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Exact First-Order Ordinary Differential Equation


Consider a first-order ODE in the slightly different form

 p(x,y)dx+q(x,y)dy=0.
(1)

Such an equation is said to be exact if

 (partialp)/(partialy)=(partialq)/(partialx).
(2)

This statement is equivalent to the requirement that a conservative field exists, so that a scalar potential can be defined. For an exact equation, the solution is

 int_((x_0,y_0))^((x,y))p(x,y)dx+q(x,y)dy=c,
(3)

where c is a constant.

A first-order ODE (◇) is said to be inexact if

 (partialp)/(partialy)!=(partialq)/(partialx).
(4)

For a nonexact equation, the solution may be obtained by defining an integrating factor mu of (◇) so that the new equation

 mup(x,y)dx+muq(x,y)dy=0
(5)

satisfies

 partial/(partialy)(mup)=partial/(partialx)(muq),
(6)

or, written out explicitly,

 p(partialmu)/(partialy)+mu(partialp)/(partialy)=q(partialmu)/(partialx)+mu(partialq)/(partialx).
(7)

This transforms the nonexact equation into an exact one. Solving (7) for mu gives

 mu=(q(partialmu)/(partialx)-p(partialmu)/(partialy))/((partialp)/(partialy)-(partialq)/(partialx)).
(8)

Therefore, if a function mu satisfying (8) can be found, then writing

P(x,y)=mup
(9)
Q(x,y)=muq
(10)

in equation (◇) then gives

 P(x,y)dx+Q(x,y)dy=0,
(11)

which is then an exact ODE. Special cases in which mu can be found include x-dependent, xy-dependent, and y-dependent integrating factors.

Given an inexact first-order ODE, we can also look for an integrating factor mu(x) so that

 (partialmu)/(partialy)=0.
(12)

For the equation to be exact in mup and muq, the equation for a first-order nonexact ODE

 p(partialmu)/(partialy)+mu(partialp)/(partialy)=q(partialmu)/(partialx)+mu(partialp)/(partialx)
(13)

becomes

 mu(partialp)/(partialy)=q(partialmu)/(partialx)+mu(partialp)/(partialx).
(14)

Solving for partialmu/partialx gives

(partialmu)/(partialx)=mu(x)((partialp)/(partialy)-(partialq)/(partialx))/q
(15)
=f(x,y)mu(x),
(16)

which will be integrable if

f(x,y)=((partialp)/(partialy)-(partialq)/(partialx))/q
(17)
=f(x),
(18)

in which case

 (dmu)/mu=f(x)dx,
(19)

so that the equation is integrable

 mu(x)=e^(intf(x)dx),
(20)

and the equation

 [mup(x,y)]dx+[muq(x,y)]dy=0
(21)

with known mu(x) is now exact and can be solved as an exact ODE.

Given an exact first-order ODE, look for an integrating factor mu(x,y)=g(xy). Then

 (partialmu)/(partialx)=(partialg)/(partialx)y
(22)
 (partialmu)/(partialy)=(partialg)/(partialy)x.
(23)

Combining these two,

 (partialmu)/(partialx)=y/x(partialmu)/(partialy).
(24)

For the equation to be exact in mup and muq, the equation for a first-order nonexact ODE

 p(partialmu)/(partialy)+mu(partialp)/(partialy)=q(partialmu)/(partialx)+mu(partialp)/(partialx)
(25)

becomes

 (partialmu)/(partialy)(p-y/xq)=((partialq)/(partialx)-(partialp)/(partialy))mu.
(26)

Therefore,

 1/x(partialmu)/(partialy)=((partialq)/(partialx)-(partialp)/(partialy))/(xp-yq)mu.
(27)

Define a new variable

 t(x,y)=xy,
(28)

then partialt/partialy=x, so

 (partialmu)/(partialt)=(partialmu)/(partialy)(partialy)/(partialt)=((partialq)/(partialx)-(partialp)/(partialy))/(xp-yq)mu(t)=f(x,y)mu(t).
(29)

Now, if

 f(x,y)=((partialq)/(partialx)-(partialp)/(partialy))/(xp-yq)=f(xy)=f(t),
(30)

then

 (partialmu)/(partialt)=f(t)mu(t),
(31)

so that

 mu=e^(intf(t)dt)
(32)

and the equation

 [mup(x,y)]dx+[muq(x,y)]dy=0
(33)

is now exact and can be solved as an exact ODE.

Given an inexact first-order ODE, assume there exists an integrating factor

 mu=f(y),
(34)

so partialmu/partialx=0. For the equation to be exact in mup and muq, equation (◇) becomes

 (partialmu)/(partialy)=((partialq)/(partialx)-(partialp)/(partialy))/pmu=f(x,y)mu(y).
(35)

Now, if

 f(x,y)=((partialq)/(partialx)-(partialp)/(partialy))/p=f(y),
(36)

then

 (dmu)/mu=f(y)dy,
(37)

so that

 mu(y)=e^(intf(y)dy),
(38)

and the equation

 mup(x,y)dx+muq(x,y)dy=0
(39)

is now exact and can be solved as an exact ODE.

Given a first-order ODE of the form

 yf(xy)dx+xg(xy)dy=0,
(40)

define

 v=xy.
(41)

Then the solution is

 {lnx=int(g(v)dv)/(c[g(v)-f(v)])+c   for g(v)!=f(v); xy=c   for g(v)=f(v).
(42)

If

 (dy)/(dx)=F(x,y)=G(v),
(43)

where

 v=y/x,
(44)

then letting

 y=xv
(45)

gives

 (dy)/(dx)=xdv/dx+v
(46)
 x(dv)/(dx)+v=G(v).
(47)

This can be integrated by quadratures, so

 lnx=int(dv)/(f(v)-v)+c    for f(v)!=v
(48)
 y=cx    forf(v)=v.
(49)

See also

First-Order Ordinary Differential Equation, Ordinary Differential Equation

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References

Boyce, W. E. and DiPrima, R. C. Elementary Differential Equations and Boundary Value Problems, 4th ed. New York: Wiley, 1986.Ross, C. C. §3.3 in Differential Equations. New York: Springer-Verlag, 2004.Zwillinger, D. Ch. 62 in Handbook of Differential Equations. San Diego, CA: Academic Press, 1997.

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Exact First-Order Ordinary Differential Equation

Cite this as:

Weisstein, Eric W. "Exact First-Order Ordinary Differential Equation." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/ExactFirst-OrderOrdinaryDifferentialEquation.html

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