integrating factor method this method applies to differential equations of the form dy/dx +g(x)y = h(x) here, g(x)=cot x, h(x)=sin2x a) determine the integrating factor p=e^(int(g(x)dx)) = e^( int (cotx)dx)) =e^(ln(sinx)dx) =sinx b) rewrite the differential equation as d/dx(p(x)y) =p(x)h(x) >>>>>d/dx( y.sinx)= sinx.sin2x c)integrate this last equation to obtain y.sinx =int(sinx.sin2x)dx = int(sinx.2sinx.cosx)dx =2.int((sinx)^2(cosx))dx =2.int(1-(cosx)^2)cosx)dx =2.int(cosx-(cos)^3)dx =2.sinx -2(sinx-((sinx)^3)/3)+C =(2.(sinx)^3)/3 +C d) y.sinx = (2.(sinx)^3)/3 +C hence, general solution is y = (2.(sinx)^2)/3 + C/sinx we are given y=1 at x= pi/4 >>>>> sin (pi/4) = 1/(2)^(1/2) substitute into general solution 1 = 1/3 + root(2).C >>>>> C = root (2)/3 the particular solution is y = (2(sinx)^2)/3 + root(2)(cosecx)/3 = (2(sinx)^2+root(2)(cosecx))/3 where root is the square root (0