Show that the differential equation is homogeneous. Find the particular solution of this equation when x = 1 and y =π / 2. Share with your friends Share 0 Varun.Rawat answered this The given differential equation is,xdydxsinyx + x - y sinyx = 0⇒dydx = yx - cosecyx ......1Let Fx,y = yx - cosecyxNow, Fλx, λy = λyλx - cosecλyλx =yx - cosecyx = λ0 Fx,yThus, Fx,y is a homogeneous function of degree 0. Therefore the given differential equation is a homogeneous differential equation.Consider 1, and put y = vx⇒dydx = v + xdvdxNow, from 1, we get v + xdvdx = v - cosec v⇒xdvdx = - cosec v⇒-sin v dv = dxxIntegrating both sides, we get-∫sin v dv = ∫dxx ⇒cos v = log x + C⇒cosyx = log x + C .......2put x = 1 and y = π2 in 1, we getcosπ2 = log 1 + C⇒0 = 0 + C⇒C = 0Now, put C =0 in 2, we getcosyx = log x 2 View Full Answer Jeevani answered this x dy/dx sin y/x + x-y sin y/x =0dividing both sides by x sin y/x,dy/dx + cosec y/x - y/x =0dy/dx = y/x- cosec y/x ............... (i)let F(x,y) = y/x - cosec y/x F(#x, #y) = #y/#x - cosec #y/#x = #0 [y/x - coesc y/x] = #0 F(x,y)hence, differential equation is homogeneous.let y=vxy/x = vdy/dx = v+x dv/dxnow equation (1) becomes,v+x. dv/dx = vx/x - cosec vx/xv+x .dv/dx = v- cosec vx.dv/dx = -cosec v-sinv dv = dx/xintegrating both sides,cos v = log|x| + ccos y/x = log |x| + cgiven y =pi/2 , x=1cos pi/2 = log 1 + c0= 0 + cc = 0hence particular solution iscos y/x = log |x| + 0cos y/x = log | x| 0