Page 127 - DMTH202_BASIC_MATHEMATICS_II

Basic Mathematics-II

Notes Separating the variables, we get


1 y 1 x
dy  dx  0
y x
Integrating both sides, we get
1 y 1 x



 y dy   x dx c.
or, logy + y + log x + x = a, a constant
or, log xy + x + y = a

which is the required solution.
2  dy   dy 
x
Example: Solve (  y )  x  y   xy  1 
 dx   dx 
Solution:
Put x + y = u, then differentiating w.r.t. x and y, we get respectively

dy du dy dv
1  and x  y 
dx dx dx dx
Substituting these values in the given equation,
2 dv du dv du
u  v or,  2
dx dx v u
Integrating both sides, we have
dv du 1
    C or, log  v  C
v u 2 u
1 1
or, log xy    C or, log xy   . C
x  y x  y
ydx  xdy dx
Example: Solve 2 
x
(  y ) 2 1 x 2
Solution:

(  xdy ydx ) dx

2 2
 1  y x 2 2 1 x
 
  x
y xdy  ydx
Put  , v 2  dv .
x x
Substituting these values in the given equation,
dv dx
  , Integrating we get
(1 v ) 2 2 1 x 2
1 1  1
  sin x c

1 v 2

x 1  1

or,  sin x C is the required solution.
y x 2

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