Ordinary Differential Equation

Presented by:-1.Patel Nirmal 1508601020202.Patel Yash 150860102021 3.Prajapati vivek 1508601020224.Savant Nikhil 1508601020235.Tandel mehar 1508601020266.Vaghdodiya kalpesh 150860102027

Definitions Differential equation is an equation involving an

unknown function and its derivatives.Ordinary Differential equation is differential

equation involving one independent variable and its differentials are ordinary.

Partial Differential equation is differential equation involving two or more independent variables and its differentials are partial.

Order of Differential equation is the order of the highest derivative appearing in the equation.

Degree of Differential equation is the power of highest derivative appearing in the equation.

particular solution of a differential equation is any one solution.

The general solution of a differential equation is the set of all solutions.

Solutions of First Order Differential Equations

1- Separable Equations2- Homogeneous

Equation3- Exact Equations4- Linear Equations5- Bernoulli Equations

1- Separable Equations (separation variable)

General form of differential equation is (x ,y) dx + (x ,y) dy = 0By separation variable Then 1 (x) 2 (y) dx + 1(x) 2(y) dy = 0

by integrating we find the solution of this equation.

Ex) find general solution for

0)()(

)()(

2

2

1

1 dyyydx

xx

cxy

dyy

dxx

dyxxydx

lnln

nintegratioby 1102

2- Homogeneous EquationThe condition of homogeneous function is

f (x , y) = f (x ,y)and n is Homogeneous degree

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

and , is Homogeneous function and have the same degree

so the solution is put y = xz , dy = x dz + z dx and

substituting in the last equation

the equation will be separable equation, so separate variables and then integrate to find the solution.

n

3- Exact Equations

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

only. timeonefactor repeated take: note

issolution general its and

isequation exact be oequation t ofcondition required The

cdydx

yx

dyy

dxx

yx

xy

)( exp)(

)( exp)(

1

1

. following asfactor integralby it multiply exact be it toconvert to

exact,not be illequation w The

if yx

integral factor is

Examples i ) (2x + 3cosy) dx + (2y – 3x siny) dy = 0Solutionit is exact so, (2x + 3cosy) dx = x + 3x cosy (2y - 3x siny) dy = y + 3x cosyThe solution is x + 3x cosy + y = c

ii) (1 – xy) dx + (xy – x ) dy = 0

exactnot its So,

2)()1( 2

xyy

xxyxxxy

dxxyxxThen

dxx

yxy

xy

)2( exp)(

)( exp)( Since

)(1

1

cyxyx

cdyxydxy

dyxydxyx

x

xxdx

x

x

2ln

)(

exact isequation s thi0)(1

ii)equation by value thisgmultiplyin(by 1

lnexplnexpexp

1

11

note :- we took the repeated factor one time only

4- Linear Equations Linear Equation form is

the integral factor that convert Linear Equations to exact equation is :-

= exp p(x) dx by multiplying integral factor by Linear Equation

form

so the general solution is :- y = Q dx + c

)()( xQ y x Pdxdy

exact isequation his t)()( xQ y xP dxdy

cxx

dxxxx

dxxxyxx

cdxQy

issolutiongeneralxx

xxxdx

xxQxxpxQyxpdxdy

solutiondxdyyxxyyEx

sin21sin

)cossin(cos

cos).tan(sec)tan(sec

tansec

tanseclnexpsecexp

cos)( , sec)( )()(

, cossec )

2

2

2

5- Bernoulli Equation Bernoulli Equation form is

nyxQ y x Pdxdy )()(

before. told weassolution its andequation linear is this

)()1()()1(

)()()1(

1

)1( then Put b)

)()(1yover Equation Bernoulli Divide a)

Equation Bernoulli solve To

)1(

)1(

n

xQnzxpndxdz

xQxpdxdz

n

dxdyyn

dxdzyz

xQ yx Pdxdy

y

n

nn

note :- if n = 0 the Bernoulli Equation will be linear equation.

if n = 1 Bernoulli Equation will be separable equation

the general solution will be

2

22

1

2

lnexpln2expexpexp

linear isequation thissin62

2

.sin3

solution

sin3 )

x

xxdxpdx

x xz

dxdz

dxdyy

dxdzthen y and put z

y xxy –

dxdy

xxy –

dxdyy Ex

x

c x x x x y x

cxdx x y x

)cossincos(6

sin622

22

Solution of 1st order and high degree differential equation :-

1- Acceptable solution on p. 2- Acceptable solution on y. 3- Acceptable solution on x. 4- Lagrange’s Equation. 5- Clairaut’s Equation. 6- Linear homogeneous differential Equations with

Constant Coefficients. 7- Linear non-homogeneous differential Equations

with Constant Coefficients.

