Question

If xy – log_e y = 1 satisfies the equation x(yy₂ + y₁²) – y₂ + λyy₁ = 0, then λ = 

A. –3 

B. 1 

C. 3 

D. none of these

Answer 1

Correct Answer is (C) 3

Given:

xy - log_ey = 1

xy = log_ey + 1

Differentiate w.r.t. ‘x’ on both sides;

\(y+x\frac{dy}{dx}=\frac{1}{y}\frac{dy}{dx}\)

\(\frac{dy}{dx}(\frac{1}{y}-x)=y\)

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

\((\frac{dy}{dx})^2=\begin{bmatrix}\frac{y^2}{(1-xy)}\end{bmatrix}^2\)

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

\(\frac{d}{dx}[\frac{dy}{dx}]=\frac{d}{dx}[\frac{y^2}{(1-xy)}]\)

\(=\frac{1}{(1-xy)^2}\begin{Bmatrix}2y\frac{dy}{dx}(1-xy)-y^2(-y+x\frac{dy}{dx})\end{Bmatrix}\)

Since,

So,

λ = 3