It turns out that theoretically, this is the most general method for solving equations
m (x,y)dx + n(x,y)dy = 0
I want to stress the word “theoretically” however. If the above equation is not exact, the idea is to multiply
by a function μ which will make it exact. Thus it would be sufficient to have
(μm )y = (μn)x
The function μ is called an integrating factor. In other words, it is required that
μym + μmy = μxn +μnx (23.5)
(23.5)
This is called a first order linear partial differential equation and we don’t know how to solve them.
However, we don’t need to find all solutions, just one which works. The idea is to look for μ = μ
(x )
or
μ = μ
(y)
. For us, if there is no such easy solution, the method has failed. So what would happen if there is
a solution μ = μ
(x)
? then you would have
μ′(x) = μ(x) my(x,y)−-nx-(x,y)
n(x,y)
and so there will be such an integrating factor if
my (x,y) − nx (x,y)
-----------------
n (x,y)
depends only on x. Similarly, there will be an integrating factor μ = μ
(y)
if
nx −-my-
m
depends only on y.
Example 23.6.1Find the solutions to
( 3 ) ( 2 )
2y + 2y dx + 3xy + x dy = 0
The equation is clearly not exact so we look for an integrating factor.
Thus μ = x is also an integrating factor. Which would you rather use? Multiply by x. The equation is
now
( 3 ) ( 22 2)
2xy + 2yx dx + 3x y + x dy = 0
and it is an exact equation so you are in the situation of the preceding section. You find a scalar potential.
The manipulations explained in the last section yield x^{2}y^{3} + x^{2}y as a scalar potential. Then the solutions
are
23 2
xy + x y = C
All of these are the same. You begin with 23.5 and look for solutions. In particular you look for
solutions that depend on only one variable. If you can find one, then the problem has been reduced to that
of the preceding section. If you can’t find such a solution, then you give up. Under general conditions, it
can be proved that solutions exist but as usual in mathematics, there is a big gap between knowing
something exists and finding it. However, here is something nice which was discovered by Euler back
in the 1700s. It is called Euler’s identity along with the more famous one involving complex
numbers.^{3}
Lemma 23.6.2A function M
(x,y)
is homogeneous of degree α if M
(tx,ty)
= t^{α}M
(x,y)
. For such afunction,
αM (x,y) = x∂M--(x,y) +y ∂M-(x,y)
∂x ∂y
Proof:You use the chain rule to differentiate both sides of the equation
M (tx,ty) = tαM (x,y)
with respect to t. Thus
αtα−1M (x,y) = x∂M-(tx,ty)+ y∂M--(tx,ty)
∂x ∂y
Now let t = 1. ■
The reason this is pretty nice is that if you have the equation
M (x,y)dx + N (x,y)dy = 0
and both M and N are homogeneous of degree α, then
----1----
xM +yN
is an integrating factor. Here M_{x} =
∂M
∂x-
. We verify this next. It is so if
( N ) ( M )
--------- = ---------
xM + yN x xM + yN y
By the quotient rule, this will be so if and only if
Nx(xM + yN)− N (M + xMx + yNx) =
My (xM + yN )− M (xMy + N + yNy )
In both sides of the above equation, some terms cancel and it follows that the desired result follows if and
only if
xM Nx − (N M + xMxN ) = yN My − (M N + yM Ny )
and this happens if and only if
xM Nx − xMxN = yN My − yM Ny
which happens if and only if
M xNx + M yNy = N yMy + N xMx
if and only if
M (xNx + yNy ) = N (yMy + xMx )
But this is true because by Euler’s identity, xN_{x} + yN_{y} = αN and yM_{y} + xM_{x} = αM so the above is just
αNM = αNM. Of course it is assumed that xM + yN≠0 in the above.
Example 23.6.3Find the integral curves for
(x2 + xy)dx + (y2 + x2)dy = 0
Of course this can be written as a homogeneous equation and the technique for solving
these can be used. However, let’s use this new technique which says that an integrating factor
is
Unfortunately, it is too complicated for me to solve this conveniently. However, knowing that it is exact
allows the use of the formula derived in showing that if M_{y} = N_{x} then the equation was exact. Thus the
integral curves are of the form
∫ x
M (t,y)dt+ N (0,y)
0
∫ x t2 + ty 1
= -3----2----3dt+ --= C
0 t + 2t y+ y y
Now we consider an easier one.
Example 23.6.4Find the integral curves for
( )
xy+ y2 dx + x2dy = 0
The integrating factor is
-----1----
xy (2x+ y)
and so the equation to solve is
1 x
---------(x+ y)dx + --------dy = 0
x (2x+ y) y(2x + y)
Then integrating the first term with respect to x, the scalar potential is of the form
1 || 1 ||
f (x,y) = ln|x|− 2 ln ||x + 2y||+ g(y)
Then differentiating with respect to y,
1 x
− 2(2x+-y)-+ g′(y) = y(2x+-y)-
′ 1-
g (y) = 2y
and so g
(y)
=
12
ln
|y|
will work. Thus the integral curves are of the form
Look for an integrating factor μ which is a function of x alone. You do this if
My-−-Nx-
N
does not depend on y. In this case, you solve
( )
μ′(x) = μ(x) My-−-Nx-
N
which is a separable equation. Solve and choose constant to satisfy initial condition. If this doesn’twork,
Look for an integrating factor μ which is a function of y alone. You do this if
Nx-−-My-
M
does not depend on x. In this case, you solve
′ Nx − My
μ (y) =---M----μ (y)
which is a separable equation. Solve and choose constant to satisfy initial condition.
If neither of these work, check to see if M,N are both homogeneous of the same degree. If they are,you could use either the methods of homogeneous equations or Euler’s formula for the integratingfactor
---1----.
xM + yN
If none of the above works, give up. You don’t know how to do it. The integrating factor exists, butyou don’t know how to find it.