Transcript Document

Mathematics
Session
Differential Equations - 3
Session Objectives

Linear Differential Equations

Applications of Differential Equations
 Differential Equations of Second Order

Class Exercise
Linear Differential Equations
The standard form of a linear differential equation of
dy
+Py = Q
first order and first degree is
dx
where P and Q are the functions of x, or constants.
Examples : 1
3
dy
+ 2y = 6ex ;
dx
dx y
+ = y2 etc.
dy x
2 
dy
+ ytanx = cosx ;
dx
Linear Differential Equations type-1
Rule for solving
dy
+Py = Q
dx
where P and Q are the functions of x, or constants.
Pdx
Integrating factor (I..F.) = e
The solution is y I.F. =
 Q×(IF) dx + C
Example – 1
Solve the differential equation
dy
+ 2y = 6ex .
dx
dy
+2y = 6ex .
dx
Solution: The given differential equation is
dy
It is a linear equation of the form dx +Py = Q
Here P = 2 and Q = 6ex
Pdx
2dx
I.F.= e
= e
= e2x

The solution is given by y I.F. = Q I.F. dx + C
  =  6e
2x
y e
x
2x
×e
dx + C



d

2x
2x  dy
Note
:
ye
=
e
+2y
 dx


dx



Solution Cont.

=6 e

 ye2x = 6e3x dx + C
 ye2x
3x
 ye2x
dx + C
e3x
= 6×
+C
3
 ye2x = 2e3x +C
 y=
2e3x + C
e2x
 y = 2ex +C×e-2x is the required solution.
Example -2
Solve the following differential equation:
dy
tan-1x
(1+ x )
+y =e
dx
CBSE 2002
2
Solution: The given differential equation is
(1+ x2 )
tan-1x
dy
dy
1
e
+ y = etan x 
+
.y =
dx
dx 1+ x2
1+ x2
-1
It is a linear differential equation of the form
Here, P =
1
1+ x2
and Q =
e
tan-1x
1+ x2
dx
+ Py = Q
dy
Solution Cont.
I.F = e
 Pdx

1
2
= e 1+x
dx
tan-1x
=e
The solution is given by
y × I.F. =  Q × I.F.dx + C
tan-1x
e
 y ×etan x =  etan x ×
dx + C
2
1+ x
-1
tan-1x
 ye
=
tan-1x
 2ye
-1
2tan-1x
e
2
2tan-1x
=e
+C
+C is the required solution.
Example – 3
Solve the differential equation
dy
+  secx  y = tanx.
dx
Solution: The given differential equation is
dy
+  secx  y = tanx.
dx
It is a linear differential equation of the form
dy
+Px = Q
dx
Here P = secx and Q = tanx
Pdx
secx dx
log secx + tanx
I.F.= e
= e
=e e
= secx + tanx
Solution Cont.
The solution is given by
y × IF  =
 Q × IF  dx + C

 y  secx + tanx  = tanx  secx + tanx  dx + C


 y  secx + tanx  = secxtanx dx + tan2 x dx + C
 y  secx + tanx  = secx +
sec2x -1 dx + C


 y  secx + tanx  = secx + tanx - x +C
Linear Differential Equations type – 2
Rule for solving
dx
+Px = Q
dy
where P and Q are the functions of y, or constants.
Pdy
Integrating factor (I.F.) = e
The solution is xI.F. =
 Q×(IF) dy + C
Example - 4
Solve the following differential equation:

2x - 10y3

dy
+y=0
dx
Solution: The given differential equation is

2x - 10y3

dy
dx 2
+y =0 
+ x = 10y2
dx
dy y
dy
+Px = Q
It is a linear differential equation of the form dx
Here, P =
2
and Q = 10y2
y
Solution Cont.
2
 dy
2loge y
Pdy
logey 2
y

I.F = e
=e
=e
=e
= y2
The solution is given by
x × I.F. =  Q × I.F.dy + C
 x×y2 = 10y2 ×y2 dy+C  xy2 =10 y4 dy+C
 xy2 = 2y5 +C
 x = 2y3 +Cy-2 is the required solution.
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
dt
20
P
20
 100 

dP
1
=
dt
P
20 
 logeP =
1
t +C
20
[Integrating both sides]
Solution Cont.
At t = 0, P = P0
 logeP0 =
1× 0
+ C  C = logeP0
20
P 
1
 logeP =
t +loge P0  t = 20 loge  
20
 P0 
When P = 2P0 , then
 2P0 
1
 t = 20 loge 
loge 2 years
=
 P0  20
Hence, the population is doubled in 20 loge 2 years.
Example - 6
x
.
y
If the curve passes through the point (3, -4), find the equation
of the curve.
The slope of the tangent at a point P(x, y) on a curve is -
Solution: The slope of the curve at P(x, y) is

dy
dx
dy
x
= -  ydy = -xdx
dx
y
  ydy = - xdx
Integrating both sides
y2
x2

=+ C  x2 + y2 = 2C
2
2
... i
Solution Cont.
The curve passes through the point (3, –4).
 3 + -4
2
2
25
= 2C  C =
2
25
 x + y = 2×
2
2
2
 x2 + y2 = 25 is the required equation of the curve.
Differential Equations of Second Order
Differential equation of the form:
dy

= ƒ  x  dx + C1
dx

d2 y
2
dx
 d2 y
d  dy 
=

 dx  and
2
dx


 dx
= ƒ x 

dy 
d  dy 
dx
=

dx  dx 
dx 
dy

= F  x  +C1 , where F  x  = ƒ  x  dx
dx


 y = F  x  dx + C1x + C2
is the required general solution of the
given differential equation.
Example -7
Solve the differential equation:
d2 y
dx2
=
1
+ 6.
x
Solution: The given differential equation is
d2 y
dx2



=
dy  dy  1
1
+6 
= +6


x
dx  dx  x
dy  dy 
dx =
dx  dx 


1
dx + 6 dx
x
dy
= loge x +6x +C1 , where C1 is constant of integration.
dx
Solution Cont.


 dy 
 dx  dx =



 y = xloge x -
loge x.1 dx + 6


x dx + C1

dx
1
x2
.x dx + 6.
+ C1x + C2 , where C2 is constant of integration.
x
2
 y = xloge x - x +3x2 +C1x +C2
 y = xloge x +3x2 +x C1 -1 +C2 , is the required solution.
Example –8
Solve the differential equation
d2 y
2
dx
= xcosx.
Solution: The given differential equation is

d  dy 
= xcosx


dx  dx 
d2 y
2
dx
= xcosx.
… (i)
Integrating (i), we get

d  dy 
dx 


dx  dx 

x cos xdx  C1
dy

= xsinx + cosx +C1
dx
… (ii)
Integrating by parts
Solution Cont.
Again integrating both sides of (ii), we get

 dy 
 dx  dx 
 


 x sin x  cos x  C1  dx


 y = xsinxdx + cosxdx + C1dx + C2
 y = -xcosx + sinx + sinx + C1x + C2
 y = -xcosx +2sinx + C1x + C2
Thank you