The solution of the differential equation 2 x 2 ydy -1- y 2 x 2 y 2- y 2-1 dx -0 -Where c is a constant-A- None of theseB- x2 y2-c x-11-y2C- x2 y2-c x-11-y2D- x2 y2-c x-11-y2

The correct option is D x^2 y^2=(cx 1)(1 y^2) 2y/(1 y^2)^2.dy/dx+y^2/1 y^21/x=1/x^3 Put y^2/1 y^2=t ⇒2y/(1 y^2)dy/dx=dt/dx ⇒dt/dx+t/x=1/x^3 ⇒ t.x=∫1/x^2dx+c ⇒ ...

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Standard XII

Mathematics

Integrating Factor

The solution ...

Question

The solution of the differential equation $2 x^{2} y d y + (1 - y^{2}) (x^{2} y^{2} + y^{2} - 1) d x = 0$ [Where c is a constant]

$x^{2} y^{2} = (c x + 1) (1 - y^{2})$

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$x^{2} y^{2} = (c x + 1) (1 + y^{2})$

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$x^{2} y^{2} = (c x - 1) (1 - y^{2})$

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None of these

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Solution

The correct option is C
$x^{2} y^{2} = (c x - 1) (1 - y^{2})$

$\frac{2 y}{(1 - y^{2})^{2}} . \frac{d y}{d x} + \frac{y^{2}}{1 - y^{2}} \frac{1}{x} = \frac{1}{x^{3}}$
Put $\frac{y^{2}}{1 - y^{2}} = t \Rightarrow \frac{2 y}{(1 - y^{2})} \frac{d y}{d x} = \frac{d t}{d x}$
$\Rightarrow \frac{d t}{d x} + \frac{t}{x} = \frac{1}{x^{3}} \Rightarrow t . x = \int \frac{1}{x^{2}} d x + c \Rightarrow x^{2} y^{2} = (c x - 1) (1 - y^{2}) .$

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The solution of the differential equation 2x2ydy+(1−y2)(x2y2+y2−1)dx=0 [Where c is a constant]

The correct option is C x2y2=(cx−1)(1−y2) 2y(1−y2)2.dydx+y21−y21x=1x3 Put y21−y2=t⇒2y(1−y2)dydx=dtdx ⇒dtdx+tx=1x3⇒t.x=∫1x2dx+c⇒x2y2=(cx−1)(1−y2).

The solution of the differential equation $2 x^{2} y d y + (1 - y^{2}) (x^{2} y^{2} + y^{2} - 1) d x = 0$ [Where c is a constant]