p1

Example. Solve the differential equation

2x+ y2 + 2xyy′ = 0.

The equation is neither linear nor separable.

� Exact differential equation:

We say that the given equation

M(x, y) +N(x, y)y′ = 0 or M(x, y) +N(x, y) · dydx

= 0,

is an exact differential equation if there exists a function ψ(x, y)

such that

∂ψ

∂x(x, y) =M(x, y),

∂ψ

∂y(x, y) = N(x, y).

p2

Remark. Let the equation

M(x, y) +N(x, y)y′ = 0

be an exact differential equation. Then there is a function ψ(x, y)

such that

∂ψ

∂x(x, y) =M(x, y),

∂ψ

∂y(x, y) = N(x, y).

Thus, the equation ψ(x, y) = C defines y = φ(x) implicitly as a

differentiable function of x. Then

p3: continuation p2

d

dxψ(x, y) =

d

dx(C)

⇒∂ψ

∂x(x, y) · d

dx(x) +

∂ψ

∂y(x, y) · d

dx(y) = 0

⇒M(x, y) +N(x, y) · dydx

= 0

Therefore, Solutions of

M(x, y) +N(x, y)y′ = 0 ord

dx[ψ(x, y)] = 0

are given implicitly by

ψ(x, y) = C,

where C is an arbitrary constant.

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 or M(x, y) +N(x, y) · dydx

= 0,

is an exact differential equation in R if and only if

My(x, y) = Nx(x, y)

at each point of R.

p5: continuation p4

Proof.

(⇒) : There is a function ψ(x, y) such that

ψx(x, y) =M(x, y), ψy(x, y) = N(x, y).

Then

My(x, y) =∂

∂y(ψx(x, y)) = ψxy(x, y)

and

Nx(x, y) =∂

∂x(ψy(x, y)) = ψyx(x, y),

and since My, Nx are continuous on R, it follows that

ψxy(x, y), ψyx(x, y) are also continuous on R. Thus

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 ∫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) =∂

∂x(Q(x, y) + h(y)) = Qx(x, y) + 0 =M(x, y).

p7: continuation p6

Next, if it is possible to find a function h(y) of only y satisfying

ψy(x, y) =∂

∂y(Q(x, y) + h(y)) = N(x, y).

We obatin

N(x, y) =∂Q

∂y(x, y) + h′(y).

Then, solving for h′(y), we have

h′(y) = N(x, y)− ∂Q

∂y(x, y).

p8: continuation p7

∂

∂x

(N(x, y)− ∂Q

∂y(x, y)

)=∂N

∂x(x, y)− ∂

∂x

∂Q

∂y(x, y)

By interchanging the order of differentiation

∂N

∂x(x, y)− ∂

∂y

∂Q

∂x(x, y) =

∂N

∂x(x, y)− ∂

∂y(M(x, y))

=∂N

∂x(x, y)− ∂M

∂y(x, y)

since ∂Q∂x (x, y) =M(x, y). Since Qyx = Qxy =My, then

∂

∂x

(N(x, y)− ∂Q

∂y(x, y)

)=∂N

∂x(x, y)− ∂M

∂y(x, y) = 0

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∫ (N(x, y)− ∂Q

∂y(x, y)

)dy,

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+ y2 + 2xyy′ = 0.

Sol.

M(x, y) = 2x+y2, N(x, y) = 2xy ⇒My(x, y) = 2y,Nx(x, y) = 2y

So the differential equation is an exact.∫M(x, y)dx =

∫(2x+y2)dx = x2+xy2+C1 ⇒ Q(x, y) = x2+xy2

and Qy(x, y) = 2xy. Compute∫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 + xy2 + 0 = x2 + xy2.

Therefore,

x2 + xy2 = C

where C is an arbitrary constant, is an equation that define

solutions of equation 2x+ y2 + 2xyy′ = 0 implicitly.

Example. Solve the differential equation

(y cosx+ 2xey) + (sinx+ x2ey − 1)y′ = 0.

Sol.

