This document presents an overview of first order ordinary differential equations and applications. It contains: 1) The standard form of a linear first order differential equation and examples of solving three types of equations. 2) Applications of differential equations to model population growth and finding the equation of a curve given its slope at a point. 3) Solutions to the examples and applications in 3 sentences or less.

1. First Order Ordinary Differential
Equations and Applications
161310109046 SOLANKI RAJVIR
161310109047 RAKSHA AGARWAL
161310109048 RAMI NIRAV
161310109050 RAVAL ARPIT
161310109051 ROHAN KAUSHIK
161310109052 ROHIT CHAVDA
Presented By:

2. 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 dy
Examples: 1 +2y = 6e ; 2 + ytanx = cosx ;
dx dx
ydx
3 + = y etc.
dy x

3. Linear Differential Equations type-1
dy
Rule for solving +Py = Q
dx
where P and Q are the functions of x, or constants.
Pdx
Integratingfactor (I..F.)=e
   The solution is y I.F. = Q×(IF) dx +C


4. Example – 1
xdy
Solve the differential equation +2y = 6e .
dx
xdy
Solution: The given differential equation is +2y = 6e .
dx
It is a linear equation of the form
dy
+Py = Q
dx
x
Here 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 2x
y e = 6e ×e dx +C
  2x 2xd dy
Note: ye = e +2y
dx dx
  
  
  

5. Solution Cont.
2x 3x
ye = 6e dx +C

2x 3x
ye = 6 e dx +C

3x
2x e
ye = 6× +C
3

2x 3x
ye = 2e +C
3x
2x
2e +C
y =
e

x -2x
y = 2e +C×e is the required solution.

6. Example -2
Solve the following differential equation:
 
-1
2 tan xdy
(1+ x ) + y = e CBSE 2002
dx
Solution: The given differential equation is
It is a linear differential equation of the form dy
+Py = Q
dx
-1
tan x
2 2
1 e
Here, P = and Q =
1+x 1+x
-1
-1 tan x
2 tan x
2 2
dy dy 1 e
(1+x ) +y = e + .y =
dx dx 1+x 1+x


7. Solution Cont.
The solution is given by
-1
-1 2tan x
tan x e
ye = +C
2

-1 -1
tan x 2tan x
2ye = e +C is the required solution.
-12
1
dx
Pdx tan x1+xI.F = e = e = e


y × I.F. = Q × I.F.dx + C
-1
-1 -1 tan x
tan x tan x
2
e
y×e = e × dx+C
1+x
 

8. Example – 3
 
dy
Solve the differential equation + secx y = tanx.
dx
 
dy
Solution: The given differential equation is + secx y = tanx.
dx
It is a linear differential equation of the form
dy
+Px = Q
dx
Here P = secx and Q= tanx
ePdx secx dx log secx + tanx
I.F.= e = e = e = secx + tanx 

9. Solution Cont.
   y × IF = Q ×IF dx+C

   y secx + tanx = tanx secx + tanx dx +C

The solution is given by
  2
y secx + tanx = secxtanx dx + tan x dx +C
 
   2
y secx+ tanx = secx+ sec x -1 dx+C

 y secx+tanx = secx+ tanx - x+C

10. Applications of Differential Equations
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]
e
1
log P = t +C
20

dP 1
= dt
P 20
  

11. Solution Cont.
At t = 0, P = P0 e 0 e 0
1×0
log P = + C C = log P
20
 
e e 0 e
0
1 P
log P = t+log P t =20log
20 P
 
   
 
0When P = 2P , then
0
e e
0
2P 1
t =20log = log 2 years
P 20
 
  
 
Hence, the population is doubled in e20log 2 years.

12. 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 equation
of the curve.
Solution: The slope of the curve at P(x, y) is
dy
dx
dy x
= - ydy = -xdx
dx y
 
ydy = - xdx Integrating both sides    
 
2 2
2 2y x
= - + C x + y = 2C ... i
2 2
 

13. Solution Cont.
The curve passes through the point (3, –4).
   2 2 25
3 + -4 = 2C C =
2
 
2 2
x + y = 25 is the required equation of the curve.
2 2 25
x + y = 2 ×
2
