This is a Linear Differential Equation of First Order , usually
taken as, #dy/dx+yP(x)=Q(x)#.
To find its General Soln. , we need to multiply it by the Integrating
Factor, i.e., IF , given by, IF #=e^(intP(x)dx)#.
In our Example, #P(x)=-3cotx, so, intP(x)dx=int{-3cotx}dx#
#= -3lnsinx=lnsin^(-3)x rArr e^(P(x)dx)=e^(lnsin^(-3)x)=1/sin^3x#.
Multiplying the Diff. Eqn., by #1/sin^3x#, we get,
#sin^(-3)xdy/dx-3ycotx/sin^3x=(sin2x)/sin^3x#
#:. sin^(-3)xdy/dx-3ycosxsin^(-4)x=(2sinxcosx)/sin^3x=(2cosx)/sin^2x#
#:. sin^(-3)xd/dx(y)+yd/dx(sin^(-3)x)=2cotxcscx#
#:.d/dx{ysin^(-3)x}=2cotxcscx#
Integrating, #ysin^(-3)x=2intcotxcscx+c=-2cscx+c#
#:. y+2sin^2x=csin^3x,# is the reqd. Gen. Soln.
To find the Particular Soln. , we utilise the Initial Condition :
#y=2," when "x=pi/2#.
Sub.ing in the Gen. Soln., we get, #2+2=c=4#
Hene, the P.S. # : y+2sin^2x=4sin^3x#
Enjoy Maths.!
