Question

$I$f $u=\frac{e^{x/y}\sin \left(y/x\right)}{x^3}$

then $x^2\frac{{\partial }^{2}u}{\partial x^2}+2xy\frac{{\partial }^{2}u}{\partial x\partial y}+y^2\frac{{\partial }^{2}u}{\partial y^2}=?$

• A. $0$
• B. $2u$
• C. $12u$
• D. $4u$

Answer 1

The correct option is C $12u$

Given, $u=\frac{e^{\left(\frac{x}{y}\right)}.\sin \frac{y}{x}}{x^3}$

$u{x}^3=e^{\left(\frac{x}{y}\right)}.\sin \frac{y}{x}$ $\to$(1)

Partially differentiating both sides w.r.t $x$

$3x^{2}u+x^3\frac{\partial u}{\partial x}=\frac{1}{y}{e}^{\left(\frac{x}{y}\right)}\sin \left(\frac{y}{x}\right)-\frac{y}{x^2}{e}^{\left(\frac{x}{y}\right)}\cos \left(\frac{y}{x}\right)$

$⟹3x^{2}u+x^3\frac{\partial u}{\partial x}=\frac{u{x}^3}{y}-\frac{y}{x^2}{e}^{\left(\frac{x}{y}\right)}\cos \left(\frac{y}{x}\right)$ $\to$ (2)

Partially differentiating (1) both sides wrt $y$

$x^3\frac{\partial u}{\partial y}=\frac{-x}{y^2}{e}^{\left(\frac{x}{y}\right)}\sin \left(\frac{y}{x}\right)-\frac{1}{x}{e}^{\left(\frac{x}{y}\right)}\cos \left(\frac{y}{x}\right)$

$⟹x^3\frac{\partial u}{\partial y}=\frac{-x^{4}u}{y^2}-\frac{1}{x}{e}^{\left(\frac{x}{y}\right)}\cos \left(\frac{y}{x}\right)$ $\to$ (3)

(2)$+$(3)$\times \frac{y}{x}$

$⟹3x^{2}u+x^3\frac{\partial u}{\partial x}+x^{2}y\frac{\partial u}{\partial y}$

$⟹\frac{u{x}^3}{y}-\frac{y}{x^2}{e}^{\left(\frac{x}{y}\right)}\cos \left(\frac{y}{x}\right)-\frac{u{x}^3}{y}+\frac{y}{x^2}{e}^{\left(\frac{x}{y}\right)}\cos \left(\frac{y}{x}\right)$

$\therefore 3x^{2}u+x^3\frac{\partial u}{\partial x}+x^{2}y\frac{\partial u}{\partial y}=0$

$⟹x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}+3u=0$

Partially differentiating both sides wrt $x$ and $y$ separately,

$⟹x\frac{{\partial }^{2}u}{\partial x^2}+\frac{\partial u}{\partial x}+y\frac{{\partial }^{2}u}{\partial x\partial y}+3\frac{\partial u}{\partial x}=0$ $\to$ (4)

$⟹x\frac{{\partial }^{2}u}{\partial x\partial y}+\frac{\partial u}{\partial y}+y\frac{{\partial }^{2}u}{{\partial }^{2}y}+3\frac{\partial u}{\partial y}=0$ $\to$ (5)

$\left(4\right)$ $\times x+\left(5\right)$ $\times y$

$⟹x^2\frac{{\partial }^{2}u}{\partial x^2}+2xy\frac{{\partial }^{2}u}{\partial x\partial y}+y^2\frac{{\partial }^{2}u}{\partial y^2}+4(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y})=0$

$\therefore ⟹x^2\frac{{\partial }^{2}u}{\partial x^2}+2xy\frac{{\partial }^{2}u}{\partial x\partial y}+y^2\frac{{\partial }^{2}u}{\partial y^2}=12u$