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b . If equation (1) is not exact and where Q is a function of y only. Then eq (1) has anintegrating factor which also depends on y.

is solution of

c . If eq (1) is homogeneous and

then is the I.F of eq (1)

d . If eq (1) is of the form

and then is the IF

=Qdy

e y)(

0 yN xM

yN xM +1

0),(),( =dy y x xf dx y x yf

0 yN xM

yN xM

1

Q M

M N y x =

)( y )( y

Qdyd

Qdy

d =

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e . When the DE

(a, b, A and B being constants) Is not exact, then

is an integrating factor of the DE where K is anyconstant whose value is to be determined.,Example The eq is not exact.

But if we multiply it by it becomes an exactDE and its solution is

)( 11 bkBakA y x

0)( = xBdy yAdx y x ba

0 ydx xdy

2

1

x c

x y

x y

d x

ydx xdy =0}{02

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Note: The number of I F of an eq may be infinite.The following differential formulas are useful in

the calculation of certain Exact equation byinspection or Regrouping.1.

2.

3.

4.

)( xyd ydx xdy =

)(21)](

21[ 2222 y x d y x d ydx xdy

][2 x y

d x ydx xdy

=][][2 y

x d

y x

d y

ydx xdy

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5.

6.

7.

8.

][ln x y

d xy

ydx xdy =

}]{[tan 122 x y

d y x ydx xdy =

+

}]{21

[22 y x y x

d y x ydx xdy

+=

}]{[sin 122 x

yd y x x

ydx xdy =

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9.

10.

11.

12.

)][ln( y x d y x dydx +

++

]1

[22 xy

d

y x

ydx xdy

}]{21

[)( 2 y x

y x d

y x xdy ydx

+=

+

}]{21

[)( 2 y x

y x d

y x ydx xdy

+=

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Solve (1)

Soln:

(a function of x alone)

02)34( 2 = xydydx y x

2y.....6 ==

x N

and y y

M xy N y x M Here 2....34.... 2 =

x xy y y

N x

N

y

M

2 2

26 ==

2lnln22 2

x eee IF x x x

dx

==

exactnotiseq..so x N

y M

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Multiply eq (1) by IF we obtain ..(2)

Eq (2) become exact, therefore soln is

02)34( 3223 = ydy x dx y x x

y x N and y x x M Now 3223 2....34... =

y6x.....x6 22 ==

x N

and y y

M

,x2

xfromFreeconstanty

c Ndy Mdx ==

0)34(xfromFreeconstanty

223 cdydx y x x ==

234 c y x x =

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Solve

Soln: The given eq may be written as

)( 22 dx y x ydx xdy +

22 dx y x

ydx xdy

=+

2

22

2

dx x

y x x

ydx xdy

=+

12

2

dx

x

y x

ydx xdy

=+

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}]{[tan 1 dx

x

yd =

Integrate it

c}{tan1

+

x x y

c)tan( + x x

y

c)tan( + x x y

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Solve (1)

Soln:

a function of alone

0)2( 22 =dy x dx y xy

2x.....22 +

x N

and y x y

M

22 ....2.... x N y xy M Here

y x y y

x y

x y y y x

xy y y y x

M y

M

x

N

2

)2(

)2(2

)2(42

2222 2

=+

+=

+=+ =

exactnotiseq..so x N

y M

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Multiply eq (1) by IF we obtain

..(2)

Eq (2) become exact

0)2

1(2

2

= y

dy x dx

y

x

y x

x N

and y x

y M 2

.....2

,y1

2

22lnln2

2)( 12

y yeeee IF y y y

dydy y g ===

2

2

....2

1.. y x

N and y x

M Now

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Therefore soln is

xfromFreeconstanty

c Ndy Mdx ==

0)2

1(xfromFree

constanty

cdydx y x =

=

2

c y

x x =

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Solve

Soln:

Here M = y-x and N = -(x+y)

eq is not exact

dx dy

y x dx dy

x y +

)( dx dy y x x y + 0)()( =dy y x dx x y

-1.....1 ==

x N

and y

M

)( 2222 y x y xy x xy Ny Mx +

)(

122

y x

IF

+

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0])()[()(

122

=+ dy y x dx x y y x

0)

2222 =++

+

dy y x y x

dx y x

x y

0222222 =++

++dy

y x y x

y x xdx

y x ydx

2222y.constant

++ y x xdx y x ydx Mdx

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Now

]2

2

1[

1][

]1

[

22

2

2constanty ++

== y x

xdx

y x

dx y

Mdx

)ln(21

][tan221

constanty y x y

x Mdx +

= )(ln][tan 221

constanty

y x

y

x Mdx +

=

.xfromFree

cdyo Ndy =

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The solution is

)ln(21

][tan 221 c y x y x =

xfromFreeconstanty

c Ndy Mdx ==

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Solve

Soln:

eq is not exact, so

0)()2( 2222 =dy y x xy x dx y x xy y

2222 32.....62 y x xy x N

and y x xy y

M +

03

233

33223322

+

y x

y x y x y x y x Ny Mx

0)()2(232322

=dy y x y x dx y x xy

)(..

)2(..232

322

y x y x N and

y x xy M Here

+

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3331 y x

IF =

0)(3

1

)2(3

1

23233

32233

=

+

dy y x y x y x

dx y x xy y x

0)11()21( 22 =dy y y x dx

x y x

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On integration, we get

02

22 = ydy

x dx

y x dy

y x dx

02

22 = ydy

x dx

y x xdy ydx

02)()( 2 = ydy

x dx xyd xy

lnln21

c y x xy

= ln12

c y

x xy

=

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Solve

Soln:

Eq-I + Eq-II=0Solving eq-I, a=1, b=0, A=4 and B=2

IF

0)32()43( 22 =dy y x x dx xy y

0324322

= ydy x xdydx xy ydx 03324 22 = ydy x ydx xdydx xy

0)33()24( 22 =dy x dx ydydx y x

0)33()24( 210201 =dy x dx y x dydx y y x

122401211411 = k k k k bkBakA y x y x y x

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Solving eq-II, a=0, b=1, A=3 and B=3so IF is

Now

And

Solving eq (i) and eq (ii), we get

So IF

231311301311 111111

=k k k k b Bk a Ak

y x y x y x

)...(1341324 11 i k k k k =

)...(1322312 11 ii k k k k

11 k k =

y x 2=

0)32()43( 2222 =dy y x x y x dx xy y y x

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Solve

Solve

0)()4( 222 =dy y x x dx y y

0)csc2cot( =dy y y yedx e x x