$\begin{array}{1 1} (A)\quad(4x+6y+5)\;dy-(3y+2x+4)\;dx = 0\\(B)\quad(xy)\;dx-(x^3+y^3)\;dy = 0\\(C)\quad(x^3+2y^2)dx+2xy\;dy=0\\(D)\quad y^2dx+(x^2-xy-y^2)\;dy=0\end{array} $

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- A differential equation of the form $dy/dx = F(x,y)$ is said to be homogenous, if $F(x,y)$ is a homogenous function of degree zero.

Using the information in the tool box, if the equations are expressed in the form of $F(kx,ky)$, we get only for option D as a function of zero.

Writing the equation of option D as $dy/dx = y^2/(x^2 -xy-y^2)$

i.e;$ F(x,y) = y^2/(x^2 -xy -y^2)$

$F(kx,ky) =k^2x^2/(k^2x^2 - kxky - k^2y^2) = k^0.F(x,y)$

Hence it is a homogenous function with degree zero.

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