Let y = y(x) be the solution of the differential equation . If y(0) = 0, then y(2) is equal to [2024]

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Solution

Q.
Let y = y(x) be the solution of the differential equation ${(x^{2} + 4)}^{2} d y + (2 x^{3} y + 8 x y - 2) d x = 0$ . If y(0) = 0, then y(2) is equal to [2024]

1 $2 π$

2 $\frac{π}{16}$

3 $\frac{π}{8}$

4 $\frac{π}{32}$

Ans.
(4)

We Have, ${(x^{2} + 4)}^{2} d y + (2 x^{3} y + 8 x y - 2) d x = 0$

$\Rightarrow \frac{d y}{d x} + \frac{y (2 x^{3} + 8 x)}{{(x^{2} + 4)}^{2}} = \frac{2}{{(x^{2} + 4)}^{2}}$

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

$I F = e^{\int \frac{2 x}{x^{2} + 4} d x} = e^{\log_{e} (x^{2} + 4)} = x^{2} + 4$

$∴$ Solution is $y (x^{2} + 4) = \int \frac{2}{{(x^{2} + 4)}^{2}} \times (x^{2} + 4) d x + C$

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

$\Rightarrow y (x^{2} + 4) = 2 \times \frac{1}{2} \tan^{- 1} (\frac{x}{2}) + C$

$\Rightarrow y (x^{2} + 4) = \tan^{- 1} (\frac{x}{2}) + C$ ... (i)

Now. We have y(0) = 0

$\Rightarrow 0 \cdot (0 + 4) = \tan^{- 1} (0) + C \Rightarrow C = 0$

$∴ y (x^{2} + 4) = \tan^{- 1} \frac{x}{2}$

So, at x = 2

$y (4 + 4) = \tan^{- 1} (\frac{2}{2}) \Rightarrow y = \frac{1}{8} \times \tan^{- 1} 1 = \frac{1}{8} \times \frac{π}{4}$

$\Rightarrow y = \frac{π}{32}$

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