y=cos(m arc sinx)

y'= -sin(m arc sinx)*m/(1-x^2)^1/2

y''= -cos(marc sinx)*m^2/(1-x^2)^1/2  +  sin(m arc sinx)*mx/(1-x^2)^3/2

thus takin 1-x^2 on d other side

u get ....

(1-x^2)*y''=y'*x -m^2 *y

thus d ansewr is

(1-x^2)*y2 -xy1+m^2*y=0

hope u get it....