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

is equal to :

Option 1)
$-\frac{2}{5}$

Option 2)
$-\frac{4}{5}$

Option 3)
$\frac{2}{5}$

Option 4)
$\frac{4}{5}$

Answers (1)

As we learnt in

Linear Differential Equation -

$\frac{dy}{dx}+Py= Q$

- wherein

P, Q are functions of x alone.

$\frac{dy}{dx}.x=y+xy^{2}$

$\Rightarrow \frac{1}{y^{2}}\frac{dy}{dx}=\frac{1}{y.x}+1\Rightarrow put\frac{1}{y}=t$

$\Rightarrow -\frac{dt}{dx}=\frac{t}{x}+1\:\:\:\:\:\:\:\:\:\:\:\:\:-\frac{1}{y^{2}}\frac{dy}{dx}=\frac{dt}{dx}$

$\therefore \frac{dt}{dx}+\frac{t}{x}=-1$

$P=\frac{1}{x},Q=-1$

$\int P.dx=logx$

$I.F.=e^{logx}=x$

$\therefore t.x=\int -xdx=-\frac{x^{2}}{2}+C$

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

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

Put $x=-\frac{1}{2}$

$\Rightarrow -\frac{1}{2y}=-\frac{1}{8}-\frac{1}{2}$

$\Rightarrow \frac{1}{y}=\frac{1}{4}+1=\frac{5}{4}$

$\therefore y= \frac{4}{5}$

Option 1)

$-\frac{2}{5}$

This solution is incorrect

Option 2)

$-\frac{4}{5}$

This solution is incorrect

Option 3)

$\frac{2}{5}$

This solution is incorrect

Option 4)

$\frac{4}{5}$

This solution is correct

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