M(x, y) = y cosx+ 2xey, N(x, y) = sinx+ x2ey − 1

p12: continuation p11

My(x, y) = cosx+ 2xey = Nx(x, y)

so the given equation is exact. Compute∫M(x, y)dx =

∫(y cosx+ 2xey)dx = y sinx+ x2ey + C1,

we define Q(x, y) = y sinx+ x2ey, and then

Qy(x, y) = sinx+ x2ey. Next, compute∫N(x, y)−Qy(x, y)dy =

∫(sinx+ x2ey − 1)− (sinx+ x2ey)dy

=

∫(−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 sinx+ x2ey − y.

Hence solutions of an equation

(y cosx+ 2xey) + (sinx+ x2ey − 1)y′ = 0 are given implicitly by

y sinx+ x2ey − y = C

where C is an arbitrary constant.

p14

Quiz: Determine whether the equation

(2xy2 + 2y) + (2x2y + 2x)y′ = 0

is exact. If it is exact, find the solution.

Exercise.

(a) Determine whether the equation

(y/x+ 6x)dx+ (lnx− 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 + y2) + (x2 + xy)y′ = 0.

Sol.

M(x, y) = 3xy + y2, 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 + y2, ψy(x, y) = N(x, y) = x2 + xy.

p16: continuation p15

Compute∫M(x, y)dx =

∫3xy + y2dx = 3

2x2y + xy2 + C1, and

we set

ψ(x, y) =3

2x2y + xy2 + h(y),

where h is an arbitrary function of y only. Compute ψy and set it

equal to N(x, y), obtaining

ψy(x, y) =3

2x2 + 2xy + h′(y) = x2 + xy = N(x, y)

or

h′(y) = −1

2x2 − xy.

Since the right side of eqaution h′(y) = −12x

2 − 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

dx= 0

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,

uyM + uMy = uxN + uNx ⇒Muy −Nux + (My −Nx)u = 0.

p18

That is, if we can find u(x, y) satisfying

Muy −Nux + (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

Muy −Nux + (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 +Ndu

dx

and

(uM)y = uMy = uNx +Ndu

dx= (uN)x

p20

it is necessary thatdu

dx=My −Nx

Nu.

That is, ifMy−Nx

N is a function of x only, then there is an

integrating factor u that aslo depends only on x.

Exercise. Show that ifNx−My

M = Q, where Q is a function of y

only, then the differential equation

M +Ny′ = 0

has an integrating factor of the form

u(y) = exp(

∫Q(y)dy).

p21

Example Find an integrating factor for the equation

(3xy + y2) + (x2 + xy)y′ = 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 + y2 and N(x, y) = x2 + xy

My −Nx

N=

(3x+ 2y)− (2x+ y)

x2 + xy=

1

x.

Thus there is an integrating factor u that is a function of x only,

and it satisfies the differential equation

du

dx=

1

x· u⇒ 1

udu =

1

xdx (separable)

p22: continuation p21

Hence, we have a solution u(x) = x, and obtain the equation

x(3xy+y2)+x(x2+xy)y′ = 0⇒ (3x2y+xy2)+(x3+x2y)y′ = 0

is exact.∫3x2y + xy2dx = x3y +

1

2x2y2 + C1 ⇒ Q(x, y) = x3y +

1

2x2y2.

Define ψ(x, y) = Q(x, y) + h(y) satisfying ψy = N(x, y), then∫N(x, y)−Qy(x, y)dy =

∫(x3 + x2y)− (x3 + x2y)dy = C2.

p23: continuation p22

Hence, we have a function h(y) = 0 and obtain

ψ(x, y) = x3y + 12x

2y2 satisfying ψx =M and ψy = N . Then

solutions of the differential equation are given implicitly by

x3y +1

2x2y2 = C.

Quiz: Find an integrating factor and solve the equation

ydx+ (2xy − e−2y)dy = 0.

p24

Exercise.

(a) Show that u(x, y) = 1xy(2x+y) is also a integrating factor of

equation (3xy + y2) + (x2 + xy)y′ = 0.

(b) Solve the differential equation

(3xy + y2) + (x2 + xy)y′ = 0

using the integrating factor u(x, y) = 1xy(2x+y) . Verify that

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 +Ny′ = 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(3x+

6

y

)+

(x2

y+ 3

y

x

)dy

dx= 0

Hint: By (a)