Question

The equation of the curve satisfying the equation (xy – x²) $\frac{dy}{dx}$= y² and passing through the point (–1, 1), is –

• A. y = (log y – 1) x
• B. y = (log y + 1) x
• C. x = (log x – 1) y
• D. x = (log x + 1) y

Answer 1

Answer: (A)

y = (log y – 1) x

Answer 2

We have, (xy – x²) $\frac{dy}{dx}$= y² Ž y²$\frac{dx}{dy}$ = xy – x²

Ž$\frac{1}{x^{2}}$ $\frac{dx}{dy}$ – $\frac{1}{x}$ . $\frac{1}{y}$ = $\frac{–1}{y^{2}}$

Let $\frac{1}{x}$ = V Ž – $\frac{1}{x^{2}}$ $\frac{dx}{dy}$ = $\frac{dV}{dy}$, we obtain

$\frac{dV}{dy}$ + $\frac{V}{y}$ = $\frac{1}{y^{2}}$, which is linear.

I. F. = $e^{\int_{}^{}{\frac{1}{y}dy}}$= e^{log y} = y.

\ The solution is

Vy = $\int_{}^{}\frac{1}{y^{2}}$ . y dy + c

Ž $\frac{y}{x}$ = log y + c

Ž y = x (log y + c).

This passes through the point (–1, 1).

\ 1 = –1 (log 1 + c) Ž c = –1.

Hence (1) is the correct answer