Mathematics

Session

Differential Equations - 3

Session Objectives

Linear Differential Equations

Differential Equations of Second Order

Applications of Differential Equations

Class Exercise

Linear Differential Equations

The standard form of a linear differential equation of

first order and first degree is

where P and Q are the functions of x, or constants.

dy+Py = Q

dx

x

2

dy dyExamples : 1 +2y = 6e ; 2 +ytanx = cosx ;

dx dxydx

3 + = y etc.dy x

Linear Differential Equations type-1

dyRule for solving +Py = Q

dx

where P and Q are the functions of x, or constants.

PdxIntegrating factor (I ..F.) = e

The solution is y I .F. = Q×(IF) dx +C

Example – 1

xdySolve the differential equation +2y = 6e .

dx

xdySolution: The given differential equation is +2y = 6e .

dx

It is a linear equation of the formdy

+Py = Qdx

xHere P = 2 and Q = 6e

Pdx 2dx 2x.F. = e = e = eI

The solution is given by y I.F. = Q I.F. dx +C

2x x 2xy e = 6e ×e dx +C 2x 2xd dyNote: ye =e +2y

dx dx

Solution Cont.

2x 3xye = 6e dx +C 2x 3xye = 6 e dx +C

3x2x e

ye = 6× +C3

2x 3xye = 2e +C

3x

2x

2e +Cy =

e

x -2xy = 2e +C×e is the required solution.

Example -2

Solve the following differential equation:

-12 tan xdy

(1+x ) +y = e CBSE 2002dx

Solution: The given differential equation is

It is a linear differential equation of the form dy+Py=Q

dx

-1tan x

2 21 e

Here, P = and Q =1+x 1+x

-1-1 tan x

2 tan x2 2

dy dy 1 e(1+x ) +y = e + .y =

dx dx 1+x 1+x

Solution Cont.

The solution is given by

-1-1 2tan x

tan x eye = +C

2

-1 -1tan x 2tan x2ye = e +C is the required solution.

-121

dxPdx tan x1+xI .F = e = e = e

y ×I.F. = Q ×I.F. dx + C-1

-1 -1 tan xtan x tan x

2e

y×e = e × dx+C1+x

Example – 3

dySolve the differential equation + secx y = tanx.

dx

dySolution: The given differential equation is + secx y = tanx.

dx

It is a linear differential equation of the formdy

+Px =Qdx

Here P = secx and Q = tanx

ePdx secx dx log secx + tanxI .F. = e = e = e = secx +tanx

Solution Cont.

y × IF = Q ×IF dx +C y secx +tanx = tanx secx +tanx dx +C

The solution is given by

2y secx +tanx = secxtanx dx + tan x dx +C

2y secx +tanx = secx + sec x - 1 dx +C y secx +tanx = secx + tanx - x +C

Linear Differential Equations type – 2

dxRule for solving +Px = Q

dy

PdyIntegrating factor (I .F.) = e

The solution is x I .F. = Q×(IF) dy +C

where P and Q are the functions of y, or constants.

Example - 4

Solve the following differential equation:

3 dy2x - 10y + y = 0

dx

Solution: The given differential equation is

3 2dy dx 22x - 10y + y = 0 + x = 10y

dx dy y

It is a linear differential equation of the form dx+Px=Q

dy

22Here, P = and Q =10y

y

Solution Cont.

The solution is given by

2 5xy = 2y +C

3 -2x = 2y +Cy is the required solution.

2e e

2dy

2log yPdy log y 2yI .F = e = e = e = e = y

x ×I.F. = Q ×I.F. dy + C

2 2 2 2 4x×y = 10y ×y dy+C xy =10 y dy+C

Applications of Differential Equations

Differential equations are used to solve problems of science and engineering.

Example - 5

A population grows at the rate of 5% per year. How long does it take for the population to double? Use differential equation for it.

Solution: Let the initial population be P0 and let the population after t years be P, then

dP 5 dP P dP 1= P = = dt

dt 100 dt 20 P 20

[Integrating both sides]

e1

log P = t +C20

dP 1= dt

P 20

Solution Cont.

At t = 0, P = P0 e 0 e 01×0

log P = + C C = log P20

e e 0 e0

1 Plog P = t+log P t =20 log

20 P

0When P = 2P , then

0e e

0

2P 1t =20 log = log 2 years

P 20

Hence, the population is doubled in e20 log 2 years.

Example - 6

The slope of the tangent at a point P(x, y) on a curve is x

- .y

If the curve passes through the point (3, -4), find the equationof the curve.

Solution: The slope of the curve at P(x, y) is dydx

dy x= - ydy = -xdx

dx y

ydy = - xdx Integrating both sides

2 2

2 2y x= - + C x + y = 2C ... i

2 2

Solution Cont.

The curve passes through the point (3, –4).

2 2 253 + -4 = 2C C =

2

2 2x + y = 25 is the required equation of the curve.

2 2 25x + y = 2 ×

2

Differential Equations of Second Order

is the required general solution of the given differential equation.

2

2

d yDifferential equation of the form: = ƒ x

dx

2

1 2

dy d y dy dy dyd d= ƒ x dx +C = and dx =

dx dx dx dx dx dxdx

1dy

=F x +C , where F x = ƒ x dxdx

1 2y = F x dx +C x +C

Example -7

Solve the differential equation: 2

2

d y 1= +6.

xdx

Solution: The given differential equation is

e 1 1dy

=log x +6x +C , where C is constant ofi ntegration.dx

2

2

d y dy dy1 1= +6 = +6

x dx dx xdx

dy dy 1dx = dx +6 dx

dx dx x

Solution Cont.

2e 1 2y = xlog x - x +3x +C x +C

e 1dy

dx = log x.1 dx +6 xdx +C dxdx

2

e 1 2 21 x

y = xlog x - .x dx +6. +C x +C , where C is constant ofi ntegration.x 2

2e 1 2y = xlog x +3x +x C - 1 +C , is the required solution.

Example –82

2

d ySolve the differential equation = xcosx.

dx2

2

d ySolution: The given differential equation is = xcosx.

dx

dyd= xcosx …(i)

dx dx

Integrating (i), we get

1dyd

dx x cos xdx Cdx dx

1dy

= xsinx +cosx +C … (ii) Integrating by partsdx

Solution Cont.

1dy

dx x sin x cos x C dxdx

1 2y = xsinxdx + cosxdx + C dx +C 1 2y = -xcosx +sinx +sinx +C x +C

1 2y = -xcosx +2sinx +C x +C

Again integrating both sides of (ii), we get

Thank you