Chapter 2: First Order DE 2.6 Exact DE and Integrating...

Sample I (Ex. 1-12)Sample II (Ex. 1-12)Sample III (Ex. 1-12)

Integrating Factors

Chapter 2: First Order DE

§2.6 Exact DE and Integrating Factor

Satya Mandal, KU

Satya Mandal, KU Chapter 2: First Order DE §2.6 Exact DE and Integrating Facto


Integrating Factors

First Order DE

◮ Recall the general form of the First Order DEs (FODE):

dy

dx= f (x , y) (1)

(In this section x is the independent variable; not t.)

◮ This can also be written as

M(x , y) + N(x , y)dy

dx= 0 (2)

with variety of choices of M(x , y),N(x , y).



Integrating Factors

Exact FODE

Sometimes, we may be lucky:

◮ Suppose there is a function ψ(x , y) such that

∂ψ

∂x(x , y) = M(x , y) and

∂ψ

∂y(x , y) = N(x , y). (3)

◮ In that case, the equation (2) would reduce to

dψ

dx=∂ψ

∂x+∂ψ

∂y

dy

dx= M+N

dy

dx= 0 =⇒ ψ(x , y) = c

where c is an arbitrary constant.



Integrating Factors

Exact FODE

◮ Therefore, any solution y = ϕ(x) of (2) would satisfy theequation ψ(x , ϕ(x)) = c .

◮ If we can find such a function ψ(x , y), DE (2) would becalled an Exact DE.

◮ Question: When or how can we tell that a DE of the type(2) is exact? The Answer: Theorem 2.6.1



Integrating Factors

Exact FODE

Theorem 2.6.1(page 96):

◮ Assume M(x , y),N(x , y), ∂M∂y, ∂N∂x

are continuous on

the rectangle R :=

{

(x , y) :α < x < β

γ < y < δ

}

◮ Then the DE (2) is EXACT in R ⇐⇒

∂M

∂y=∂N

∂xat each (x , y) ∈ R . (4)



Integrating Factors

Continued

◮ That means there is a function ψ(x , y) satisfying equation

(3) :∂ψ

∂x= M,

∂ψ

∂y= N ⇐⇒ M,N satisfy equation (4).

◮ Proof. See page 97.

◮ For solving problems, it is important to follow the steps,as in the following samples or the examples in thetextbook.



Integrating Factors

Sample I Exercise 2

Consider the DE

(2x + 4y) + (2x − 2y)dy

dx= 0 (5)

◮ Determine, if it is exact.

◮ If yes, solve it.



Integrating Factors

Continued: Solution

◮ Here M = (2x + 4y) and N = 2x − 2y .

◮ So, ∂M∂y

= 4 and ∂N∂x

= 2

◮ So, ∂M∂y

6= ∂N∂x

. So, equation (5) is not exact.

◮ In the next Sample, I change equation (5) slightly, tomake it exact.



Integrating Factors

Sample I: Exercise 2 (edited)

Consider the DE

(2 + 4yx) + (2x2 − 2y)dy

dx= 0 (6)





Integrating Factors

Exercise 2 (edited): Solution

◮ Here M = 2 + 4yx and N = 2x2 − 2y .

◮ So, ∂M∂y

= 4x and ∂N∂x

= 4x

◮ So, ∂M∂y

= ∂N∂x

= 4x . So, equation (6) is exact.



Integrating Factors

Continued: Step 1

◮ We set

∂ψ

∂x= M = 2 + 4yx ,

∂ψ

∂y= N = 2x2 − 2y (7)

◮ Integrating the first one, with respect to x :

ψ(x , y) =

∫

(2 + 4yx)dx + h(y)

where h(y) is function of y , to be determined. So,

ψ(x , y) = 2x + 2x2y + h(y) (8)



Integrating Factors

Continued: Step 2

◮ Differentiating equation (8), then combining with thesecond part of the equation (7)

∂ψ

∂y= 2x2 +

dh(y)

dy= 2x2 − 2y (9)

◮ (we h(y) to be independent of x) and we have:

dh(y)

dy= −2y =⇒ h(y) = −y 2

(We do not need to have a constant).◮ Combining with equation (8), we have

ψ(x , y) = 2x+2x2y+h(y) =⇒ ψ(x , y) = 2x+2x2y−y 2



Integrating Factors

Exercise 2 (edited): Answer

◮ So, the general solution to the DE (6) is

ψ(x , y) = 2x + 2x2y − y 2 = c


◮ For each value of c , we get a solution of (6).



