# Solve the differential equation $x.\cos\bigg(\large\frac{y}{x} \bigg).\frac{dy}{dx}$$=y\cos \bigg( \large\frac{y}{x} \bigg)$$+x$

Toolbox:
• A function is said to be homogenous function in degree n if $F(kx,ky) = k^nF(x,y)$ for any non zero constant $k.$
• To solve these type of homogenous functions we make the substitution $y = vx$ hence $\large\frac{dy}{dx}$$= v+x\large\frac{dv}{dx} Step 1: \large\frac{dy}{dx}=\frac{y\cos\big(\Large\frac{y}{x}\big)+x}{x\cos\big(\Large\frac{y}{x}\big)} Clearly this is a homogeneous differential equation. Let us now substitute y=vx Differentiating with respect to x we get, \large\frac{dy}{dx}$$=v+x\large\frac{dv}{dx}$-----(1)
Step 2:
Substituting for $y$ and $\large\frac{dy}{dx}$ in equ(1) we get,
$v+x\large\frac{dv}{dx}=\frac{v\cos v+1}{\cos v}$
$x\large\frac{dv}{dx}=\frac{v\cos v+1}{\cos v}$$-v x\large\frac{dv}{dx}=\frac{1}{\cos v} Step 3: Now separating the variables we get, \cos v dv=\large\frac{dx}{x} Integrating on both sides we get, \int\cos vdv=\int\large\frac{dx}{x} \Rightarrow \sin v=\log\mid x\mid+\log c \Rightarrow \sin v=\log cx Step 4: Replace v by \large\frac{y}{x} \sin\big(\large\frac{y}{x}\big)$$=\log cx$
This is the required solution for the given equation.