Consider a first-order ODE in the slightly different form
![\begin{displaymath}
p(x,y)\,dx+q(x,y)\,dy = 0.
\end{displaymath}](o_658.gif) |
(1) |
Such an equation is said to be exact if
![\begin{displaymath}
{\partial p\over\partial y} = {\partial q\over \partial x}.
\end{displaymath}](o_582.gif) |
(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
![\begin{displaymath}
\int^{(x,y)}_{(x_0,y_0)} p(x,y)\,dx+q(x,y)\,dy = c,
\end{displaymath}](o_659.gif) |
(3) |
where
is a constant.
A first-order ODE (1) is said to be inexact if
![\begin{displaymath}
{\partial p\over \partial y} \not= {\partial q\over\partial x}.
\end{displaymath}](o_660.gif) |
(4) |
For a nonexact equation, the solution may be obtained by defining an Integrating Factor
of (6) so that the
new equation
![\begin{displaymath}
\mu p(x,y)\,dx+\mu q(x,y)\,dy = 0
\end{displaymath}](o_661.gif) |
(5) |
satisfies
![\begin{displaymath}
{\partial \over \partial y} (\mu p) = {\partial \over \partial x} (\mu q),
\end{displaymath}](o_662.gif) |
(6) |
or, written out explicitly,
![\begin{displaymath}
p{\partial \mu \over \partial y} + \mu {\partial p\over \par...
...tial \mu \over \partial x} + \mu {\partial p\over \partial x}.
\end{displaymath}](o_663.gif) |
(7) |
This transforms the nonexact equation into an exact one. Solving (7) for
gives
![\begin{displaymath}
\mu={q{\partial \mu\over \partial x}-p{\partial \mu\over \pa...
... {\partial p
\over \partial y}-{\partial q\over \partial x}}.
\end{displaymath}](o_664.gif) |
(8) |
Therefore, if a function
satisfying (8) can be found, then writing
in equation (5) then gives
![\begin{displaymath}
P(x,y)\,dx+Q(x,y)\,dy = 0,
\end{displaymath}](o_669.gif) |
(11) |
which is then an exact ODE. Special cases in which
can be found include
-dependent,
-dependent, and
-dependent integrating factors.
Given an inexact first-order ODE, we can also look for an Integrating Factor
so that
![\begin{displaymath}
{\partial \mu\over\partial y} = 0.
\end{displaymath}](o_671.gif) |
(12) |
For the equation to be exact in
and
, the equation for a first-order nonexact ODE
![\begin{displaymath}
p{\partial\mu\over\partial y} +\mu {\partial p\over\partial ...
...q{\partial\mu\over\partial x} +\mu {\partial p\over\partial x}
\end{displaymath}](o_674.gif) |
(13) |
becomes
![\begin{displaymath}
\mu{\partial p\over\partial y}= q {\partial\mu\over\partial x}+\mu {\partial p\over\partial x}.
\end{displaymath}](o_675.gif) |
(14) |
Solving for
gives
![\begin{displaymath}
{\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),
\end{displaymath}](o_677.gif) |
(15) |
which will be integrable if
![\begin{displaymath}
f(x,y)\equiv {{\partial p\over\partial y} - {\partial q\over\partial x}\over q}=f(x),
\end{displaymath}](o_678.gif) |
(16) |
in which case
![\begin{displaymath}
{d\mu\over\mu} = f(x)\,dx,
\end{displaymath}](o_679.gif) |
(17) |
so that the equation is integrable
![\begin{displaymath}
\mu (x) = e^{\int f(x)\,dx},
\end{displaymath}](o_680.gif) |
(18) |
and the equation
![\begin{displaymath}[\mu p(x,y)]dx+[\mu q(x,y)]dy = 0
\end{displaymath}](o_681.gif) |
(19) |
with known
is now exact and can be solved as an exact ODE.
Given in an exact first-order ODE, look for an Integrating Factor
. Then
![\begin{displaymath}
{\partial\mu\over\partial x} = {\partial g\over\partial x} y
\end{displaymath}](o_683.gif) |
(20) |
![\begin{displaymath}
{\partial\mu\over\partial y} = {\partial g\over\partial y} x.
\end{displaymath}](o_684.gif) |
(21) |
Combining these two,
![\begin{displaymath}
{\partial\mu\over\partial x} = {y\over x} {\partial\mu\over\partial y}.
\end{displaymath}](o_685.gif) |
(22) |
For the equation to be exact in
and
, the equation for a first-order nonexact ODE
![\begin{displaymath}
p{\partial\mu\over\partial y} +\mu {\partial p\over\partial ...
