Order of a diff. equation

DIFF. EQUATIONSDefinition; A diff equation is an equation

involving differential coefficient and constants

2

3

or examples

cot , sin

y = x ( ) +( )

F

dyx v

dx u vdy dy

dx dx

A diff. equation which involve only one diff.coefficient and one

independent variable is called an ordinary diff. equation

for example

Ordinary Diff. equations

= sinxdy

dx2

2 sindy

x xdx

2 x 2dy

xy xdx

Order of a diff. equation The order of a diff. equation is the highest ordered

derivative present in the diff. equation

for example,

(1) sin order =1 dy

xdx

2

2 (2) = 2xy order = 2

d y

dx2

2 (3) = 2xy - x

d y dy

dx dx

Degree of a diff. equation The degree of a diff. equation is the degree of the highest

ordered derivative present in the diff. equation

for example,

(1) sin order =1 degree = 1dy

xdx

2

2 (2) = 2xy order = 2 degree =1

d y

dx

2

2 (3) = 2xy - x order = 2 degree = 1

d y dy

dx dx

32 2

2

2

1

(4) = order = 2 degree = 2

dydx

d ydx

Formation of diff. equations Diff. equation are formed by eleminating arbitrary

constants or functions present in a equation

For example

y = a cos x

a sinx = - sinxcos

dy y

dx x

cos + y sinx = 0dy

xdx

Example : y = mx

on diff.,we get dy

mdx

eliminating m y = x dy

dx

Example

y = e a cos x sinx b x

w.r.t. xDiff

e a cos x sin e - a sin x cosx xdyb x b x

dx

= y + e - a sin x cosx b x

e - a sin x cosxdyy b x

dx

w.r.t. xAgain Diff

2

2+ - y

d y dy dyy

dx dx dx

2

2 +e - a sin x cos - e a cos x sinx xd y dy

b x b xdx dx

2

22 - 2y is the req. sol

d y dy

dx dx

; y = aex xExample be On diff. we get

aex xdybe

dx

2

2

Again diff. we get

aex xd ybe y

dx

2

2 d y

ydx

222

2

2

33 2

3 2

1: Find the order and degree of the following equations.

(a) 0

(b) y = x 1

(c) 4 sin

(d)

Q

d y dyx

dx dx

dy dya

dx dx

d y d y dyxy x

dx dx dx

32 22

2

2 32

2

1 5

(e) 2 4 0

dy d y

dx dx

d y dy

dx dx

2 2 2

2 2

2 2

2 2

2 : Find the diff. equations in the following cases.

(a)

(b) y = e sin 2

(c) 1

(d) 1

(e) y

x

Q

x a y b r

a x

x y

a b

x y

a b

a b x

Equ. of first order and first degree

A diff. equ .of the form ,

or Mdx +Ndy = 0 where M ,N are function

of x,y is called a diff.equ. of first order

and first degree

dyf x y

dx

for ex.

2

or (x+y)dx + dy = 0

dyx y

dx

x

y

A diff. equation can be exprsed as all function of x and dx

on one side and all function of y and dy on another side

then we say the diff.equ.is variable seperable

for example

Variable seperable

1 = dy x

dx y

ydy = xdx

2 2 x 2dx ydy

To solve these diff. equation integrate both sides.

