Differential Equations

Unit-7

Exact Differential Equations: 0M d x N d y+ =

Verify the condition M N

y x

∂ ∂=

∂ ∂

Then integrate M d x with respect to x as if y were constants,

then integrate the terms in N d y which do not contain terms in x and equate sum of

these two integrals to a constant.

Example: Solve

2 2 2

2 2 2 2 2

( 2 ) ( ) 0

2 ; ( ) 2

2 2 ; 2 2

a xy y dx x y dy

M a xy y N x y x y xy

M N M Nx y x y

y x y x

− − − + =

= − − = − + = − − −

∂ ∂ ∂ ∂∴ = − − = − − ∴ =

∂ ∂ ∂ ∂

∴ The given equation is an exact differential equation.∴ Solution is given by

2 2 2( 2 ) ( ) 0a x y y d x y d y− − + − =∫ ∫

2 2 2 33 3 3 3a x x y y x y c⇒ − − − =

Example:

Solve ( 1 ) c o s s i n 0y y

e d x e x d y+ + =

cos ; cosy yM N

e x e xy x

∂ ∂= =

∂ ∂

Solution is : ( )1 cosy

e x dx c+ =∫

( ) cxey =+ sin1

Integrating Factors:

Sometimes a differential equations which is not exact may become exact on multiplication by a

suitable function known as an Integrating Factor.

For example the differential equation 0ydx xdy− =

is not exact, but when multiplied by 2 2

1 1 1o r o r b y

y x x y

it becomes on exact equation. The number of integrating factor is infinite for a given equation.

Rules For Finding The Integrating Factors::

Rule I: when Mx + N y ≠ 0 and the equation is homogeneous then NyMx +

1 is an Integrating

factor of Mdx + Ndy = 0

Example: Solve: ( ) 0332 =+− dyyxydxx

Solution: Mx+Ny=-y4 I.F.=-1/y4

Multiply Given equation By I.F. 0

14

3

3

2

=

++− dy

yy

xdx

y

x

On integration we have cyy

x=+− log

3 3

3

Rule II: When Mx - Ny ≠ 0 and the equation has the form f1(xy)ydx+f2(xy)xdy=0

then Integrating factor is NyMx −

1

Example: Solve ( ) ( ) 011 =−++ xdyxyydxxy

Mx - Ny = 2x2y2 I.F.=1/2x2y2

Multiply given equation by I.F. we get 22yx

xdyydx

y

dy

x

dx ++− on intergration

cxyy

xlog

1log =−

Rule III: WhenN

x

N

y

M

∂

∂−

∂

∂

is a function x alone say f(x) then ∫ dxxf

e)(

is an Integrating factor

Problem:

Solve (x2+y2+2x)dx + 2ydy = 0

x

y

ye

N

x

N

y

M

=∫===∂

∂−

∂

∂

1dx

2

2eI.F. 1

Multiplying given equation by I.F. and it becomes exact

Solution of given equation is given by ( ) cdxxyxe x =++∫ 222

ceyexxx =+ 22

Rule IV: WhenM

y

M

x

N

∂

∂−

∂

∂

is a function y alone say F(y) then ∫ dyyF

e)(

is an Integrating

factor

Example: solve ( ) ( ) 0422 434 =−+++ dyxyxydxyy

( )( ) yyy

y

M

y

M

x

N

3

2

232

2 −=

+

+−=

∂

∂−

∂

∂

3

)( 1..

−=∫=

yeFI

dyyF

Multiply given equation by I.F. and it becomes

04

22

32=

−++

+ dy

y

xyxdx

yy

Solution is given by cyxy

y =+

+ 2

2

2

Linear Differential Equation:

A differential equation is said to be linear if the dependent variable and its derivatives appear

only in the first degree.

The first order linear differential equation is of the form QPydx

dy=+ where P and Q are

functions of x only or constants.

Solution of Linear differential equation: ( ) ( )

∫=

+= ∫Pdx

eI.F.

....

where

cdxFIQFIy

Example: Solve 241

=+ ydx

dy

xgiven the boundary conditions x =0 when y = 4.

Solution : Rearranging gives xxydx

dy24 =+

Integrating factor

22 xpdx

ee =∫.

Solution gives: ( ) .2

12

222 222cedxxeye

xxx +== ∫

Using B.C. Obtain C=7/2.Hence particular solution gives ( ).712

1 22 xey

−+=

Example: Solve the differential equation θθθ

tansec yd

dy+=

given the boundary conditions y = 1 when θ = 0.

Solution: Rearranging gives ( ) θθθ

sectan =− yd

dy

Integrating factor ( ) θθ coscosln ==∫ ee

pdx

.

Solution gives: ( ) .seccoscos cdy +== ∫ θθθθθ

When θ = 0, y = 1, thus c=1. Hence particular solution is ( )

.cos

1

θ

θ +=y

Example: Solve the differential equation 2xx

y

dx

dy=+

Solution:

Integrating factor xee xpdx

==∫ ln.

