This is a Linear Differential Equation of First Order , usually
taken as, dy/dx+yP(x)=Q(x)dydx+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)=e∫P(x)dx.
In our Example, P(x)=-3cotx, so, intP(x)dx=int{-3cotx}dxP(x)=−3cotx,so,∫P(x)dx=∫{−3cotx}dx
= -3lnsinx=lnsin^(-3)x rArr e^(P(x)dx)=e^(lnsin^(-3)x)=1/sin^3x=−3lnsinx=lnsin−3x⇒eP(x)dx=elnsin−3x=1sin3x.
Multiplying the Diff. Eqn., by 1/sin^3x1sin3x, we get,
sin^(-3)xdy/dx-3ycotx/sin^3x=(sin2x)/sin^3xsin−3xdydx−3ycotxsin3x=sin2xsin3x
:. 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.!
