62I tried to think.. but could not work out the solution :(
1sin x dydx= 2y2 cosxsinx + 2y cosx +2cosxsinx
sinxdy-ycosxdxsin2x=1sin2x[ 2y2cosxsinxdx+ycosxdx+2sinx cosxdx]
d[ysinx]= cosxsin2x[2y2dxsinx + ydx + 2sinx dx ]
d[ysinx]= 2cotx[ y2sin2x+ y2sinx+ 1]dx
d[ysinx]= 2cotx [(ysinx+14)2+ 1516]dx
put t= ysinx
∫ dt[(t+1/4)2+(√15/4)2= ∫ 2cot x dx
2√15tan-1[4ysinx+1√15]= ln/sinx/ + c
where /./ is mod and c is a constant
106theres an even shorter solution
take sinx=t
ull get it of homogenous nature then u can do watever is reqd.. (simple variable substitution)
btw surbhi's answer is correct
1Solve the differential equation dy
dx = 2y2 sin x.
