Lesson 7 Non-Exact DE - Lecture notes Non-Exact DE PDF

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Course: MATH 021

TOPIC # 7 Non- Exact Differential Equation; Integrating Factors Introduction: This part is an extension of the exact differential equations. The concept of integrating factors is introduced here, most especially its essential role in the reduction of the special cases of non-exact equations to an exact equations. 1. Identify non-exact differential equation. 2. Solve non-exact differential equation. Definition. Non-Exact Differential Equation In the differential equation represented by, M(x,y)dx + N(x,y)dy = 0, if ∂M ∂N ≠ ∂y ∂x the equations is said to be non-exact. However, we can try to find so-called integrating factor, which is a function of μ(x, y) such that the equation becomes exact after multiplication by this factor. Determination of Integrating factors � Case 1.When the integrating factor is a function of x alone, μ = e ∫ f(x)dx ∂M ∂N ∂y− ∂x

N

where: f(x) =

Case 2. When the integrating factor is a function of y alone, μ = e∫ g(y)dy ∂N ∂M

where: g(y) =

∂x

−

∂y

M

Case 3. When the integrating factor is the product of powers of the variables x and y, or μ = xmyn where m and n can be determined by comparison using the equation, Mn y

Nm x

−

=∂x

∂N

− ∂y

∂M

Derivation of the integrating factors � Case 1. μ is a function of x alone, (Mdx + Ndy = 0) μ μMdx + μNdy = 0 , ∂(μM)

=

∂(μN)

∂y ∂M ∂μ

μ N

∂M

,

apply product rule of differentiation

∂x

+M

∂y

μ ∂μ

∂μ

= μ

∂y

∂N

∂μ

) ∂x

−

∂y

→ solve for μ

∂M ∂N ∂y ∂x

μ

; ∂μ = 0

∂x

μ (∂y − ∂x ) = N ∂x = (

+ ∂x

+N

∂x

∂M

∂N

∂x

∂N

=μ

∂y

∂μ

test for exactness

→ integrate both side

N

∂M ∂N ∂y ∂x

lnμ = ∫ (

−

) ∂x

N

∂M ∂N

∂M ∂N

μ= e∫

(

∂y

−

N

∂x

)

∂x

=

∴ μ = e∫ f(x)dx

Let f(x)

∂y

− N

∂x

Case 2. μ is a function of y alone, (Mdx + Ndy = 0) μ μMdx + μNdy = 0

test for exactness

∂(μM) ∂(μN) ∂y

apply product rule of differentiation

∂M ∂μ

μ N

=

∂x

+M

∂y

μ ∂N

∂M

∂μ

= μ

∂y

+M

∂y

∂μ

∂N ∂M ∂x ∂y

μ

M

; ∂μ = 0 ∂x

→ solve for μ

∂x

∂M

= ( −

∂x

=μ

μ (∂x − ∂y ) = M ∂y ∂μ

+

∂x

∂y

∂N

∂N

∂μ

) ∂y

∂N ∂M ∂x ∂y

lnμ = ∫ ( −

→ integrate both side

) ∂y

M

∂N ∂M

∂N ∂M

μ= e∫

(

∂x

−

∂y

M

)

∂y

∂x

Let g(y)

−

∂y

M

=

∴ μ = e∫ g(y)dy Case 3. μ is in the form of μ = xmyn (Mdx + Ndy = 0) μ μMdx + μNdy = 0

test for exactness

∂(xmynM) ∂(xmynN) ∂y

apply product rule of differentiation

x m [(yn x myn

∂M

= ∂x ∂M

) + (ny n−1M)] = y n [(x m

∂y

+ nxmyn−1M = xmyn

∂y ∂M ∂y

+

Mn ∂N y = ∂x

∂N ∂x

Nm

+x

∂N

) + (mx m−1 N)]

∂x

+ mxm−1ynN

multiply both side by

1

x myn

∴

Mn y

−

Nm x

=∂x

∂N

−∂y

∂M

Remarks: Use comparison technique to solve for the values of m and n.

Sample Problems: Find the solution of the given differential equation. 1. (2y2 + 3x)dx + 2xydy = 0 Solution: ∂M

M = 2y2 + 3x

∂y

∂N

N = 2xy

∂x

= 4y

= 2y(1) = 2y

Try the solution of Case 1 ∂M

−

∂N

∂y ∂x

f(x) =

N

(x) = f4y−2y

2y

2xy

= 2x y

1

f(x) = (Case 1 is possible) x

Solve for the integrating factor μ ∫ f(x)dx

μ=

1

∫ dx

eμ

Important Note: 1. elnx = x 2. lnxn = nln x 3. eln x2 = x2

= elnx = x

x

Proof of eq (1) and eq (3) ��� � = ���� Take ln both side of the eq. 𝑙𝑙𝑙 = 𝑙𝑙𝑙 = =1 𝑙𝑙𝑙 = Since they are equal, ∴ = ∴ =

=e

Multiply the given DE by x , (2xy2 + 3x2)dx + 2x2ydy = 0 Which can now be shown to be an exact equation M = 2xy2 + 3x2

∂M ∂y

N = 2x 2y

∂N ∂x

= 2x(2y) = 4xy

= 2y(2x) = 4xy

Since the two partial derivatives are now equal, the equation is exact.

