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Variable Separable Method

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Question

Let the solution curve $y = y (x)$ of the differential equation $(4 + x^{2}) d y - 2 x (x^{2} + 3 y + 4) d x = 0$ pass through the origin. Then $y (2)$ is equal to

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Solution

$(4 + x^{2}) d y - 2 x (x^{2} + 3 y + 4) d x = 0$
$\Rightarrow \frac{d y}{d x} = (\frac{6 x}{x^{2} + 4}) y + 2 x$

$\Rightarrow \frac{d y}{d x} - (\frac{6 x}{x^{2} + 4}) y = 2 x$

I.F. $= e^{- 3 ln (x^{2} + 4)} = \frac{1}{{(x^{2} + 4)}^{3}}$

So $\frac{y}{{(x^{2} + 4)}^{3}} = \int \frac{2 x}{{(x^{2} + 4)}^{3}} d x + c$

$\Rightarrow y = - \frac{1}{2} (x^{2} + 4) + c {(x^{2} + 4)}^{3}$

When $x = 0, y = 0$ gives $c = \frac{1}{32}$ ,

So, for $x = 2, y = 12$

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