Math 3120 Differential Equations

withBoundary Value Problems

Chapter 2: First-Order Differential Equations

Section 2-4: Exact Equations

Exact Equations

Consider a first order ODE of the form

Suppose there is a function such that

and such that (x,y) = c defines y = (x) implicitly. Then

and hence the original ODE becomes

Thus (x,y) = c defines a solution implicitly. In this case, the ODE is said to be exact.

0),(),( yyxNyxM

),(),(),,(),( yxNyxyxMyx yx

)(,),(),( xxdx

d

dx

dy

yxyyxNyxM

0)(, xxdx

d

Suppose an ODE can be written in the form

where the functions M, N, My and Nx are all continuous in the rectangular region R: (x, y) (, ) x (, ). Then Eq. (1) is an exact differential equation iff

That is, there exists a function satisfying the conditions

iff M and N satisfy Equation (2).

)1(0),(),( yyxNyxM

)2(),(),,(),( RyxyxNyxM xy

)3(),(),(),,(),( yxNyxyxMyx yx

Theorem 2.4.1

Write the DE in the form

Check if the eq. (1) is an exact differential equation.

Write down the system of eq.

Integrate either the first equation with respect of the variable x or the second with respect of the variable y. The choice of the equation to be integrated will depend on how easy the calculations are. Let us assume that the first equation was chosen, then we get

Use the second equation of the system to find the derivative of g(y).

Integrate to find

Write down the function

All the solutions are given by the implicit equation

If you are given an IVP, plug in the initial condition to find the constant C.

)1(0),(),( yyxNyxM

),(),(),,(),( yxNyxyxMyx yx

Method to solve Exact Equations

),( yxfy

)(),(),( ygdxyxMyx

),()('),(),( yxNygdxyxMy

yxy

),( yx)(' yg

Cyx ),(

Example 1: Exact Equation (1 of 4)

Consider the following differential equation.

Then

and hence

From Theorem 2.4.1,

Thus

0)4()4(4

4

yyxyxyx

yx

dx

dy

yxyxNyxyxM 4),(,4),(

exact is ODE),(4),( yxNyxM xy

yxyxyxyx yx 4),(,4),(

)(42

14),(),( 2 yCxyxdxyxdxyxyx x

Example 1: Solution (2 of 4)

We have

and

It follows that

Thus

By Theorem 2.6.1, the solution is given implicitly by

yxyxyxyx yx 4),(,4),(

)(42

14),(),( 2 yCxyxdxyxdxyxyx x

kyyCyyCyCxyxyxy 2

2

1)()()(44),(

kyxyxyx 22

2

14

2

1),(

cyxyx 22 8

Example 1: Direction Field and Solution Curves (3 of 4)

Our differential equation and solutions are given by

A graph of the direction field for this differential equation,

along with several solution curves, is given below.

cyxyxyyxyxyx

yx

dx

dy

22 80)4()4(4

4

Example 1: Explicit Solution and Graphs (4 of 4)

Our solution is defined implicitly by the equation below.

In this case, we can solve the equation explicitly for y:

Solution curves for several values of c are given below.

cxxycxxyy 222 17408

cyxyx 22 8

Example 2: Exact Equation (1 of 3)

Consider the following differential equation.

Then

and hence

From Theorem 2.4.1,

Thus

0)1(sin)2cos( 2 yexxxexy yy

1sin),(,2cos),( 2 yy exxyxNxexyyxM

exact is ODE),(2cos),( yxNxexyxM xy

y

1sin),(,2cos),( 2 yy

yx exxNyxxexyMyx

)(sin2cos),(),( 2 yCexxydxxexydxyxyx yyx

Example 2: Solution (2 of 3)

We have

and

It follows that

Thus

By Theorem 2.6.1, the solution is given implicitly by

1sin),(,2cos),( 2 yy

yx exxNyxxexyMyx

)(sin2cos),(),( 2 yCexxydxxexydxyxyx yyx

kyyCyC

yCexxexxyx yyy

)(1)(

)(sin1sin),( 22

kyexxyyx y 2sin),(

cyexxy y 2sin

Example 2: Direction Field and Solution Curves (3 of 3)

Our differential equation and solutions are given by

A graph of the direction field for this differential equation,

along with several solution curves, is given below.

cyexxy

yexxxexyy

yy

2

2

sin

,0)1(sin)2cos(

Example 3: Non-Exact Equation

Consider the following differential equation.

Then

and hence

0)2()3( 32 yxxyyxy

32 2),(,3),( xxyyxNyxyyxM

exactnot is ODE),(3223),( 2 yxNxyyxyxM xy

It is sometimes possible to convert a differential equation that is not exact into an exact equation by multiplying the equation by a suitable integrating factor (x, y):

For this equation to be exact, we need

This partial differential equation may be difficult to solve. If is a function of x alone, then y = 0 and hence we solve

provided right side is a function of x only. Similarly if is a function of y alone.

Integrating Factors

0),(),(),(),(

0),(),(

yyxNyxyxMyx

yyxNyxM

0 xyxyxy NMNMNM

,N

NM

dx

d xy

Given a first ODE by

If is a function of x alone, then an integrating factor for (1) is

If is a function of y alone, then an integrating factor for (1) is

Summarize

)1( 0),(),( yyxNyxM

dxN

xNyM

ex)(

NNM xy /)(

MMN yx /)(

dyM

xNyM

ey)(

Example 4: Non-Exact Equation

Consider the following non-exact differential equation.

Seeking an integrating factor, we solve the linear equation

Multiplying our differential equation by , we obtain the exact equation

which has its solutions given implicitly by

0)()3( 22 yxyxyxy

xxxdx

d

N

NM

dx

d xy

)(

,0)()3( 2322 yyxxxyyx

cyxyx 223

2

1