If a curve y = f(x) passes through the point (1, – 1) and satisfies the differential equation y (1 + xy) dx = x dy, then is equal to

Engineering

Mathematics

Exact Differential Equation

Question

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

$\frac{2}{5}$

$\frac{4}{5}$

$\frac{- 2}{5}$

$\frac{- 4}{5}$

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y (1 + xy) dx = x dy

ydx – xdy = – xy² dx

$\frac{ydx - xdy}{y^{2}} = - x dx$

$d (\frac{x}{y}) = x dx$

$\frac{x}{y} = \frac{- x^{2}}{2} + C$

$- 1 = \frac{- 1}{2} + C$ $\Rightarrow C = \frac{- 1}{2}$

$\frac{x}{y} = \frac{- x^{2}}{2} - \frac{1}{2} \Rightarrow \frac{- 2 x}{y} = x^{2} + 1 \Rightarrow y = \frac{- 2 x}{x^{2} + 1}$

$f (\frac{- 1}{2}) = \frac{- 2 (\frac{- 1}{2})}{\frac{1}{4} + 1} = \frac{4}{5} .$