p1
Example. Solve the differential equation
2x + y 2 + 2xyy 0 = 0.
The equation is neither linear nor separable.
Exact differential equation:
We say that the given equation
dy
= 0,
dx
is an exact differential equation if there exists a function ψ(x, y)
such that
∂ψ
∂ψ
(x, y) = M (x, y),
(x, y) = N (x, y).
∂x
∂y
M (x, y) + N (x, y)y 0 = 0
or
M (x, y) + N (x, y) ·
p2
Remark. Let the equation
M (x, y) + N (x, y)y 0 = 0
be an exact differential equation. Then there is a function ψ(x, y)
such that
∂ψ
(x, y) = M (x, y),
∂x
∂ψ
(x, y) = N (x, y).
∂y
Thus, the equation ψ(x, y) = C defines y = φ(x) implicitly as a
differentiable function of x. Then
p3: continuation p2
d
d
ψ(x, y) =
(C)
dx
dx
d
∂ψ
d
∂ψ
(x) +
(x, y) ·
(y) = 0
⇒ (x, y) ·
∂x
dx
∂y
dx
dy
⇒M (x, y) + N (x, y) ·
=0
dx
Therefore, Solutions of
M (x, y) + N (x, y)y 0 = 0 or
are given implicitly by
ψ(x, y) = C,
where C is an arbitrary constant.
d
[ψ(x, y)] = 0
dx
p4
Theorem 3. Let the functions M, N, My , and Nx , where
subscripts denote partial derivatives, be continuous in the
rectangular region R : α < x < β, γ < y < δ. Then
M (x, y) + N (x, y)y 0 = 0
or
M (x, y) + N (x, y) ·
is an exact differential equation in R if and only if
My (x, y) = Nx (x, y)
at each point of R.
dy
= 0,
dx
p5: continuation p4
Proof.
(⇒) : There is a function ψ(x, y) such that
ψx (x, y) = M (x, y),
Then
My (x, y) =
ψy (x, y) = N (x, y).
∂
(ψx (x, y)) = ψxy (x, y)
∂y
and
∂
(ψy (x, y)) = ψyx (x, y),
∂x
and since My , Nx are continuous on R, it follows that
ψxy (x, y), ψyx (x, y) are also continuous on R. Thus
Nx (x, y) =
ψxy (x, y) = ψyx (x, y) ⇒ My (x, y) = Nx (x, y).
p6: continuation p5
(⇐) : Claim: My = Nx ⇒ there is a function ψ(x, y) satisfying
ψx (x, y) = M (x, y) and ψy (x, y) = N (x, y).
Since M (x, y) is continuous, then there is a function Q(x, y) such
that
Z
M (x, y)dx = Q(x, y) + C1 (y).
where C1 (y) is an arbitrary function of y only. We define ψ(x, y)
by
ψ(x, y) = Q(x, y) + h(y),
where h(y) is a differentiable function of y only, then
ψx (x, y) =
∂
(Q(x, y) + h(y)) = Qx (x, y) + 0 = M (x, y).
∂x
p7: continuation p6
Next, if it is possible to find a function h(y) of only y satisfying
ψy (x, y) =
∂
(Q(x, y) + h(y)) = N (x, y).
∂y
We obatin
N (x, y) =
∂Q
(x, y) + h0 (y).
∂y
Then, solving for h0 (y), we have
h0 (y) = N (x, y) −
∂Q
(x, y).
∂y
p8: continuation p7
∂
∂x
∂Q
∂N
∂ ∂Q
N (x, y) −
(x, y) =
(x, y) −
(x, y)
∂y
∂x
∂x ∂y
By interchanging the order of differentiation
∂N
∂ ∂Q
∂N
(x, y) −
(x, y) =
(x, y) −
∂x
∂y ∂x
∂x
∂N
=
(x, y) −
∂x
since
∂Q
∂x (x, y)
∂
(M (x, y))
∂y
∂M
(x, y)
∂y
= M (x, y). Since Qyx = Qxy = My , then
∂Q
∂N
∂M
∂
N (x, y) −
(x, y) =
(x, y) −
(x, y) = 0
∂x
∂y
∂x
∂y
since My = Nx .
p9: continuation p8
That is, N (x, y) − ∂Q
∂y (x, y) is a function of only y. Therefore, we
can determine h(y) by computing indefinite integral
Z ∂Q
N (x, y) −
(x, y) dy,
∂y
and ψ(x, y) = Q(x, y) + h(y) satisfies
ψx (x, y) = M (x, y),
ψy (x, y) = N (x, y).
p10
Solve the differential equation
2x + y 2 + 2xyy 0 = 0.
