Question

Solve the differential equation:

`(xdy - ydx)  ysin(y/x) = (ydx + xdy)  xcos(y/x)`.

Find the particular solution satisfying the condition that y = π when x = 1.

Answer 1

`(xdy - ydx)ysin(y/x) = (ydx + xdy)xcos(y/x)`.

`\implies (xdy - ydx) (sin(y/x))/(cos(y/x)) = ((ydx + xdy)x)/y`

`\implies ((xdy - ydx))/x^2 tan(y/x) = ((ydx + xdy)x)/(x^2y)`

`\implies ((xdy - ydx))/x^2 tan(y/x) = ((ydx + xdy))/(xy)`

`\implies tan(y/x)*d(y/x) = (d(xy))/(xy)`,

which is variable separable form where variables are `y/x` and xy 

Integrating both sides,

`\implies int tan  y/x * d(y/x) = int(d(xy))/(xy)`

`\implies log|sec(y/x)| = log(xy) + logc`

`\implies sec(y/x) = c(xy)`.

Given, when x = 1, y = π `\implies` c = `-1/π`.

Hence required particular solution is `sec(y/x) = -1/π(xy)`.

Answer 2

`(xdy - ydx)  ysin(y/x) = (ydx + xdy)  xcos(y/x)`.

`\implies dy/dx = ((xy)cos(y/x) + y^2sin(y/x))/((xy)sin(y/x) - x^2cos(y/x))`

`\implies dy/dx = (1 + (y/x)tan(y/x))/(tan(y/x) - 1/((y/x))`,

which is homogenous differential equation of type `dy/dx = f(y/x)`.

Let y = vx

`\implies dy/dx = v + x(dv)/dx`

`\implies v + x(dv)/dx = (1 + vtanv)/(tanv - 1/v)` 

`\implies (tan v - 1/v)dv = 2dx/x`,

which is variable separable form where variables are v and x

Integrating both sides,

`\implies int(tanv - 1/v)dv = 2intdx/x`

`\implies log|secv| - log|v| = 2log|x| + logc`

`\implies log|secv| = log|v.x^2.c|`

`\implies secv = c(xy)`

`\implies sec(y/x) = c(xy)`.

Given, when x = 1, y = π `\implies` c = `-1/π`.

Hence required particular solution is `sec(y/x) = -1/π(xy)`.