Question

The equation of a curve passing through (0, 1) and having gradient $\frac{–(y + y^{3})}{1 + x + xy^{2}}$at (x, y) is-

• A. xy + tan^–1 y = $\frac{\pi}{2}$
• B. xy + tan^–1 y = $\frac{\pi}{4}$
• C. xy – tan^–1 y = $\frac{\pi}{2}$
• D. xy – tan^–1y = $\frac{\pi}{4}$

Answer 1

Answer: (B)

xy + tan^–1 y = $\frac{\pi}{4}$

Answer 2

We have, $\frac{dy}{dx}$ = $\frac{–(y + y^{3})}{1 + x(1 + y^{2})}$

or $\frac{dx}{dy}$ = – $\frac{1 + x(1 + y^{2})}{y + y^{3}}$

= – $\left\lbrack \frac{1}{y(y^{2} + 1)} + \frac{x(1 + y^{2})}{y(1 + y^{2})} \right\rbrack$

Ž $\frac{dx}{dy}$ + $\frac{x}{y}$ = $\frac{- 1}{y(1 + y^{2})}$.

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

Hence the solution is,

x . y = – $\int_{}^{}\frac{1}{y(1 + y^{2})}$ . y dy + C

= – $\int_{}^{}\frac{dy}{1 + y^{2}}$ + C = – tan^–1 y + C

or xy + tan^–1 y = C.

This passes through (0, 1), therefore, tan^–1 1 = C i.e. C = $\frac{\pi}{4}$.

Thus, the equation of the curve is

xy + tan^–1 y = $\frac{\pi}{4}$