Question

If $u = {\tan ^{ - 1}}\left( {\dfrac{{{x^3} + {y^3}}}{{x - y}}} \right)$ , then $x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}}$ is equal to
1. $- \sin 2u$
2. $\sin 2u$
3. $\cos 2u$
4. $- \cos 2u$

Answer 1

Here we need to apply the famous Euler’s theorem to obtain the value of the given partial differential equation. Euler’s theorem for two variables states that if $u = f\left( {x,y} \right)$ is a homogeneous function of degree $n$ , then $x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}} = nu$

Formula to be used:
a)$\dfrac{\partial }{{\partial x}}\tan x = {\sec ^2}x$
b) $\tan x = \dfrac{{\sin x}}{{\cos x}}$
c) $\sec x = \dfrac{1}{{\cos x}}$
d) $\sin 2x = 2\sin x\cos x$

Complete step-by-step answer:
It is given that $u = {\tan ^{ - 1}}\left( {\dfrac{{{x^3} + {y^3}}}{{x - y}}} \right)$
We need to find the value of $x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}}$
Let us consider the given equation $u = {\tan ^{ - 1}}\left( {\dfrac{{{x^3} + {y^3}}}{{x - y}}} \right)$
Now, we shall bring the inverse of the tangent to the left-hand side of the equation.
Thus, we have $\tan u = \left( {\dfrac{{{x^3} + {y^3}}}{{x - y}}} \right)$
Let us assume that $\tan u = z$
Now, we shall partially differentiate the above equation with respect to x.
So, we get $\dfrac{\partial }{{\partial x}}\left( {\tan u} \right) = \dfrac{{\partial z}}{{\partial x}}$
$\Rightarrow {\sec ^2}u\dfrac{{\partial u}}{{\partial x}} = \dfrac{{\partial z}}{{\partial x}}$ (Here we applied the formula $\dfrac{\partial }{{\partial x}}\tan x = {\sec ^2}x$ )
$\Rightarrow \dfrac{{\partial z}}{{\partial x}} = {\sec ^2}u\dfrac{{\partial u}}{{\partial x}}$ ………………$\left( 1 \right)$
Again, we shall partially differentiate the above equation with respect to y.
So, we get $\dfrac{\partial }{{\partial y}}\left( {\tan u} \right) = \dfrac{{\partial z}}{{\partial y}}$
$\Rightarrow {\sec ^2}u\dfrac{{\partial u}}{{\partial y}} = \dfrac{{\partial z}}{{\partial y}}$
$\Rightarrow \dfrac{{\partial z}}{{\partial y}} = {\sec ^2}u\dfrac{{\partial u}}{{\partial y}}$ ………………$\left( 2 \right)$
Let us consider $\tan u = z$obtaining the z with respect to x.
Since we have found that $\tan u = \left( {\dfrac{{{x^3} + {y^3}}}{{x - y}}} \right)$, we need to substitute it in the above equation.
Thus, we have $z = \left( {\dfrac{{{x^3} + {y^3}}}{{x - y}}} \right)$
$\Rightarrow z = \dfrac{{{x^3}}}{x}\left( {\dfrac{{1 + \dfrac{{{y^3}}}{{{x^3}}}}}{{1 - \dfrac{y}{x}}}} \right)$
$\Rightarrow z = {x^2}\left( {\dfrac{{1 + {{\left( {\dfrac{y}{x}} \right)}^3}}}{{1 - \dfrac{y}{x}}}} \right)$
Hence, we can come to a decision that z is a homogeneous function of the form ${x^n}f\left( {\dfrac{y}{x}} \right)$ in the above equation.
When we compare $z = {x^2}\left( {\dfrac{{1 + {{\left( {\dfrac{y}{x}} \right)}^3}}}{{1 - \dfrac{y}{x}}}} \right)$ with ${x^n}f\left( {\dfrac{y}{x}} \right)$, we can get $n = 2$
Now, by Euler’s theorem, we have $x\dfrac{{\partial z}}{{\partial x}} + y\dfrac{{\partial z}}{{\partial y}} = nz$ ………$\left( 3 \right)$
We have found $\dfrac{{\partial z}}{{\partial x}} = {\sec ^2}u\dfrac{{\partial u}}{{\partial x}}$, $\dfrac{{\partial z}}{{\partial y}} = {\sec ^2}u\dfrac{{\partial u}}{{\partial y}}$ and $n = 2$. Now, we shall substitute these values in the equation $\left( 3 \right)$
Thus, we get $x{\sec ^2}u\dfrac{{\partial u}}{{\partial x}} + y{\sec ^2}u\dfrac{{\partial u}}{{\partial y}} = 2\tan u$ (We assumed $\tan u = z$)
$\Rightarrow {\sec ^2}u\left( {x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}}} \right) = 2\tan u$
$\Rightarrow x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}} = 2\dfrac{{\tan u}}{{{{\sec }^2}u}}$
$\Rightarrow x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}} = 2\dfrac{{\sin u}}{{\cos u}}{\cos ^2}u$ (We applied $\tan x = \dfrac{{\sin x}}{{\cos x}}$ and $\sec x = \dfrac{1}{{\cos x}}$ )
$\Rightarrow x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}} = 2\sin u\cos u$
$\Rightarrow x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}} = \sin 2u$ (Here we applied $\sin 2x = 2\sin x\cos x$ )
Therefore, $x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}}$ is equal to $\sin 2u$ and hence option 2) is the answer.

So, the correct answer is “Option 2”.

Note: We should be able to identify the difference between an ordinary differential equation and a partial differential equation. The ordinary differential equation contains only one independent variable whereas the partial differential equation contains several independent variables. Here, we are given differentials with respect to two independent variables.