Sol.
M (x, y) = 2x+y 2 ,
N (x, y) = 2xy ⇒ My (x, y) = 2y, Nx (x, y) = 2y
So the differential equation is an exact.
Z
Z
M (x, y)dx = (2x+y 2 )dx = x2 +xy 2 +C1 ⇒ Q(x, y) = x2 +xy 2
and Qy (x, y) = 2xy. Compute
R
R
N (x, y) − Qy (x, y))dy = 2xy − 2xydy = C2 . We choose
h(y) = 0 and
p11: continuation p10
ψ(x, y) = Q(x, y) + h(y) = x2 + xy 2 + 0 = x2 + xy 2 .
Therefore,
x2 + xy 2 = C
where C is an arbitrary constant, is an equation that define
solutions of equation 2x + y 2 + 2xyy 0 = 0 implicitly.
Example. Solve the differential equation
(y cos x + 2xey ) + (sin x + x2 ey − 1)y 0 = 0.
Sol.
M (x, y) = y cos x + 2xey , N (x, y) = sin x + x2 ey − 1
p12: continuation p11
My (x, y) = cos x + 2xey = Nx (x, y)
so the given equation is exact. Compute
Z
Z
M (x, y)dx = (y cos x + 2xey )dx = y sin x + x2 ey + C1 ,
we define Q(x, y) = y sin x + x2 ey , and then
Qy (x, y) = sin x + x2 ey . Next, compute
Z
Z
N (x, y) − Qy (x, y)dy = (sin x + x2 ey − 1) − (sin x + x2 ey )dy
Z
= (−1)dy = −y + C2
p13: continuation p12
We define h(y) = −y and ψ(x, y) is given by
ψ(x, y) = Q(x, y) + h(y) = y sin x + x2 ey − y.
Hence solutions of an equation
(y cos x + 2xey ) + (sin x + x2 ey − 1)y 0 = 0 are given implicitly by
y sin x + x2 ey − y = C
where C is an arbitrary constant.
p14
Quiz: Determine whether the equation
(2xy 2 + 2y) + (2x2 y + 2x)y 0 = 0
is exact. If it is exact, find the solution.
Exercise.
(a) Determine whether the equation
(y/x + 6x)dx + (ln x − 2)dy = 0,
x>0
is exact. If it is exact, find the solution.
(b) Solve the initial value problem
(2x − y)dx + (2y − x)dy = 0,
y(1) = 3
and determine at least approximately where the solution is
valid.
p15
Example. Solve the differential equation
(3xy + y 2 ) + (x2 + xy)y 0 = 0.
Sol.
M (x, y) = 3xy + y 2 ,
N (x, y) = x2 + xy
⇒My (x, y) = 3x + 2y, Nx (x, y) = 2x + y
since My 6= Nx , the given equation is not exact. Assume that we
can find a function ψ(x, y) such that
ψx (x, y) = M (x, y) = 3xy + y 2 ,
ψy (x, y) = N (x, y) = x2 + xy.
p16: continuation p15
Compute
we set
R
M (x, y)dx =
R
3xy + y 2 dx = 23 x2 y + xy 2 + C1 , and
3
ψ(x, y) = x2 y + xy 2 + h(y),
2
where h is an arbitrary function of y only. Compute ψy and set it
equal to N (x, y), obtaining
3
ψy (x, y) = x2 + 2xy + h0 (y) = x2 + xy = N (x, y)
2
or
1
h0 (y) = − x2 − xy.
2
Since the right side of eqaution h0 (y) = − 12 x2 − xy depends on x
as well as y, it is impossible to solve this equation for h(y). Thus
there is no ψ(x, y) satisfying above both equations.
p17
Let a given equation
M (x, y)dx + N (x, y)dy = 0 or M (x, y) + N (x, y)
dy
=0
dx
be not an exact differential equation.
