If $V = z e^{a x + b y}$ and $z$ is a homogeneous function of degree $n$ in $x$ and $y$ , prove that $x \frac{\partial V}{\partial x} + y \frac{\partial V}{\partial y} = (a x + b y + n) V$

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Solution

Taking $log$ on both sides.
$log V = log (z e^{a x + b y})$ $= log z + (a x + b y)$ $log e = log z + a x + b y$
Differentiating with respect to $x$
$\frac{1}{V} \frac{\partial V}{\partial x} = \frac{1}{z} \frac{\partial z}{\partial x} + a$
$\Rightarrow \frac{\partial V}{\partial x} = \frac{V}{z} \frac{\partial z}{\partial x} + a V$
$x \frac{\partial V}{\partial x} = \frac{x V}{z} \frac{\partial z}{\partial x} + a x V$ .... (1)
Similarly, Diff, w.r. to $y$
$y \frac{\partial V}{\partial y} = \frac{y V}{z} \frac{\partial z}{\partial x} + b y V$ ..... (2)
Add (1) and (2)
$x \frac{\partial V}{\partial x} + y \frac{\partial V}{\partial y} = \frac{V}{z} (x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y}) a x V + b y V$
Since $z$ is a function of $x$ and $y$ of degree $n$
$x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = n z$ .........(3)
Substituting in (3), we get
$x \frac{\partial V}{\partial x} + y \frac{\partial V}{\partial y} = \frac{V}{z} (n z) + a x V + b y V$
$= (a x + b y + n) V$

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