Show that the differential equation is homogeneous. Find the particular solution of this equation when x = 1 and y =​π / 2.

The given differential equation is,xdydxsinyx + x - y sinyx = 0dydx = 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 = vxdydx = v + xdvdxNow, from 1, we get v + xdvdx = v - cosec vxdvdx = - cosec v-sin v dv = dxxIntegrating both sides, we get-sin v dv = dxx cos v = log x + Ccosyx = log x + C   .......2put x = 1 and y = π2 in 1, we getcosπ2 = log 1 + C0 = 0 + CC = 0Now, put C =0 in 2, we getcosyx = log x

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x dy/dx sin y/x + x-y sin y/x =0

dividing both sides by x sin y/x,

dy/dx + cosec y/x - y/x =0

dy/dx = y/x- cosec y/x      ............... (i)

let F(x,y) = y/x - cosec y/x
 
 F(#x, #y) = #y/#x - cosec #y/#x
 
                = #[y/x - coesc y/x]

               = #0  F(x,y)
hence, differential equation is homogeneous.

let y=vx

y/x = v

dy/dx = v+x dv/dx

​now equation (1) becomes,

v+x. dv/dx = vx/x - cosec vx/x

v+x .dv/dx = v- cosec v

x.dv/dx = -cosec v

-sinv dv = dx/x

integrating both sides,

cos v = log|x| + c

cos y/x = log |x| + c

given y =pi/2 , x=1

cos pi/2 = log 1 + c

0= 0 + c

c = 0

hence particular solution is

cos y/x = log |x| + 0

cos y/x = log | x|




 
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