If we can find an integrating factor u(x, y) such that equation
u(x, y)M (x, y)dx + u(x, y)N (x, y)dy = 0
is exact. Then by Theorem 3,
u(x, y)M (x, y)dx + u(x, y)N (x, y)dy = 0 is exact if and only if
(uM )y = (uN )x .
Thus,
uy M + uMy = ux N + uNx ⇒ M uy − N ux + (My − Nx )u = 0.
p18
That is, if we can find u(x, y) satisfying
M uy − N ux + (My − Nx )u = 0,
then the equation
u(x, y)M (x, y)dx + u(x, y)N (x, y)dy = 0
is exact.
Note: Unfortunately, the equation
M uy − N ux + (My − Nx )u = 0, which determines the integrating
factor u(x, y), is ordinarily at least as hard to solve as the original
equation M (x, y)dx + N (x, y)dy.
p19
However, the most important situations in which simple integrating
factors can be found occur when u is a function of only one of the
variables x or y, instead of both.
Assume that there an integrating factor u that depends on x only
such that
u(x)M (x, y)dx + u(x)N (x, y)dy = 0
is exact, thus we have
(uM )y = uMy ,
(uN )x = uNx + N
and
(uM )y = uMy = uNx + N
du
dx
du
= (uN )x
dx
p20
it is necessary that
My − Nx
du
=
u.
dx
N
M −N
That is, if yN x is a function of x only, then there is an
integrating factor u that aslo depends only on x.
N −M
Exercise. Show that if xM y = Q, where Q is a function of y
only, then the differential equation
M + N y0 = 0
has an integrating factor of the form
Z
u(y) = exp( Q(y)dy).
p21
Example Find an integrating factor for the equation
(3xy + y 2 ) + (x2 + xy)y 0 = 0
and then solve the equation.
Sol. Let us determine whether it has an integrating fator that
depends on x only. M (x, y) = 3xy + y 2 and N (x, y) = x2 + xy
My − Nx
(3x + 2y) − (2x + y)
1
=
= .
2
N
x + xy
x
Thus there is an integrating factor u that is a function of x only,
and it satisfies the differential equation
1
1
1
du
= · u ⇒ du = dx
dx
x
u
x
(separable)
p22: continuation p21
Hence, we have a solution u(x) = x, and obtain the equation
x(3xy + y 2 ) + x(x2 + xy)y 0 = 0 ⇒ (3x2 y + xy 2 ) + (x3 + x2 y)y 0 = 0
is exact.
Z
1
1
3x2 y + xy 2 dx = x3 y + x2 y 2 + C1 ⇒ Q(x, y) = x3 y + x2 y 2 .
2
2
Define ψ(x, y) = Q(x, y) + h(y) satisfying ψy = N (x, y), then
Z
Z
N (x, y) − Qy (x, y)dy = (x3 + x2 y) − (x3 + x2 y)dy = C2 .
p23: continuation p22
Hence, we have a function h(y) = 0 and obtain
ψ(x, y) = x3 y + 12 x2 y 2 satisfying ψx = M and ψy = N . Then
solutions of the differential equation are given implicitly by
1
x3 y + x2 y 2 = C.
2
Quiz: Find an integrating factor and solve the equation
ydx + (2xy − e−2y )dy = 0.
p24
Exercise.
1
(a) Show that u(x, y) = xy(2x+y)
is also a integrating factor of
2
2
equation (3xy + y ) + (x + xy)y 0 = 0.
(b) Solve the differential equation
(3xy + y 2 ) + (x2 + xy)y 0 = 0
1
. Verify that
using the integrating factor u(x, y) = xy(2x+y)
the solution is the same as that obtained in above example
with a different integrating factor u(x) = x.
p25
Exercise
(a) Show that if (Nx − My )/(xM − yN ) = R, where R depends
on the quantity xy only, then the differential equation
M + N y0 = 0
has an integrating factor of the form u(xy). Find a general
formula for this integrating factor.
(b) Find an integrating factor and solve the equation
2
6
y dy
x
3x +
+3
=0
+
y
y
x dx
Hint: By (a)