exact solution to differential equation

Question

y'= (3-x)/xy  , x>0, y(1)=2

Answer 1

y'y= 3/x-1

Integrate to get

 

y^2/2= 3 ln x -x +C

y^2= 6lnx -2x +C2   (C2=2C)

y(1)=2

4= 6ln 1 -2 +C2

C=6

y= sqrt(6lnx-2x+6)

Answer 2

dy/dx=(3-x)/xy

y dy ={(3-x)/x } dx

y dy = (3/x) dx - dx   : as (3-x)/x = 3/x -1

 

On integrating we will get

  y²/2 = 3 ln x - x +c

 Apply initial condition find out c and replace the value of c to get exact solution ..

 

If you have any difficulty or query in further steps feel free to ask..