The correct option is
A a homogeneous but not linear differential equation.
Given,Differential equation as xdydx=5y×ln(xy)
dydx=5(yx)ln(xy)
Let f(x,y)=dydx
⇒f(x,y)=5(yx)ln(xy)
For f(x,y) to be homogenous,f(kx,ky)=knf(x,y) (where n is its order)
Consider,f(kx,ky)=5(kykx)ln(kxky)=5(yx)ln(xy)=f(x,y)
∴The given equation is homogenous equation.
Consider given Differential equation, xdydx=5y×(lnx−lny)
⇒xdydx=5y×lnx−5y×lny
For equation to be linear,
a)The powers of dependent variable and its derivative to be power of one.
b)The coefficient of dependent variable should be either function of independent variable or constant.
Since in our equation the co-efficient (for the term -5ylny) term is not function of x.The given equation is non-linear.
∴Thegivenequationishomogeneousbutnon−linear