+(xy)=C. And I don't know where to go from here to find C.
  6. Jul 25, 2005 #5


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    Thanks Qbert. That's very interesting. Can I offer some help and kindly correct me if I'm wrong because this is new to me too.

    Physicsss, look carefully at those two equations Qbert gave us. Very carefully. Isn't P=M, R is something, but look at the middle part:

    Isn't Q equal to the derivative of M (which is P) plus the antiderivative of N' which is R? You know, if R is equal to N' then N is equal to the antiderivative of R. So one can look at an equation:


    and ask immediately: is the center function equal to the derivative of the first function plus the antiderivative of the third function? Is this correct Qbert?

    Now, how about solving this equation which I'm working on too using this method:

    [tex]x^2 y^{''}+(2x+\frac{1}{4} x^4)y^{'}+x^3y=0[/tex]

    Any initial condtions, say y(0.1)=0 and y'(0.1)=1 (stay away from singularity at x=0) or whatever initial conditions are easy.

    Edit: I mean you Physicsss, not you Qbert.
    Last edited: Jul 25, 2005
  7. Jul 25, 2005 #6


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    Unless you have some additional conditions you can't- C is a "constant of integration". If you solve the first order equation y'+ xy= C, then you get another constant. That's what you would expect- when you solve a second order differential equation, you expect the general solution to involve two undetermined constants.

    If the problem were y"+ xy'+ y= 0 with initial conditions: y(0)= 1, y'(0)= 0, then, taking x= 0 in y'+ xy= C, we get 0+ 0(1)= C or C= 0. That makes the equation y'+ xy= 0 which is easy (I cleverly chose y'(0)= 0 to get that!).
    dy= -xy dx or dy/y= -xd so ln|y|= -x2+ D (D is the new constant of integration). We can write that as y= E exp(-x2 (E= exp(D)).
    At x= 0 that becomes y= 1= E e0= E so E= 1 and the solution to this intial value problem is y(x)= exp(-x2).
  8. Jul 25, 2005 #7


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    Physicsss, where you at? Try and follow this one and then I have one for you to try if you want. The technique Qbert suggested is new to me so this is my first attempt at using such and applying it to an IVP:

    For the equation:


    If it's exact we can immediately write:

    [tex](Py^{'})^{'}+(y\int R)^{'}=0[/tex]


    [tex]x^2y^{''}+(2x+\frac{1}{4}x^4)y^{'}+x^3y=0;\quad y(0.1)=0,\quad y^{'}(0.1)=1[/tex]

    Noting that the middle function is the sum of the derivative of the first function and the antiderivative of the last function, we can then say:


    Taking antiderivatives:

    [tex]\int (x^2y^{'})^{'}+\int(\frac{1}{4}x^4y)^{'}=\int 0[/tex]



    Setting up this equation in the standard form to determine the integrating factor, we divide by [itex]x^2[/itex]


    The integrating factor is:

    [tex]\sigma=Exp\left(\int 1/4 x^2dx\right)=e^{x^3/12}[/tex]

    Multiplying the ODE by [itex]\sigma[/itex] converts the LHS to an exact differential:



    [tex]\int_{\xi=0.1,\zeta=0}^{\xi=x,\zeta=y} d(e^{\xi^3/12}\zeta)=c_1\int_{0.1}^{x}t^{-2}e^{t^3/12}dt[/tex]

    This yields:

    [tex]e^{x^3/12}y(x)=c_1\int_{0.1}^x t^{-2}e^{t^3/12}dt[/tex]


    [tex]y(x)=c_1e^{-x^3/12}\int_{0.1}^x t^{-2}e^{t^3/12}dt[/tex]

    So how do we find the constant? What about differentiating y(x) using Leibnitz's rule for the integral:

    [tex]y^{'}(x)=c_1\left[e^{-x^3/12}\frac{d}{dx}\left(\int_{0.1}^x t^{-2}e^{t^3/12}dt\right)+\left(\int_{0.1}^x t^{-2}e^{t^3/12}dt\right)\frac{d}{dx}e^{-x^3/12}\right][/tex]

    I know it's messy but try to follow it alright?

    Now, we know from the IVP that the derivative at 0.1 is 1. Now look at the second sum. When we insert x=0.1 into the integral, that will be zero right? We're left with then:




    We have [itex]c_1=0.01[/itex]

    Thus the solution is:

    [tex]y(x)=0.01e^{-x^3/12}\int_{0.1}^x t^{-2}e^{t^3/12}dt[/tex]

    Now, there's nothing wrong with leaving it that way. Perfectly fine. You're comfortable with functions like Sin(x), Tan(x) right? What about a Sal(x)?


    [tex]Sal(x)=\int_{0.1}^x t^{-2}e^{t^3/12}dt[/tex]

    Then the solution is simply:


    No different than using Sin(x) ok?

    A plot is attached.

    Now try this one:

    [tex]6y^{''}+(\cos x) y^{'}-(\sin x) y=0; \quad y(0)=0,\quad y^{'}(0)=1[/tex]

    I have no idea how it will turn out. :smile:

    Attached Files:

    Last edited: Jul 25, 2005
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