f dy = g dx

f dy g dx

y x

y x c for example,

(1) sin dy = sinx dx dy

xdx

Integrate both side we get

dy = sinx dx

y = - cos x + c

3 2 2 2 (2) x y ydye x e

dx

3 2 2

Sol: The given equ. can be wrriten as

x ydye x e

dx

2 3 2 e dy = y xor e x dx

2 3 2

in tegrate both side we get

e dy = +cy xe x dx 2 3 3e e

= + + c2 3 3

y x x

2 (3) xy 1 1y dy x dx

2


1 y 1

xy dy dx

x

2


1 y 1 dy = +c

xy dx

x

2 1or dy = +cy y x dx

x

3 2 2

= log x + +c3 2 2

y y x

1 cos 2

(4) 01 cos 2

ydy

dx x


1 cos 2

1 cos 2

ydy

dx x

2 2


sec = - cosec xdx + cydy

2 2or = -

cos sin

dy dx

y x

tan y = cot x + c

2 2or sec = - cosec xdxydy

2

2

2cos

2sin

y

x

2

2

cos

sin

y

x

Reduction to variable seperable

Sol: Let = z

then 1+

x y

dy dz

dx dx

or - 1dy dz

dx dx

and the equ. can be wrritten as

- 1 = cos zdz

dx

(1) cosdy

x ydx

2


1 sec dz = dx + c

2 2

z

21 or sec dz = dx

2 2

z

or 1 cos

dzdx

z

replace the value of z we get

tan = x + c2

x y

tan = x + c2

z

Exercise

2

2

Q1: x

Q2: 0

logQ3:

cos

sin sin 2Q4:

2 log 1

Q5: xy 1

x x

x

dyy y

dx

dy xy y

dx xy x

dy xe x e

dx x y

e x xdy

dx y y

dyx y xy

dx

, A diff. equ .of the form

,

f x ydy

dx g x y

2

for ex. 2

x ydy

dx x y

Homogeneous equations

,where f

,

is called a homogenious diff.equ.

nf x y yx

g x y x

or x ydy

dx x y

Step 1 : Put y = vx so that + xdy dv

vdx dx

Method of solutions

Step 2 : Change all y to vx

Step 3 : Solve the equ.as seperable variable

Step 4 : Replace the value of v as y

x

2 2 2 (1) Solve xdy

x xy ydx

2 2

2


x

x xy ydy

dx

Put y = vx so that + xdy dv

vdx dx

2 we get + x 1+ v + vdv

vdx

2 or x 1+ vdv

dx

2

integrate both side we get

= +c1+ v

dv dx

x

1 tan logv x c

2 or

1+ v

dv dx

x

1 tan logy

x cx

y tan logx x c

2 2 (2) xdy - ydx = x y dx

2 2


y x ydy

dx x

Put y = vx so that + xdy dv

vdx dx

2 2 2

we get + xdv vx x x v

vdx x

2 x 1dv

vdx

21v v

2 or

1+ v

dv dx

x

2


= +c1+ v

dv dx

x 2 log 1 log logv v x c 2 log 1 logv v xc

2 1v v xc

2 2 2y x y cx

Reduction to homogenious

where a,b,c,A,B,C are different constants.

If the given equ is in form

then we can reduce it to homogenious equ.

ax by cdy

dx Ax By C

we can reduce it to homogenious equ.

in two diff cases

1Case 2 when = a constant

a bsay

A B m

Case 1 whena b

A B

put x = X - h and y = Y - k

where h, k are arbitrary constant such that ah + bk + c = 0

and Ah + Bk + C = 0

put A = am and B = bm

where m is a arbitrary constant

to solve put ax +by = z

(1) Solve 2 = dy + dx x y dx dy


2 1 2 1 0x y dx x y dy

2 1 or

2 1

x ydy

dx x y

here a = 1 , A = 1 and b = -2 , B = -2

so we have a b

A B 1 2

= =1 1 2

put x - 2y = z

1 1 1 and the equ. reduced to - =

2 2 1

dz z

dx z

so that 1- 2 dy dz

dx dx

3 1 or =

1

dz z

dx z

1 or =2 +1

1

dz z

dx z

integrate both side we get

1 dz = dx +c

3 1

z

z

1 4 1

dz = dx +c3 3 3 1z

1 4 z + log z

3 9x c

1dz = dx

3 1

z

z

3x - 3y + c = 2log 3 6 1x y

Exercise

2 2

2 2

2 2 2

Q1: x 2

Q2: x log log 1

Q3: tan

Q4: xdy-ydx

Q5: x +xy x +y

dyxy y

dxdy

y y xdxdy y y

dx x x

x y dx

dy

dx

A diff. equ .of the form dy

Py Qdx

2 for ex. 2 3dy

x x y xdx

Linear diff. equ.

where P,Q are the function of x only or constant

is called a linear diff.equ.

or 5 2 dy

x ydx

Step 1 : write the diff.equ. in standered form.

