If a curve y=f(x) passes through the point (1,-1) and satisfies the differential equation, y(1+xy) dx=x dy, then f\left( { - {1 \over 2}} \right) is equal to :

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Answer: B. {4 \over 5} y\left( {1 + xy} \right)dx = xdy {{xdy - ydx} \over {{y^2}}} = xdx \Rightarrow \int { - d\left( {{x \over y}} \right) = \int {xdx} } \,\,\,\,\,\,\,\,\,\,\,\,\,\, - {x \over y} = {{{x^2}} \over 2} + C\,\, as \,\,\,y\left( 1 \right) = - 1 \Rightarrow C = {1 \over 2} Hence, y = {{ - 2x} \over {{x^2} + 1}} \Rightarrow f\left( {{{ - 1} \over 2}} \right) = {4 \over 5}

If a curve $$y=f(x)$$ passes through the point $$(1,-1)$$ and satisfies the differential equation, $$y(1+xy) dx=x$$ $$dy$$, then $$f\left( { - {1 \over 2}} \right)$$ is equal to :

Answer

Explanation

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