The solution is F(x,y) = C F = ∫ M(x, y) ∂x + f(y)

F = ∫ N(x, y) ∂y +

F = ∫(2xy2 + 3x2) ∂x + f(y)

g(x) F = ∫ 2x2y ∂y +

x

x

F = 2y2 ( 2) + 3 ( 3) + f(y) 2

2

2

3

g(x) y

3

F = 2x2 ( 2) + g(x)

F = x y + x + f(y)

2

F = x2y2 + g(x) Combine the two result such that no function will be repeated and equate it to C. So the solution is ∴x2y2 + x3 = C 2. 6xydx + (4y + 9x2)dy = 0 Solution: ∂M

M = 6xy

= 6x(1) = 6x

∂y

∂N

N = 4y + 9x2

∂x

= 18x

Try the solution of Case 2 ∂N ∂M

g(y) = g(y) 6x

∂x

−

∂y

M 18x −

=

6xy

12x

= 6xy

2

g(y) = (Case 2 is possible) y

Solve for the integrating factor μ μ = e∫ g(y)dy 2

μ= e

∫ dy y

= e2lny

= eln y2

= y2

Multiply the given DE by y2 , 6xy3dx + (4y3 + 9x2y2)dy = 0

Which can now be shown to be an exact equation ∂M

M = 6xy3

∂y ∂N

N = 4y3 + 9x 2y2

= 6x(3y2) = 18xy2

= 9y2(2x) = 18xy2

∂x

Since the two partial derivatives are now equal, the equation is exact. The solution is F(x,y) = C F = ∫ M(x, y) ∂x +

F = ∫ N(x, y) ∂y + g(x)

f(y) F = ∫ 6xy3 ∂x +

F = ∫(4y3 + 9x2y2) ∂y + g(x)

f(y)

y

3

y

F = 4 ( 4 ) + 9x2 ( 3) + g(x)

x

F = 6y ( 2 ) + f(y) 2

4

2

F = 3x y3 + f(y)

3

F = y4 + 3x2y3 + g(x)

Combine the two result such that no function will be repeated and equate it to C. So the solution is ∴ 3x2y3 + y4 = C 3. (5xy2 − 2y)dx + (3x2y − x)dy = 0 Solution: ∂M

M = 5xy2 − 2y

∂y

∂N

N = 3x2y − x

∂x

= 5x(2y) − 2 = 10xy − 2

= 3y(2x) − 1 = 6xy − 1

Try the solution of Case 3 Mn Nm ∂N ∂M =∂x − y − x ∂y (5xy −2y )n (3x y−x )m

2 y

−

2

= (6xy − 1 ) − (10xy − 2)

x

(5xy − 2)n − (3xy − 1)m = −4xy + 1

Compare the left side to right side of the equation to solve for m and n

xy

5n − 3m = −4

k :

−2n + m = 1

solving for m & n Apply system of linear equations(i.e substitution method) m = 1 + 2n 5n − 3(1 + 2n) = −4 n=1 m = 1 + 2n = 1 + 2(1) = 3 μ = x3y Multiply the given DE by x3y , (5x4y3 − 2x3y2)dx + (3x5y2 − x4y)dy = 0 Which can now be shown to be an exact equation ∂M

M = 5x4y3 − 2x 3y2

∂y

∂N

N = 3x 5y2 − x4y

∂x

= 5x4(3y 2) − 2x3(2y) = 15x4y2 − 4x3y

= 3y2(5x4) − y(4x3) = 15x4y2 − 4x 3y

Since the two partial derivatives are now equal, the equation is exact. The solution is F(x,y) = C F = ∫ M(x, y) ∂x + f(y)

F = ∫ N(x, y) ∂y + g(x)

F = ∫(5x4y3 − 2x3y2 ) ∂x + f(y)

F = ∫(3x5y2 − x4y) ∂y + g(x)

x

x

F = 5y3 ( 5) − 2y2 ( 4) + f(y) 5

F = x5y3 −

4

4 x2y

2

+ f(y)

y

y

F = 3x5 ( 3) − x 4 ( 2) + g(x) 3

F = x5y3 −

2

4 x2y

2

+ g(x)

Combine the two result such that no function will be repeated and equate it to C. So the solution is

x y

∴ x5y3 – 4 2 = C 2...