Question

Write the descending order of $k$ values of the following statements

$A$: If $z=x^3\sin \left(\frac{x}{y}\right)$ and $x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=kz$, then $k=$
$B$: If $Z=\frac{x^3+y^3}{x-y}$ then $\frac{x\frac{{\partial }^{2}z}{\partial x^2}+y\frac{{\partial }^{2}z}{\partial x\partial y}}{\frac{\partial z}{\partial x}}=k$
$C:z=\log \left(\frac{x^3+y^3+3x{y}^2}{x+y}\right)$ and$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=k$

$D$: If $z={\sin }^{-1}\left(\frac{\sqrt{x}+\sqrt{y}}{x-y}\right)$ and $x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=k\tan z$, then $k=$

• A. $A,C,D,B$
• B. $B,A,C,D$
• C. $D,C,B,A$
• D. $A,C,B,D$

Answer 1

Answer: (D)

$A,C,B,D$

$\left(A\right)z=x^{3}sin\left(\frac{x}{y}\right)$

This is a homogenous function degree '3'

$\therefore x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial x}=3z$ $\therefore$ here k=3

$\left(B\right)Z=\frac{x^3+y^3}{x-y}$

$(x-y)z=x^3+y^3$

Partially differentiate on both sides w.r.t x and y

$z+\frac{\partial z}{\partial x}(x-y)=3x^2\to \left(1\right)$

$-z+\frac{\partial z}{\partial y}(x-y)=3y^2\to \left(2\right)$

$\left(1\right)\times x+\left(2\right)\times y\Rightarrow (x-y)z+(x-y)(x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y})=3(x-y)z$

$\therefore x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=2z$

Partially differentiate on both sides w.r.t x

$x\frac{{\partial }^{2}z}{\partial x^2}+\frac{\partial z}{\partial x}+y\frac{{\partial }^{2}z}{\partial x\partial y}=2\frac{\partial z}{\partial x}$

$\therefore \frac{x\frac{{\partial }^{2}z}{\partial x^2}+y\frac{{\partial }^{2}z}{\partial x\partial y}}{\frac{\partial z}{\partial x}}=1$ , here k=1

$\left(C\right)z=log\left(\frac{x^3+y^3+3x{y}^2}{x+y}\right)$

$(x+y)e^2=x^3+y^3+3x{y}^2$

Partially differentiate on both sides w.r.t x and y

$e^2+e^2(x+y)\frac{\partial z}{\partial x}=3x^2+3y^2\to \left(3\right)$

$e^2+e^2(x+y)\frac{\partial z}{\partial y}=3x^2+6xy\to \left(4\right)$

$\left(3\right)\times x+\left(4\right)\times y\Rightarrow e^2(x+y)+e^2(x+y)(a\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y})=3(x^3+y^3+3x{y}^2)$

$3(x+y)e^2$

$\therefore x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=2$ Here k=2

$\left(D\right)z=si{n}^{-1}\left(\frac{\sqrt{x}+\sqrt{y}}{x-y}\right)$

$sinz=\frac{\sqrt{x}+\sqrt{y}}{x-y}$

$(x-y)sinz=\sqrt{x}+\sqrt{y}$

Differentiate partially both sides w.r.t x and y

$sinz+(x-y)cosz\frac{\partial z}{\partial x}=\frac{1}{2\sqrt{x}}\to \left(5\right)$

$-sinz+(x-y)cosz\frac{\partial z}{\partial y}=\frac{1}{2\sqrt{y}}\to \left(6\right)$

$\left(5\right)\times x+\left(6\right)\times y\Rightarrow (x-y)sinz+(x-y)cosz(z\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y})=\frac{\sqrt{x}+\sqrt{y}}{2}$

$\therefore (x-y)cosz(x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y})=\frac{-(x-y)sinz}{2}$

$\therefore x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=\frac{-1}{2}tanz$

Here k=$\frac{-1}{2}$

$\therefore A>C>B>D(\because 3>2>1>\frac{-1}{2})$

$\left(A\right)z=x^{3}sin\left(\frac{x}{y}\right)$

This is a homogenous function degree '3'

$\therefore x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial x}=3z$ $\therefore$ here k=3

$\left(B\right)Z=\frac{x^3+y^3}{x-y}$

$(x-y)z=x^3+y^3$

Partially differentiate on both sides w.r.t x and y

$z+\frac{\partial z}{\partial x}(x-y)=3x^2\to \left(1\right)$

$-z+\frac{\partial z}{\partial y}(x-y)=3y^2\to \left(2\right)$

$\left(1\right)\times x+\left(2\right)\times y\Rightarrow (x-y)z+(x-y)(x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y})=3(x-y)z$

$\therefore x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=2z$

Partially differentiate on both sides w.r.t x

$x\frac{{\partial }^{2}z}{\partial x^2}+\frac{\partial z}{\partial x}+y\frac{{\partial }^{2}z}{\partial x\partial y}=2\frac{\partial z}{\partial x}$

$\therefore \frac{x\frac{{\partial }^{2}z}{\partial x^2}+y\frac{{\partial }^{2}z}{\partial x\partial y}}{\frac{\partial z}{\partial x}}=1$ , here k=1

$\left(C\right)z=log\left(\frac{x^3+y^3+3x{y}^2}{x+y}\right)$

$(x+y)e^2=x^3+y^3+3x{y}^2$

Partially differentiate on both sides w.r.t x and y

$e^2+e^2(x+y)\frac{\partial z}{\partial x}=3x^2+3y^2\to \left(3\right)$

$e^2+e^2(x+y)\frac{\partial z}{\partial y}=3x^2+6xy\to \left(4\right)$

$\left(3\right)\times x+\left(4\right)\times y\Rightarrow e^2(x+y)+e^2(x+y)(a\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y})=3(x^3+y^3+3x{y}^2)$

$3(x+y)e^2$

$\therefore x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=2$ Here k=2

$\left(D\right)z=si{n}^{-1}\left(\frac{\sqrt{x}+\sqrt{y}}{x-y}\right)$

$sinz=\frac{\sqrt{x}+\sqrt{y}}{x-y}$

$(x-y)sinz=\sqrt{x}+\sqrt{y}$

Differentiate partially both sides w.r.t x and y

$sinz+(x-y)cosz\frac{\partial z}{\partial x}=\frac{1}{2\sqrt{x}}\to \left(5\right)$

$-sinz+(x-y)cosz\frac{\partial z}{\partial y}=\frac{1}{2\sqrt{y}}\to \left(6\right)$

$\left(5\right)\times x+\left(6\right)\times y\Rightarrow (x-y)sinz+(x-y)cosz(z\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y})=\frac{\sqrt{x}+\sqrt{y}}{2}$

$\therefore (x-y)cosz(x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y})=\frac{-(x-y)sinz}{2}$

$\therefore x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=\frac{-1}{2}tanz$

Here k=$\frac{-1}{2}$

$\therefore A>C>B>D(\because 3>2>1>\frac{-1}{2})$