Method of solutions

Step 2 : Calculate the integrating facter . .

. .Pdx

I F

I F e Step 3 : Multiply both sides by I.F.

Step 4 : Solution is y* I.F.= * I.F.dx + cQ

2 (1) Solve dy y

xdx x

Sol: The given equ. is in the form of

dy

Py Qdx

21 where P = and Q = x

x

I.F.= ePdx

1

edxx log = e x = x

3

int egrate both side we get

yx = +cx dx4

xy = 4

xc

3

now multiply both side by x we get

xdy

y xdx

2 (2) 1 2 cosdy

x xy xdx

2 2


2 cosx + y=

1 1

dy x

dx x x

2 2

2 cos where P = and Q =

1 1

x x

x x

I.F.= ePdx 2

2

1 exdx

x 2log 1 = e

x

2 = 1 x

2


y 1 = cos +cx xdx

2 1 y = sinxx c

2

2

now multiply both side by 1 we get

1 2 cos

x

dyx xy xdx

22 2

2 1 (3)

1 1

dy xy

dx x x

Sol: The given equ. is in standered form

22 2

2 1 where P = and Q =

1 1

x

x x

I.F.= ePdx 2

2

1 exdx

x 2log 1 = e

x

2 = 1 x

2

2


1 y 1 = +c

1x dx

x

2 1 1 y = tan xx c

2

2

2

now multiply both side by 1 we get

1 1 2

1

x

dyx xydx x

1

we can reduce it to linear equ.as multiply both sides

by y and put y = z .n n

If the given equ is in form

then we can reduce it to linear equ.

ndyPy Qy

dx

Reduction to Linear equations

2 (1) Solve x logdy

y y xdx

2


1 1 1 log

dyx

y dx xy x

2

1 1put = z so that

dy dz

y y dx dx

2

1 or we have

dy dz

y dx dx

1the equ.can be log

dz zx

dx x x

1 1loglog 1

the I.F.= e = e = e =dx xx x

x

1log 1

zx c

x x

2

the solution is

1 = log

zxdx c

x x

1 1log 1x c

xy x

Exercise

2 1

2

22

5

2

Q1: x 1 tan

1Q2:

11

Q3: sec tan

Q4:

Q5: x log

dyy x

dx

dy y x x

dx xx

dyy x x

dxdy y

ydx xdy

y y xdx

A diff. equ .is said to be exact if it is formed

by its primitive without any operations.

for ex. x 0

is a exact diff. equ . because it is formed by diff of

xy = c

dyy

dx

Exact diff. equ.

A diff. equ .Mdx +Ndy =0 is said to be exact if it is

satisfy the equ . M N

y x

Step 1 : write the diff.equ. in standered form.

Method of solutions

Step 2 : Campute M,N and satisfy the condition

= M N

y x

Step 3 : integrate M w.r.t.x keeping y as a constant

Step 4 : integrate only those term of N

which is not containing x

Step 5 : Solution is step 3 +step 4 = c

(1) Solve 12 5 9 5 2 4 0x y dx x y dy

Sol: The given equ. is in standerd form

now = 5 and = 5 M N

y x

hence = M N

y x

so M =12x + 5y - 9

and N = 5x + 2y - 4

2 2solution is 6 5 9 y 4x yx x y c

2

not cont x

and 5 2 4 dy = y 4Ndy x y y

2 12 5 9 6 5 9Mdx x y dx x yx x

(2) Solve 1 cos sin 0y ye xdx e xdy

Sol: The given equ. is in standerd form

now = and = y yM Ne cosx e cosx

y x

hence = M N

y x

so M = 1 cosye x

and N = sinye x

solution is 1 sinye x c

not cont x

and sin dy = 0yNdy e x

1 cos 1 siny yMdx e xdx e x

Exercise

2 2

2

2 2 2 3

2 2

Q1: sin 2 1 cos 0

Q2: cosxtany+cos sin sec cos 0

Q3: 3 6 6 4 0

Q4: xdy-ydx

y xdx y x dy

x y dx x y x y dy

x xy dx x y y dy

x y dx

Documents

Order of a diff. equation