...q{\partial\mu\over\partial x} +\mu {\partial p\over\partial x}
\end{displaymath}](o_674.gif) |
(23) |
becomes
![\begin{displaymath}
{\partial\mu\over\partial y}\left({p - {y\over x}q}\right)
...
...al p\over\partial x} - {\partial p\over\partial y}}\right)\mu.
\end{displaymath}](o_686.gif) |
(24) |
Therefore,
![\begin{displaymath}
{1\over x} {\partial\mu\over\partial y}
= {{\partial q\over\partial x} - {\partial p\over\partial y}\over xp-yq}\mu.
\end{displaymath}](o_687.gif) |
(25) |
Define a new variable
![\begin{displaymath}
t(x,y)\equiv xy,
\end{displaymath}](o_688.gif) |
(26) |
then
, so
![\begin{displaymath}
{\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).
\end{displaymath}](o_690.gif) |
(27) |
Now, if
![\begin{displaymath}
f(x,y)\equiv {{\partial q\over\partial x} - {\partial p\over\partial y}\over xp-yq} = f(xy) = f(t),
\end{displaymath}](o_691.gif) |
(28) |
then
![\begin{displaymath}
{\partial\mu\over\partial t} = f(t)\mu (t),
\end{displaymath}](o_692.gif) |
(29) |
so that
![\begin{displaymath}
\mu = e^{\int f(t)\,dt}
\end{displaymath}](o_693.gif) |
(30) |
and the equation
![\begin{displaymath}[\mu p(x,y)]\,dx+[\mu q(x,y)]\,dy = 0
\end{displaymath}](o_694.gif) |
(31) |
is now exact and can be solved as an exact ODE.
Given an inexact first-order ODE, assume there exists an integrating factor
![\begin{displaymath}
\mu = f(y),
\end{displaymath}](o_695.gif) |
(32) |
so
. For the equation to be exact in
and
, equation (7) becomes
![\begin{displaymath}
{\partial\mu\over\partial y} = {{\partial q\over\partial x} - {\partial p\over\partial y}\over p}\mu = f(x,y)\mu (y).
\end{displaymath}](o_697.gif) |
(33) |
Now, if
![\begin{displaymath}
f(x,y)\equiv {{\partial q\over\partial x} - {\partial p\over\partial y}\over p} = f(y),
\end{displaymath}](o_698.gif) |
(34) |
then
![\begin{displaymath}
{d\mu\over\mu} = f(y)\,dy,
\end{displaymath}](o_699.gif) |
(35) |
so that
![\begin{displaymath}
\mu (y) = e^{\int f(y)\,dy},
\end{displaymath}](o_700.gif) |
(36) |
and the equation
![\begin{displaymath}
\mu p(x,y)\,dx+\mu q(x,y)\,dy = 0
\end{displaymath}](o_661.gif) |
(37) |
is now exact and can be solved as an exact ODE.
Given a first-order ODE of the form
![\begin{displaymath}
yf(xy)\,dx + xg(xy)\,dy = 0,
\end{displaymath}](o_588.gif) |
(38) |
define
![\begin{displaymath}
v \equiv xy.
\end{displaymath}](o_701.gif) |
(39) |
Then the solution is
![\begin{displaymath}
\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}
\end{displaymath}](o_702.gif) |
(40) |
If
![\begin{displaymath}
{dy\over dx} = F(x,y) = G(v),
\end{displaymath}](o_703.gif) |
(41) |
where
![\begin{displaymath}
v \equiv {y\over x},
\end{displaymath}](o_704.gif) |
(42) |
then letting
![\begin{displaymath}
y \equiv xv
\end{displaymath}](o_705.gif) |
(43) |
gives
![\begin{displaymath}
{dy\over dx} =x {dv/dx} + v
\end{displaymath}](o_706.gif) |
(44) |
![\begin{displaymath}
x {dv\over dx} + v = G(v).
\end{displaymath}](o_707.gif) |
(45) |
This can be integrated by quadratures, so
![\begin{displaymath}
\ln x = \int {dv\over f(v)-v} + c \qquad {\rm for\ } f(v) \not = v
\end{displaymath}](o_708.gif) |
(46) |
![\begin{displaymath}
y = cx \qquad {\rm for} f(v) = v.
\end{displaymath}](o_709.gif) |
(47) |
References
Boyce, W. E. and DiPrima, R. C. Elementary Differential Equations and Boundary Value Problems, 4th ed.
New York: Wiley, 1986.
© 1996-9 Eric W. Weisstein
1999-05-26