1- Acceptable solution on pif we can analysis the equation then the

equation will be acceptable solution on p

equation. theofsolution general theis thisand0))((

0 0

lnln2ln lnlnln

2

02 0

02 002

023 )

22

1

22

1

21

222

cyxcxy

cyxorcxy

cxyorcxyx

dxy

dyorx

dxy

dy

ydxdyxory

dxdyx

yxporyxpyxpyxp

SolyxpypxEx

2- Acceptable solution on yIf we can not analysis the equation then the

equation will be acceptable solution on y or x

firstly , to solve the equation that acceptable solution on y

there are three steps :-1- Let y be in term alone . 2- By differentiation the equation with respect to x

and solve the differential equation .3- By deleting p from two equations (the origin

equation and the equation that we got after second step) if we can not delete it the solution called the parametric solution .

2

2

2

2

3212

32

32

xrespect to ation withdifferentiby 32

32

, 223 )

xp

dxdp

xp

dxdpxp

dxdy

xppxy

Solutiondxdy p

xppxyEx

22

2

2

2

by gmultiplyin , 222

3by gmultiplyin , 34

32

32

31

xdxdp

xpx

xpp

dxdp

xpx

xpp

3

2

2

2

2

22

322

61

equationorigin on pabout ngsubstitutiequation twofrom p delete to

ln21ln

332

2 2

2 2

02 02

02)2(

)2(2)2(

)2(22

xy

xpxpcxy

xdx

pdpdxxdy

dxdpxpx

dxdp

dxdpxporpx

dxdpxppx

dxdppxxpxp

dxdppxxppx

3- Acceptable solution on x

secondly, to solve the equation that acceptable solution on x

there are three steps :-1- Let x be in term alone . 2- By differentiation the equation with respect to y

and solve the differential equation .3- By deleting p from the two equations (the origin equation and the equation that we got after

second stepif we can not delete it the solution called the

parametric solution .

solution. parametric thethis so equations last tow thefrom p deletenot can we

equation)origin (the 43

21

)3(

)31(1

)31(1

1but , 3

y respect to ation withdifferentiby

, )

3

42

3

2

2

2

3

ppx

ppy

dpppdy

ppdydp

dydpp

p

dydx

pdydpp

dydp

dydx

dxdypppxEx

4- Lagrange’s Equation

Lagrange’s Equation form y = x g (p) + f (p)

cpxp

dppxppe

pdp

px

dpdx

px

dpdx

dxdp

px

dxdppxp

dxdppxp p

dxdpp

dxdpx p

dxdy

pxp y

p

32

2

factor integral 2 exp

equation aldifferentilinear 22

22)22(1

)22()22(2

222

2 Ex)

32

222ln2

5- Clairaut’s EquationClairaut’s Equation is special case of Lagrange’s

EquationClairaut’s Equation form :-

y = x p + f (p)

0)(0

0

)

2

2

2

paxr o

dxdp

dxdp

pax

dxdp

paxpp

dxdp

pa

dxdpx p

dxdy

pa p x y Ex

(parabola)solution single 4

22

2

2

22

2222

2

2222

axy

ax a x a , yax pa xp y

papx

pa x p y

ca x cy

xap & c p

6 - Linear homogeneous Differential Equations with Constant Coefficients

L(D) y = f (x) non-homogeneous but L(D) y = 0 homogeneous then L() = 0 assistant equation

Roots of this equation are 1 , 2 , 3 ,……,n

This roots take different forms as following:-

constant are ,.....,,, ,

)()........(

3210

22

110

n

nnnn

aaaaadxdD

xfyaDaDaDa

1- if roots are real and different each other then the complement solution is

xn

xxc

neCeCeCy .........2121

2- if roots are real and equal each other then complement solution is

3- if roots are imaginary then complement solution is

).........( 121

rn

xc xCxCCey

)sincos( 21 xCxCey xc

Examples:-

xxc eCeCCy

yDD

yy

321

321

23

3

1,1,00)1)(1(

0)1(0)(

0)(

0)1

xxc eCeCy

yDD

yyy

221

21

2

2,10)2)(1(

0)23(0)23(

023)2

axCaxCyaia

yaD

yay

c sincos

0)(

0)(

0))(3

21

22

22

xCxCeCy

i

yDD

xc sincos

, 10)1)(1(

0)22)(4

321

21

2

2