Write the correct alternative in the following:
Question:

Write the correct alternative in the following:

If $x y-\log _{e} y=1$ satisfies the equation $x\left(y y_{2}+y_{1}^{2}\right)-y_{2}+\lambda y y_{1}=0$, then $\lambda=$

A. $-3$

B. 1

C. 3

D. none of these

Solution:

Given:

$x y-\log _{e} y=1$

$x y=\log _{e} y+1$

Differentiate w.r.t. ‘ $x$ ‘ on both sides;

$\mathrm{y}+\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{d y}{d x}\left(\frac{1}{y}-x\right)=y$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}^{2}}{(1-\mathrm{xy})}$

$\left(\frac{d y}{d x}\right)^{2}=\left[\frac{y^{2}}{(1-x y)}\right]^{2}$

$=\frac{y^{4}}{(1-x y)^{2}}$

$\frac{d}{d x}\left[\frac{d y}{d x}\right]=\frac{d}{d x}\left[\frac{y^{2}}{(1-x y)}\right]$

$=\frac{1}{(1-x y)^{2}}\left\{2 y \frac{d y}{d x}(1-x y)-y^{2}\left(-y+x \frac{d y}{d x}\right)\right\}$

$=\frac{1}{(1-x y)^{2}}\left\{2 y \frac{d y}{d x}(1-x y)-y^{2}\left(-y+x \frac{d y}{d x}\right)\right\}$

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

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

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

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

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

$y \frac{d^{2} y}{d y^{2}}+\left(\frac{d y}{d x}\right)^{2}=\frac{y^{3}}{(1-x y)^{2}}\left\{3 y+x \frac{d y}{d x}\right\}+\frac{y^{4}}{(1-x y)^{2}}$

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

$=\frac{y^{3}}{(1-x y)^{2}}\left\{4 y+x \frac{d y}{d x}\right\}$

$x\left[y \frac{d^{2} y}{d y^{2}}+\left(\frac{d y}{d x}\right)^{2}\right]=\frac{y^{3} x}{(1-x y)^{2}}\left\{4 y+x \frac{d y}{d x}\right\}$

$x\left[y \frac{d^{2} y}{d y^{2}}+\left(\frac{d y}{d x}\right)^{2}\right]-\frac{d^{2} y}{d y^{2}}=\frac{y^{3} x}{(1-x y)^{2}}\left\{4 y+x \frac{d y}{d x}\right\}-\frac{y^{2}}{(1-x y)^{2}}\left\{3 y+x \frac{d y}{d x}\right\}$

$=\frac{y^{2}}{(1-x y)^{2}}\left\{x y\left(4 y+x \frac{d y}{d x}\right)-3 y-x \frac{d y}{d x}\right\}$

$=\frac{y^{2}}{(1-x y)^{2}}\left\{4 x y^{2}+x^{2} y \frac{d y}{d x}-3 y-x \frac{d y}{d x}\right\}$

$=\frac{y^{2}}{(1-x y)^{2}}\left\{y(4 x y-3)+x \frac{d y}{d x}(x y-1)\right\}$

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

$=\frac{y^{2}}{(1-x y)^{2}}\left\{y(x y-3(1-x y))-x \frac{d y}{d x} \frac{y^{2}}{d y}\right\}$

$=\frac{y^{2}}{(1-x y)^{2}}\left\{y\left(x y-3 \frac{y^{2}}{\frac{d y}{d x}}\right)-x y^{2}\right\}$

$=\frac{y^{2}}{(1-x y)^{2}}\left\{x y^{2}-3 \frac{y^{3}}{\frac{d y}{d x}}-x y^{2}\right\}$

$=-\frac{y^{2}}{(1-x y)^{2}}\left\{3 \frac{y^{3}}{\frac{d y}{d x}}\right\}$

Since $x\left[y \frac{d^{2} y}{d y^{2}}+\left(\frac{d y}{d x}\right)^{2}\right]-\frac{d^{2} y}{d y^{2}}+\lambda y \frac{d y}{d x}=0$

So, $x\left[y \frac{d^{2} y}{d y^{2}}+\left(\frac{d y}{d x}\right)^{2}\right]-\frac{d^{2} y}{d y^{2}}=-\lambda y \frac{d y}{d x}$

$-\lambda y \frac{d y}{d x}=-\frac{y^{2}}{(1-x y)^{2}}\left\{3 \frac{y^{3}}{\frac{d y}{d x}}\right\}$

$-\lambda y \frac{y^{2}}{(1-x y)}=-\frac{y^{2}}{(1-x y)^{2}}\left\{3 \frac{y^{3}}{\frac{d y}{d x}}\right\}$

$\lambda y=\frac{1}{(1-x y)}\left\{3 \frac{y^{3}}{\frac{d y}{d x}}\right\}$

$\lambda=\frac{3 y^{2}}{(1-x y) \frac{d y}{d x}}$

$\lambda=\frac{3 \frac{d y}{d x}}{\frac{d y}{d x}}$

$\lambda=3$