Integrating Factors

The Direction Field:

0 1 2 3 4 5 6 7 8 9 10

−4

−3

−2

−1

0

1

2

3

4

x

y

y ’ = − (2 + 4 y x)/(2 x2 − 2 y)



Integrating Factors

Sample II Exercise 6

Consider the DEdy

dx= −

ax − by

bx − cy(10)





Integrating Factors

Exercise 6: Solution

◮ Rewrite DE (10): (ax − by) + (bx − cy)dy

dx= 0

◮ Here M = ax − by and N = bx − cy .

◮ So, ∂M∂y

= −b and ∂N∂x

= b

◮ So, ∂M∂y

= ∂N∂x

if −b = b. So, equation (10) if b = 0.

◮ With b = 0, we rewrite DE (10) as:

ax − cydy

dx= 0, With M = ax , N = −cy . (11)



Integrating Factors

Exercise 6: Step 1

◮ We set

∂ψ

∂x= M = ax ,

∂ψ

∂y= N = −cy (12)


ψ(x , y) =

∫

axdx + h(y)


ψ(x , y) =ax2

2+ h(y) (13)



Integrating Factors

Exercise 6: Step 2

◮ Differentiate (13), with respect to y and use (12)

∂ψ

∂y=

dh(y)

dy= −cy (14)


dh(y)

dy= −cy =⇒ h(y) = −

cy 2

2

(We do not need to have a constant).



Integrating Factors

Exercise 6: Answer

◮ From (13), we have

ψ(x , y) =ax2

2+ h(y) =⇒ ψ(x , y) =

ax2

2−

cy 2

2


ψ(x , y) =ax2

2−

cy 2

2= c




Integrating Factors

Sample III Exercise 10

Consider the DE

(y

x+ 6x

)

+ (ln x − 2)dy

dx= 0 x > 0 (15)





Integrating Factors

Exercise 10: Solution

◮ Here M = y

x+ 6x and N = ln x − 2.

◮ So, ∂M∂y

= 1

xand ∂N

∂x= 1

x

◮ So, ∂M∂y

= ∂N∂x

So, DE (15) is exact.



Integrating Factors

Continued: Exercise 10: Step 2

◮ We set

∂ψ

∂x= M =

y

x+ 6x ,

∂ψ

∂y= N = ln x − 2 (16)


ψ(x , y) =

∫

(y

x+ 6x

)

dx + h(y)


ψ(x , y) = y ln x + 3x2 + h(y) x > 0 (17)



Integrating Factors

Exercise 10: Step 3

◮ Differentiate (17), with respect to y and use (12)

∂ψ

∂y= ln x +

dh

dx= ln x − 2 (18)


dh(y)

dy= −2 =⇒ h(y) = −2y

(We do not need to have a constant).



Integrating Factors

Exercise 10: Answer

◮ From (17), we have

ψ(x , y) = y ln x+3x2+h(y) =⇒ ψ(x , y) = y ln x+3x2−2y


ψ(x , y) = y ln x + 3x2 − 2y = c




Integrating Factors

Integrating Factors

Consider the FODE (2)

M(x , y) + N(x , y)dy

dx= 0 (19)

◮ Question: As in §2.1, for some FODEs, can we find anintegrating factor (IF) µ(x , y), so that when we multiplyDE (20) by µ(x , y), the new equation

µ(x , y)M(x , y) + µ(x , y)N(x , y)dy

dx= 0 (20)

becomes exact?◮ Answer: Under some conditions, we can.◮ We will not discuss IF in this section, any further.



Integrating Factors

§2.6 Assignments and Homework

◮ Read Example 1 (motivational)

◮ Read Example 2, 3 (they explain the method of solution)

◮ Suggested Problems: Exercise §2.6 (page 101) Exercise1-12

◮ Homework: §2.6 See the Homework site!


Chapter 2: First Order DE 2.6 Exact DE and Integrating...

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