The solution of dydx-2xy1-x2-0 is

The correct option is A y=c ( 1+x^2 ) dydx=2xy1+x^2 ⇒∫dyy=∫2x1+x^2dx ⇒ #160; log y=log(1+x^2)c y=c(1+x^2) ...

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Standard XII

Mathematics

Variable Separable Method

The solution ...

Question

The solution of $\frac{d y}{d x} - \frac{2 x y}{1 + x^{2}} = 0$ is

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

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y = c \sqrt{1 + x^{2}}

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y = \frac{c}{1 + x^{2}}

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y = \frac{c}{\sqrt{1 + x^{2}}}

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Solution

The correct option is A $y = c (1 + x^{2})$
$\frac{d y}{d x} = \frac{2 x y}{1 + x^{2}}$

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

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

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Similar questions

The solution of differential equation $\frac{d y}{d x} + \frac{2 x y}{1 + x^{2}} = \frac{1}{(1 + x^{2})^{2}}$ is
(a) $y (1 + x^{2}) = C + t a n^{- 1} x$
(b) $\frac{y}{1 + x^{2}} = C + t a n^{- 1} x$
(c) $y l o g (1 + x^{2}) = C + t a n^{- 1} x$
(d) $y (1 + x^{2}) = C + s i n^{- 1} x$