Question

Solve the following differential equation:

`xsin(y/x)dy = [ysin(y/x) - x]dx`

Answer 1

`xsin(y/x)dy = [ysin(y/x) - x]dx`

∴ `(dy)/(dx) = (ysin(y/x) - x)/(xsin(y/x))` ...(i)

Put y = vx

∴ `(dy)/(dx) = v + x(dv)/(dx)` and `y/x` = v

∴ Equation (i) becomes, `v + x(dv)/(dx) = (vx sinv - x)/(xsinv)`

∴ `x(dv)/(dx) = (vsinv - 1)/sinv - v`

∴ `x(dv)/(dx) = (vsinv - 1 - vsinv)/sinv = (-1)/sinv`

∴ sinv dv = `-1/x dx`

lntergrating both sides, we get

`intsinv dv = -int1/xdx`

∴ –cosv = –logx – c

∴ `cos(y/x)` = logx + c

This is the general solution.