- A$x sin y + y sin x$
- B$x e^{y/x} + y e^{x/y}$
- C$x^2 - xy$
- ✓$(B)$ or $(C)$ both
$A) f(x, y)=x \sin y+y \sin x$
Now
$f(\lambda x, \lambda y)=\lambda x \sin (\lambda y)+\lambda y \sin (\lambda x)$
$=\lambda(x \sin \lambda y+y \sin (\lambda x))$
But
$x \sin \lambda y+y \sin (\lambda x) \neq f(x, y)$
Hence it is not a homogeneous function.
$B) f(x, y)=x e^{y / x}+y e^{x / y}$
$f(\lambda x, \lambda y)=\lambda x e^{\lambda y / \lambda x}+\lambda y e^{\lambda x / \lambda y}$
$=\lambda\left(e^{x / y}+e^{y / x}\right)$
$=\lambda f(x, y)$
Hence it is a homogeneous function. $C) f(x, y)=x^{2}-x y$
$f(\lambda x, \lambda y)=\lambda^{2} x^{2}-\lambda^{2}(x y)$
$=\lambda^{2}\left(x^{2}-x y\right)$
$=\lambda^{2} f(x, y)$
Hence it is a homogeneous function.
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Statement $-1:$ The substitution $z = y^2$ transforms the above equation into a first order homogenous differential equation.
Statement $-2:$ The solution of this differential equation is ${y^2}{e^{ - {y^2}/x}} = C$.