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Consider the General Form of DE: M(x,y) dx + N(x,y) dy = 0

Suppose separation of variables does not hold, assume that there is a function F(x,y) such

thatM =F

x

andN =

F

y

. If M is differentiated with respect to y and N with respect to x, that

is

2 2

andM F F N F F

y y x y x x x y x y

but2 2F F

y x x y

Therefore,M N

y x

which is a necessary and sufficient condition to be an exact equation.

Exact differential equations may be solved using any of the four methods:

1. Integrable Combination2. Partial Derivatives3. Line integral4. Alternative Solution

1. Integrable Combinations

Integrable combinations consist of group of terms that forms an exact differential, thus it is

readily integrable. It may be obtained by rearranging the terms in the given DE until a group ofterms forms an integrable combination.

Some of the integrable combinations are listed below:

1. xdy + ydx = d(xy)2.

2

xdy ydx yd

x x

3.2

ydx xdy x d

y y

4.2

xdy ydx xd

y y

5. 12 2

tanxdy ydx y

dx y x

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6.2 2

1ln

2

xdy ydx x yd

x y x y

7. 2 22

2xydx x dy xd

y y

8. 2 22

2xydy y dx ydx x

9. mxm1yndx + nxm yn1dy = d(xmyn)10.mxm1yndxnxm yn1dy/(yn)2= m

n

xd

y

Examples:Test for exactness and find the general solution.

1. (2x + 3y5) dx + (3xy2) dy = 0 ans: 6xy + 2x210xy24y = c2.

(3x

2

y

6x) dx + ( x

3

+ 2y) dy = 0 ans: x

3

y

3x

2

+ y

2

= c

3. 3y (x21) dx + (x

3+ 8y3x) dy = 0 ans: 4y

2+ x

3y3xy = c

4. (4x3y

3+ 1/x) dx + (3x

4y

21/y) dy = 0 ans: x

4y

3+ ln x/y = c

2. Partial Der ivatives

Consider the equation: M(x,y) dx + N(x,y) dy = 0 (1)

If M Ny x

, then (1) is an exact equation.

Therefore, its solution is F = c where

FM

x

(2) and

FN

y

-(3)

Determine F from (2) by integrating both sides with respect to x, treating y as constant

where the usual arbitrary constant in indefinite integration is a function T(y) which is yet

unknown. To determine T(y), obtain F

y

, equate it to (3) and integrate. This time, no arbitrary

constant is needed in obtaining T(y) since one is being introduced on the right side in the

solution F = c.

Examples:Find the general solution of the differential equation.

1. (2x + 3y5) dx + (3xy2) dy = 0 ans: 6xy + 2x210xy24y = c

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2. (3x2y6x) dx + ( x3+ 2y) dy = 0 ans: x3y3x2+ y2= c

3. 3y (x21) dx + (x

3+ 8y3x) dy = 0 ans: 4y

2+ x

3y3xy = c

4. (4x

3

y

3

+ 1/x) dx + (3x

4

y

2

1/y) dy = 0 ans: x

4

y

3

+ ln x/y = c

3. Line Integrals

Consider x

a

y

bdyyxNdxyxM

0 00),(),(

where a, b = arbitrary constants for which M and N are defined.

If M and N are polynomial functions, set a = b = 0 since a polynomial function is always

defined at the origin except when the functions become undefined. In this case, try substituting

other values for a and b that will make the functions defined.

Note: In the first integral, treat y as constant.

In the second integral, replace x by the value of a.

This method is applicable only when M and N are either polynomial functions or

transcendental functions but never a combination of polynomial and transcendental functions.

Alternative Solution for Line Integral:

Example 7-3:Find the general solution of the differential equation.

1. (2x + 3y5) dx + (3xy2) dy = 0 ans: 6xy + 2x210xy24y = c2. (3x2y6x) dx + ( x3+ 2y) dy = 0 ans: x3y3x2+ y2= c3. 3y (x

21) dx + (x

3+ 8y3x) dy = 0 ans: 4y

2+ x

3y3xy = c

4. (4x3y

3+ 1/x) dx + (3x

4y

21/y) dy = 0 ans: x

4y

3+ ln x/y = c

4. Al ternative Solution

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Consider the equation: M(x,y) dx + N(x,y) dy = 0

For M: integrate x and treat y as a constantN: integrate terms without x

Example 7-4:Find the general solution of the differential equation.

1. (2x + 3y5) dx + (3xy2) dy = 0 ans: 6xy + 2x210xy24y = c2. (3x2y6x) dx + ( x3+ 2y) dy = 0 ans: x3y3x2+ y2= c3. 3y (x

21) dx + (x

3+ 8y 3x) dy = 0 ans: 4y

2+ x

3y3xy = c

4. (4x3y

3+ 1/x) dx + (3x

4y

21/y) dy = 0 ans: x

4y

3+ ln x/y = c