Ordinary Differential Equation--First-Order Exact

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

$${\partial p\over\partial y} = {\partial q\over \partial x}.$$ (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

$$\int^{(x,y)}_{(x_0,y_0)} p(x,y)\,dx+q(x,y)\,dy = c,$$ (3)

where $c$ is a constant.

A first-order ODE (1) is said to be inexact if

$${\partial p\over \partial y} \not= {\partial q\over\partial x}.$$ (4)

For a nonexact equation, the solution may be obtained by defining an Integrating Factor $\mu$ of (6) so that the new equation

$$\mu p(x,y)\,dx+\mu q(x,y)\,dy = 0$$ (5)

satisfies

$${\partial \over \partial y} (\mu p) = {\partial \over \partial x} (\mu q),$$ (6)

or, written out explicitly,

$$p{\partial \mu \over \partial y} + \mu {\partial p\over \par... ...tial \mu \over \partial x} + \mu {\partial p\over \partial x}.$$ (7)

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

$$\mu={q{\partial \mu\over \partial x}-p{\partial \mu\over \pa... ... {\partial p \over \partial y}-{\partial q\over \partial x}}.$$ (8)

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

$$P(x,y) = \mu p$$ (9)

$$Q(x,y) = \mu q$$ (10)

in equation (5) 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

$${\partial \mu\over\partial y} = 0.$$ (12)

For the equation to be exact in $\mu p$ and $\mu q$, the equation for a first-order nonexact ODE

$$p{\partial\mu\over\partial y} +\mu {\partial p\over\partial ... ...q{\partial\mu\over\partial x} +\mu {\partial p\over\partial x}$$ (13)

becomes

$$\mu{\partial p\over\partial y}= q {\partial\mu\over\partial x}+\mu {\partial p\over\partial x}.$$ (14)

Solving for $\partial\mu/\partial x$ gives

$${\partial\mu\over\partial x} =\mu(x) {{\partial p\over\partial y} - {\partial q\over\partial x}\over q} \equiv f(x,y)\mu(x),$$ (15)

which will be integrable if

$$f(x,y)\equiv {{\partial p\over\partial y} - {\partial q\over\partial x}\over q}=f(x),$$ (16)

in which case

$${d\mu\over\mu} = f(x)\,dx,$$ (17)

so that the equation is integrable

$$\mu (x) = e^{\int f(x)\,dx},$$ (18)

and the equation

$$[\mu p(x,y)]dx+[\mu q(x,y)]dy = 0$$ (19)

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

Given in an exact first-order ODE, look for an Integrating Factor $\mu(x,y)=g(xy)$. Then

$${\partial\mu\over\partial x} = {\partial g\over\partial x} y$$ (20)

$${\partial\mu\over\partial y} = {\partial g\over\partial y} x.$$ (21)

Combining these two,

$${\partial\mu\over\partial x} = {y\over x} {\partial\mu\over\partial y}.$$ (22)

For the equation to be exact in $\mu p$ and $\mu q$, the equation for a first-order nonexact ODE

$$p{\partial\mu\over\partial y} +\mu {\partial p\over\partial ... ...q{\partial\mu\over\partial x} +\mu {\partial p\over\partial x}$$ (23)

becomes

$${\partial\mu\over\partial y}\left({p - {y\over x}q}\right) ... ...al p\over\partial x} - {\partial p\over\partial y}}\right)\mu.$$ (24)

Therefore,

$${1\over x} {\partial\mu\over\partial y} = {{\partial q\over\partial x} - {\partial p\over\partial y}\over xp-yq}\mu.$$ (25)

Define a new variable

$$t(x,y)\equiv xy,$$ (26)

then ${\partial t/\partial y} = x$, so

$${\partial\mu\over\partial t} = {\partial\mu\over\partial y}{... ...ial p\over\partial y}\over xp-yq}\mu(t) \equiv f(x,y)\mu (t).$$ (27)

Now, if

$$f(x,y)\equiv {{\partial q\over\partial x} - {\partial p\over\partial y}\over xp-yq} = f(xy) = f(t),$$ (28)

then

$${\partial\mu\over\partial t} = f(t)\mu (t),$$ (29)

so that

$$\mu = e^{\int f(t)\,dt}$$ (30)

and the equation

$$[\mu p(x,y)]\,dx+[\mu q(x,y)]\,dy = 0$$ (31)

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

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

$$\mu = f(y),$$ (32)

so $\partial\mu/\partial x = 0$. For the equation to be exact in $\mu p$ and $\mu q$, equation (7) becomes

$${\partial\mu\over\partial y} = {{\partial q\over\partial x} - {\partial p\over\partial y}\over p}\mu = f(x,y)\mu (y).$$ (33)

Now, if

$$f(x,y)\equiv {{\partial q\over\partial x} - {\partial p\over\partial y}\over p} = f(y),$$ (34)

then

$${d\mu\over\mu} = f(y)\,dy,$$ (35)

so that

$$\mu (y) = e^{\int f(y)\,dy},$$ (36)

and the equation

$$\mu p(x,y)\,dx+\mu q(x,y)\,dy = 0$$ (37)

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

Given a first-order ODE of the form

$$yf(xy)\,dx + xg(xy)\,dy = 0,$$ (38)

define

$$v \equiv xy.$$ (39)

Then the solution is

$$\cases{ \ln x = \int{g(v)\,dv\over c[g(v)-f(v)]} + c & for $g(v) \not = f(v)$\cr xy = c & for $g(v) = f(v)$.\cr}$$ (40)

If

$${dy\over dx} = F(x,y) = G(v),$$ (41)

where

$$v \equiv {y\over x},$$ (42)

then letting

$$y \equiv xv$$ (43)

gives

$${dy\over dx} =x {dv/dx} + v$$ (44)

$$x {dv\over dx} + v = G(v).$$ (45)

This can be integrated by quadratures, so

$$\ln x = \int {dv\over f(v)-v} + c \qquad {\rm for\ } f(v) \not = v$$ (46)

$$y = cx \qquad {\rm for} f(v) = v.$$ (47)

References

Boyce, W. E. and DiPrima, R. C. Elementary Differential Equations and Boundary Value Problems, 4th ed. New York: Wiley, 1986.