Solution gives: c

xy

dxxxy

+=

= ∫

4

x

3

2

Equations Reducible to linear form Or (Bernoulli’s equation):

nQyPy

dx

dy=+ Dividing by yn

( )11 −−−−−=+ +−− QPydx

dyy nn

put v=y-1+n we get ( ) ( )

dx

dv

dx

dyyn

n =+− −+− 111

i.e ( )

( ) dx

dv

ndx

dyy n

1

1

+−=−

( )QPv

dx

dv

n=+

+− 1

1

( ) ( )QnPvndx

dv11 +−=+−+

Which is linear in v

Example: Solve ( )2y

x

y

dx

dy=+

Solution: ( ) ( ) ( )11

112 −−=+ −−

xy

dx

dyy

Put y-1=v , ( )

dx

dv

dx

dyy =− 2

1

From (1)

11

.

11

−=−

=+−

xv

dx

dvei

xv

dx

dv

which is linear in v

Integrating factor x

eedx

xpdx 1

1

=∫

=∫ −

Solution is cdxxx

v +−= ∫1

11

cxxy

+−= log1

Example: solve xyxydx

dy 22 tantan2 =−

Solution: ( )12

tantan1

22

−−=−

−−

xydx

dyy

Put y-1=v dx

dv

dx

dyy =− − 2

Fron (1) xxvdx

dv 2tantan2 −=+ which is linear in v

Integrating factor xx

exdx

epdx

e2

secseclog2tan2

==∫

=∫

∴ Solution is cxdxxxv +∫ −=2

sec2

tan2

sec

cx

xy +−=−

3

tansec

321

Orthogonal trajectory

Introduction:

The word “Orthogonal” comes from the Greek “right angle” and the word “trajectory”

comes from Latin “cut-across”. Hence the curve that curve that cut across the other at

right angle is called orthogonal trajectory.

Rules to find the equation of orthogonal trajectories of a family of cartesian

curves:

• Diff f(x ,y , c)=0 and eliminate c.

• Resulting equation gives the DE of the family f(x ,y , c)=0

• i.e.

• Replace

• The differential equation of orthogonal trajectory is

• Integrate to get the equation of the required orthogonal

trajectory

0dxdy

y.x,F =

0y,x,Fin dvdx-by =

dx

dy

dxdy

0dydxy,-x,F =

0,,F =−

dydxyx

Rules to find the equation of orthogonal trajectories of a family of polar curves:

Let the equation of given family of curves be f(r , θθθθ , c)= 0

• Diff f(r , θθθθ , c)= 0 and eliminate c.

• Resulting equation gives the DE of the family f(r , θθθθ , c)= 0

• i.e.

• Replace

• The DE of OT is

• Integrate to get the equation of the required

orthogonal trajectory

Example :Find the orthogonal trajectories of parabolas y=c x2

Solution: Differentiate given equation we get

Replace we get is the DE of OT.

On integration we get

Example: Find the orthogonal trajectories of the series of hyperbolas xy= k2

Solution: Differention given equation w.r.t x we get y + x y1=0 Replace y1 by

The differntial equation of the OT is

0ddr.r,F =

θθ

0,r,Fin drd2r-by =

θθθ

θ d

dr

ddr

0dr

d2,-rr,F =

θθ

02,,F =−

drdrr θθ

x

yx

x

yCx

dx

dy 2

22 2 ===

dydx

dxdy −= y

xdxdy

2−=

−= V.S 2 dxxdyy

2

22 cxy =+

1

1y

−

0x-or 0 =−⇒−=−⇒−==− ydyxdxydyxdxydy

dx

dy

dxxy

Which the required OT

Example: Find the orthogonal trajectories of circles

x2+(y-c)2=c2 (1)

solution: Differentiation w.r.t x , 2x + 2(y-c)y1=0

or x + (y-c)y1 = 0 (2)

From (1) x2+y2 – 2cy = 0 or c = (x2+y2) / 2y Substitute this in (2) we get

(3)

Replace y1 by The differential equation of the OT is

(4)

(5) Put y2/2 = z Diff w. r . t. x

Equation (5) becomes

linear in z I.F.=1/x solution is

1222 x

2

2

2

2ccyc

yx ==−⇒=−

012

22or 0

12

22=−+=+−

+ yyxy

xyyyx

yx

222xy

1y x

12

22 x -

12

22

yxy

yyx

yy

xy

−=⇒=

−⇒=−

1

1y

−

0221

2 12

22

22

2

1

1 =+−=−

⇒−

=− xyxyyoryxy

xy

yx

xyy

y

x

x

yxy

dx

dyxy

22dx

dy2x or 2xy by dividen 0222

−=−⇒=+−

22

2

dx

dy x

x

yy

−=−

dx

dzy =

dx

dy

2dx

dz x

x

z −=−

0222yor 22

2y

or 22

11=−++−=+−=∫ +−= cxxc

x

xc

xcdx

xz

• Or be the equation of OT

Example: The orthogonal trajectories of cardiod r=a(1+cos(θθθθ)) where a is a parameter.

Solution: Given equation r = a(1+cos(θθθθ)) Diff. w.r.t. θθθθ

Which is the DE of family r=a(1+cos(θθθθ))

Replace Integrate this we get

on simplification r = c(1- cos(θθθθ))

( ) 222 cycx =−−

θθθ

θ ddra

ddr

sin1 a sin −=⇒−=

( )

−=−

=+

−=⇒+−=2

tan

22cos2

2cos

2sin2

cos1sincos1

sin1 θ

θ

θθ

θθ

θθ

θθr

rr

ddr

ddrr

θθθθ

dr

drddr

=⇒=2

cot2

cot

cr +=

21

2sin

loglogθ