Diifferential equation akshay

DIFFERENTIAL EQUATION
• An equation which involves differential
co-efficient is called differential equation.
• Ex =
1) dy/dx=1 + /1 +
2) dy/dx= 2(x + y)
3) y/d + y = 0

ORDER OF A DIFFERENTIAL EQUATION
The ORDER of a differential equation is the order of the
highest differential co-efficient
Present in the equation.
22
1
yxdx
dy


1)( 43
2
2
 y
dx
yd
0)( 223
 ynxyxyx
For ex =
1)
2)
3)
Order=2
Order = 1
Order = 2

DEGREE OF A DIFFERENTIAL EQUATION
The DEGREE of a differential equation is the
Degree of the highest derivative after
Removing the radicals and sign and fraction.
22
1
yxdx
dy


1)( 43
2
2
 y
dx
yd
0)( 223
 ynxyxyx
For ex =
1)
2)
3)
degree=1
degree = 1
degree = 3

Exact Differential Equation
• An exact differential equation is formed by
directly differentiating its primitive (solution)
Without any other process.
Mdx +Ndy =0
Is said to be an exact differential equation if it
satisfies the following condition
0)3()3( 3223
 dyyyxdxxyx

Where ∂M/ ∂y denotes the differential co–
efficient of M with respect to y keeping x
constant.
AND
∂N/ ∂x denotes the differential co-efficient of N
with respect to x keeping y constant.
∂M/∂y = ∂N/∂x

Method for solving exact differential
equation
Step 1) integrate M with respect to x keeping y
constant
Step 2) integrate with respect to y , take only those
term of N which do not contain x
Step 3) result of step1 + result of step 2 = constant

0)3()3( 3223
 dyyyxdxxyx
Exactxy
x
N
y
M
xy
x
yyx
xy
y
xyx
,6
6
3
6
3
32
23













EXAMPLE :-
M = 23
3xyx  N = 32
3 yyx 
∂M =
∂N =

1
4
32
4
)(
3
)(
c
y
yk
yyxN
yk


)(
2
3
4
1
)()3(
)(
224
23
ykyxx
ykdxxyx
ykMdxu





In integration of dx taking y as constant
And
in integration of dy do not take those term which
contain x.

cyyxxyxu  )6(
4
1
),( 4224
Solution:-

If
N
, DE is not exact.
x
M
y
 

 
For making it exact

Equation reducible to the exact
equation
Sometimes a differential equation which
is not exact may become so, on
multiplication by a suitable function
known as the integrating factor.
INTEGRATING FACTOR

RULES FOR EXACTNESS IN
DIFFERENTIAL EQUATION
THERE ARE FIVE RULES FOR EXACTNESS IN
DIFFERENTIAL EQUATION :-
1 ) RULE1:- IF (∂M/∂y-∂N/∂x)/N is a function of x
alone , say f(x) , then I.F=
2) RULE2 :- IF (∂N/∂x - ∂M/∂y)/M is a function y
alone , say f(y) , then I.F =

3) RULE 3 :- IF M is of the form M=y f1(xy) and N is of
the form N=x f2(xy) then I.F = 1/(Mx - Ny)
4) Rule 4:- IF the given equation Mdx +Ndy =0 is
homogenous equation and Mx + Ny ≠ 0 , then
1/(Mx + Ny) is an integrating factor .
5) RULE 5 :- xᵐyᵑ(ay dx + bx dy)+ xᵐˡyᵑˡ(a'y dx + b'x dy)= 0
the integrating factor is xʰyᴷ.
where (m+h+1)/a = (n+k+1)/b, and
(mᶦ+h+1) / aˡ = (nᶦ+k+1) / bᶦ

RULE 1
IF (∂M/∂y-∂N/∂x)/N is a function of x
alone , say f(x) , then I.F=
EXAMPLE 1:-
SOLVE 2y dx + (2x logx –xy)dy = 0
M=2y ,N = 2x logx –xy
∂M/∂y=2 , ∂N/∂x = 2(1+logx) –y
Here (∂M/ ∂y - ∂N / ∂x ) /N

= [2 – {2(1 + logx) –y } ] / (2x logx - xy)
= (- 2 logx + y) / (2x logx - xy)
= - 1/x = f(x)
= I.F = = = = =
=
On multiply the given differential equation by , we get
= 2y dx + (2 log x -y ) dy = 0

= ∫ dx + ∫ - y dy = c
= 2y log x - ½ = c
EXAMPLE 2 :-
SOLVE (y - 2 ) dx + (-x + y)dy = 0
M = (y - 2 ) , N = (-x + y)
∂M/∂y = 1 , ∂N/∂x = 2xy – 1
Here (∂M/ ∂y - ∂N / ∂x ) /N

= (1 - 2xy +1)/ ( y –x)
= (-2xy + 2) / ( y –x)
= -2 ( xy - 1 ) / x ( xy - 1 )
= -
= I.F = =
=
= = = = F(x)

= On multiply the given differential equation by ,
we get
= ( y/ -2x ) dx + (- + y )dy = 0
= ∫ ( y/ - 2x ) dx + ∫ (y)dy = c
= -y/x - + /2= c

RULE 2
IF (∂N/∂x - ∂M/∂y)/M is a function y
alone , say f(y) , then I.F =
EXAMPLE 1 :-
SOLVE ( +2y)dx + (x +2 - 4x)dy = 0
M = ( +2y) , N = (x +2 - 4x)dy
∂M/∂Y = 4 +2 , ∂N/∂x = - 4

Here (∂N / ∂x - ∂M / ∂y ) / M
= [( - 4) - ( 4 + 2)] / ( + 2y)
= - 3( + 2)/y ( + 2)
=
= I.F = =
= =
= = = f(y)

