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# Show that the differential equation is homogeneous and solve $x\large\frac{dy}{dx}$$-y+x\sin\bigg(\large\frac{y}{x}\bigg)=$$0$

Toolbox:
• A differential equation of the form $\large\frac{dy}{dx }$$= F(x,y) is said to be homogenous if F(x,y) is a homogenous function of degree zero. • To solve this type of equations substitute y = vx and \large\frac{dy}{dx }$$= v + x\large\frac{dv}{dx}$
Step 1:
The given equation can be written as $\large\frac{dy}{dx} =\frac{ [y - x\sin(y/x)]}{x}$
$F(x,y) =\large\frac{ [y - x\sin(y/x)]}{x}$
$F(kx.ky) = \large\frac{[ky - kx\sin (ky/kx)]}{kx}$$= k^0.F(x,y) Hence this is an homogenous funtion of degree zero. Step 2: Using the information in the tool box, let us substitute for y and \large\frac{dy}{dx} v+x\large\frac{dv}{dx }= \frac{[vx - x\sin v]}{x} v + x\large\frac{dv}{dx }=$$ [v-\sin v]$
Cancelling $v$ on both sides we get
$x\large\frac{dv}{dx }= $$- \sin v Seperating the variables we get \large\frac{dv}{\sin v }= -\frac{dx}{x} Step 3: Integrating on both sides we get, \int cosec v dv = - \int\large\frac{ dx}{x} \log|cosec v - \cot v| = -\log x + \log C cosec v - \cot v = \large\frac{C}{x} (\large\frac{1}{\sin v}) - (\large\frac{\cos v}{\sin v}) =\frac{ C}{x} \large\frac{[1-\cos v]}{\sin v} = \frac{C}{x} 1 -\cos v = \large\frac{C\sin v}{x} Step 4: Writing v = \large\frac{y}{x } we get, 1 - \cos(\large\frac{y}{x}) = \large\frac{C\sin(\Large\frac{y}{x})}{x} x[1-\cos(\large\frac{y}{x})] =$$ C\sin(\large\frac{y}{x})$
This is the required solution.