math 3120 differential equations with boundary value problems chapter 2: first-order differential...
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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
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