= On multiply the given differential equation by ,we
get
= (y + ) dx + (x + 2y - ) dy = 0
= ∫ (y + ) dx + ∫ 2y dy = c
= xy + x + = c

EXAMPLE 2:-
SOLVE ( + y) dx + (- ) dy
= M = ( + y) , N = (- )
= ∂M/∂Y = 2 y + , ∂N /∂x = -
= Here (∂N / ∂x - ∂M / ∂y ) / M
= (- - 2 y - )/ ( + y)
= -2 ( y + ) / y( y + )

=
= I.F = =
= =
= =
= On multiply the given differential equation by ,we
get
= ( + ) dx + ( ) dy = 0

RULE 3
IF M is of the form M=y f1(xy) and N is of the form N=x
f2(xy) then I.F = 1/(Mx - Ny).
EXAMPLE 1 :-
SOLVE y(xy + 2 ) dx + x (xy- )dy = 0
= ( x + 2 )dx + ( y - ) dy = 0
= M = ( x + 2 ) , N = ( y - )
= ∂M /∂y = 2xy + 6 , ∂N /∂x = 2xy - 3

I.F = 1/ ( Mx - NY)
= 1/ {( x + 2 ) X - (y - )Y}
= 1 / (3 )
we get
= ( + ) dx + ( - ) dy = 0
= ∫ ( + ) dx + ∫ ( ) dy = C

= log x + yx + logy = C
EXAMPLE 2 :-
= SOLVE y ( 1 - xy ) dx - x ( 1 + xy ) dy
= ( y - x ) dx + ( - x - y) dy
= M = (y – x ) , N = (-x - y)
= I.F = 1/ { (y – x ) x - ( - x - y) y }
=

we get
= ( - )dx + (- - )dy = 0
= ∫ ( - ) dx + ∫ ( - ) dy = C
= log x - x - logy = C
= log - yx = C

RULE 4
IF the given equation Mdx +Ndy =0 is homogenous
equation and Mx + Ny ≠ 0 , then 1/(Mx + Ny) is
an integrating factor .
EXAMPLE 1 :-
SOLVE ( + ) dx - (x ) dy = 0
M = ( + )dx , N = ( -x )dy
I.F = 1/ { ( + ) x + (- x ) y }

= 1/ { ( + x ) + (-x )}
=
we get
= ( + ) dx - ( ) dy = 0
= ∫( + ) dx - ∫ ( 0) dy = C
= log x - = C

EXAMPLE 2:-
SOLVE ( y – 2x ) dx - ( - 3 y) dy
= M = ( y – 2x ) dx , N = (- + 3 y ) dy
= I.F = 1/ { ( y – 2x )x + (- + 3 y ) y }
=
we get
= ( y – 2x ) dx + (- + 3 y ) dy

= ( - )dx + (- + ) dy = C
= ∫ ( - ) dx + ∫ dy = C
= - 2log x + 3 log y = C

RULE 5
IF xᵐyᵑ(ay dx + bx dy)+ xᵐᶦyᵑᶦ (aᶦy dx + bᶦx dy)= 0
the integrating factor is xʰyᴷ.
where (m+h+1)/a = (n+k+1)/b, and
(mᶦ+h+1) / aᶦ = (nᶦ+k+1) / bᶦ
EXAMPLE 1 :-
SOLVE ( - 2 y )dx + ( 2x - )dy = 0
= ( dx + 2x dy ) + ( - dy - 2 y dx ) = 0
= ( y dx + 2x dy ) + ( - x dy – 2y dx ) = 0

= comparing the coefficient , we get
= xᵐyᵑ(ay dx + bx dy)+ xᵐᶦyᵑᶦ (aᶦy dx + bᶦx dy)= 0
= m= 0 , n =2 , a=1 , b=2 ,mᶦ =2 , nᶦ=0 , aᶦ = -2 , bᶦ = -1
= ( 0 +h+1)/1= ( 2+k+1)/2 and ( 2+h+1)/-2 = (0 + k+1)/ -1
= 2h + 2 = 2+ k + 1 and h + 3 = 2k + 2
= 2h – k =1 ----------------------- 1)
and
= h – 2k = -1-------------------------2)

= on solving equation 1 and equation 2 , we get
= h=1 , k=1
= I.F =
= = xy
= On multiply the given differential equation by xy ,
we get
= (x - 2 ) dx + ( 2 - y )dy = 0
= ∫ ( x - 2 ) dx = C

= - = C
EXAMPLE 2 :-
SOLVE (3y – 2x ) dx + ( 4x – 3 ) dy = 0
= (3y – 2x ) dx + ( 4x – 3 ) dy = 0
= ( 3y dx + 4x dy) + ( - 3 dy- 2x dx) = 0
= comparing with the cofficient
= xᵐyᵑ(ay dx + bx dy)+ xᵐᶦyᵑᶦ (aᶦy dx + bᶦx dy)= 0 , we get

= m= 0 , n=0 , a=3 , b=4 , mᶦ=1 , nᶦ =2 , aᶦ = -2 ,bᶦ = -3
= ( 0+ h+1 )/3 = (0+k+1)/4 and ( 1+h+1)/-2 = (2+k+1)/-3
= (h+1)/ 3 = (k+1)/ 4 and ( h+2)/ 2 = ( k+3) / 3
= 4h - 3k +1 = 0 -----------------------------1)
and
= 3h - 2k = 0 -------------------------------2)
= on solving we get h=2 , k=3

= I.F =
=
we get
= ( 3y - 2x ) dx + (4x – 3 )dy = 0
= ( 3 - 2 ) dx + (4 -3 ) dy = 0
= ∫(3 - 2 ) dx = C
= - = C

Diifferential equation akshay

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