Page 126 - DMTH202_BASIC_MATHEMATICS_II

Unit 9: Solution of Differential Equation

Integrating both sides, we get Notes



 sin y dy  y cosdy  2 x log xdx  xdx c

I II  II I
 x 1 x 2  x 2
y

 
or cosy  sin y  1.sin y dy  2 log x    dx  
 2 x 2  2
 x 2 1 2  x 2
y
or cosy  sin y  cosy  2  logx  x    c
 2 4  2
or y sin y  x 2 log x  .
c
which is the required general solution.

Example: Show that the curve in which the angle between the tangent and the radius
vector at every point is one half of the vectorial angle is a cardioid.
Solution:

If the angle between the radius vector and the tangent at any point be j, and q the vectorial angle,
then according to the given condition.


 
2
  
or tan   tan  
2
 
d  
 r  tan
dr 2
dr 
or  cot d 
r 2
Integrating, we get

  
logr  2logsin    log 2c
2
 
 
c
or logr  log 2 sin 2   
2
 
 r = c (1 – cos ),
which is the required equation of the curve. Clearly this represents a cardioid.

Example: Solve the differential equation (1 + x)ydx + (1 + y)xdy = 0
Solution:

The given equation can be written as
dy
y
(1 y )x  (1 x )  0.
dx

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