69. differential equations-1

8/6/2019 69. Differential Equations-1

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Mathematics


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Session

Differential Equations - 1


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Session Objectives

Differential Equation

Order and Degree

Solution of a Differential Equation, Generaland Particular Solution

Initial Value Problems

Formation of Differential Equations

Class Exercise


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Differential Equation

An equation containing an independentvariable x,dependent variable y and thedifferential coefficients of the dependentvariable y with respect to independentvariable x, i.e.

2

2

dy d y, ,

dx dx


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Examples

dy1 = 3xydx

2

2

d y2 + 4y = 0

dx

3

3 2

3 2

d y d y dy3 + + + 4y = si x

dxdx dx

2 24 x dx + y dy = 0


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Order of the Differential Equation

The order of a differential equation is the orderof the highest order derivative occurring in thedifferential equation.

2 32 2

3 2 2

d y dy d y dyExam le : = =dx dxdx dx

The order of the highest order derivative

2

2d y is 2 .dx

Therefore, order is 2


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Degree of the Differential Equation

The degree of a differential equation is the degree of the highestorder derivative, when differential coefficients are ade free frofractions and radicals.

3322 22 22

2 2

d y dy d y dyExa le : + 1+ = 0 = 1+

dx dxdx dx

The degree of the highest order derivative is 2.

2

2d ydx

Therefore, degree is 2.


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Example - 1

Deter ine the order and degree of the differential

equation:d d

= +a 1+d d

.

2dy dy

Sol tion: W y = x + a 1+dx dx

2dy dy

y - x = a 1+dx dx

2 22dy dyy - x = a 1+

dx dx


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Solution Cont.

2 22 2 2 2dy dy dyy - 2xy + x = a + a

dx dx dx

The order of the highest derivative isand its degree is 2.

dydx


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Example - 2

Deter ine the order and degree of the differential

equation:

324 2

4

d y dyc

dxdx

!

Solution: e have

3

322 24 42

4 4

d y dy d y dy= c + = c +

dx dxdx dx

Here, the order of the highest order is 4

4

4

d y

dx

and, the degree of the highest order is 24

4

d y

dx


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Linear and Non-LinearDifferential Equation

A differential equation in which the dependent variable y and

its differential coefficients i.e. occur only in the

first degree and are not ultiplied together is called a lineardifferential equation. Otherwise, it is a non-linear differentialequation.

2

2

dy d y, ,

dx dx


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Example - 3

is a linear differential equation of order 2 and degree .

is a non-linear differential equation because the dependent

variable y and its derivative are ultiplied together.dy

dx

2

2

d y dyi - 7y 4x

dxdx

dyii y - 4 xdx


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Solution of a Differential Equation

The solution of a differential equation is the relationbetween the variables, not taking the differentialcoefficients, satisfying the given differential equation andcontaining as any arbitrary constants as its order is.

For exa ple: y Acosx - Bsinx

is a solution of the differential equation

2

2

d y4y 0

dx


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General Solution

If the solution of a differential equation of nth order contains narbitrary constants, the solution is called the general solution.

is the general solution of the differential equation 22

d y y 0dx

y B sin x!

is not the general solution as it contains one arbitrary constant.

y Acosx - Bsinx


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Particular Solution

A solution obtained by giving particular values to the arbitraryconstants in general solution is called particular solution.

y 3 cos 2 si!

is a particular solution of the differential equation

2

2

d yy 0.

dx


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Example - 4

3 2Solution: e have y x ax bx c (i)

2dy 3x 2ax b (ii) Differentiating i w.r.t. xdx

2

2

d y6x 2a (iii) Differentiation ii w.r.t. x

dx

3 2

3

3

Verify t at y x ax bx i a olution of t e

d ydifferential e uation 6.

dx


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Solution Cont.

3

3

d y= 6 Diff r ntiating iii .r.t x

dx

3

3

d y= 6 is a diff r ntial ation of i .

dx

@


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Initial Value Problems

The proble in which we find the solution of thedifferential equation that satisfies so e prescribedinitial conditions, is called initial value proble .


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Example - 5

2x x

2

dy d ye , e

dx dx

xy e@ satisfies the differential equation2

2

d y dy- 0

dxdx

Show that is the solution of the initial value

proble

xy e

2

2

d y dy- 0 y 0 2 y' 0

dxdx

xSolution : We have y e


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Solution Cont.

0 0

x

0

dyy 0 e and e

dx

y 0 2 and y 0 1 !

x

y e@is the solution of the initial value proble .

x dyy e anddx

xe@ !


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y x

Assu e the fa ily of straightlines represented by

dy

dx

dy y

dx x

dyx ydx

@ !

!

is a differential equation of the first order.

X

Y

O

U m = tanU


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Assu e the fa ily of curves represented by

where A and B are arbitrary constants.

y = A co s x + (i)

dy A si ... iidx

@ ! [Differentiating (i) w.r.t. x]

2

2

d ya n d A co s x B

d x

! [Differentiating (ii) w.r.t. x]


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2

2

d yy

dx ! [Using (i)]

2

2

d yy 0

dx

is a differential equation of second order

Si ilarly, by eli inating three arbitrary constants, a differentialequation of third order is obtained.

Hence, by eli inating n arbitrary constants, a differentialequation of nth order is obtained.


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Example - 6

For the differential equation of the fa ily of curves

a and c being para eters. y = a sin bx + c ,

Solution: e have y = a sin bx + c

is the required differential equation.

2 22 2

2 2

d y d y= -b y + b y = 0

dx dx

[Differentiating w.r.t. x] dy = ab cos bx + cdx

[Differentiating w.r.t. x] 2

2

2

d y= -ab sin bx + c

dx


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Example - 7

Find the differential equation of the fa ily of allthe circles, which passes through the origin andwhose centre lies on the y-axis.

If it passes through (0, 0), we get c 0

2 2x y 2gx 2y 0@

This is an equation of a circle with centre (- g, - f)and passing through (0, 0).

Solution: The general equation of a circle is

2 2x y 2gx 2y c 0.


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Solution Cont.

Now if centre lies on y-axis, then g 0.

2 2x + y + 2y 0 (i)@

This represents the required fa ily of circles.

dyx y

dxdy

dx

!

dy dy2x 2y 2 0 Differentiating i w.r.t. xdx dx


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Solution Cont.

? A2 2

dyx y dxx y - 2y 0 Substituting the value of f

dy

dx

2 2 2dy dy

x y - 2xy - 2y 0dx dx

2 2 dyx - y - 2xy 0dx


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Thank you

69. differential equations-1

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