If the Solution y = y(x) of the differential equation ( x 4 + 2 x 3 + 3 x 2 + 2 x + 2 ) d y - ( 2 x 2 + 2 x + 3 ) d x = 0 satisfies y ( - 1 ) = 4 , then y(0) is equal [2024]

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Solution

Q.
If the Solution y = y(x) of the differential equation $(x^{4} + 2 x^{3} + 3 x^{2} + 2 x + 2) d y - (2 x^{2} + 2 x + 3) d x = 0$ satisfies $y (- 1) = - \frac{π}{4}$ , then y(0) is equal [2024]

1 $\frac{π}{4}$

2 $\frac{π}{2}$

3 $- \frac{π}{12}$

4 0

Ans.
(1)

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

$\frac{d y}{d x} = \frac{2 x^{2} + 2 x + 3}{x^{4} + 2 x^{3} + 3 x^{2} + 2 x + 2}$

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

$\Rightarrow \int d y = \int \frac{2 x^{2} + 2 x + 3}{(x^{2} + 1) (x^{2} + 2 x + 2)} d x$

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

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

Now, $y (- 1) = \frac{- π}{4}$

$∴$ From (i), we get $\frac{- π}{4} = \tan^{- 1} (- 1) + \tan^{- 1} (1 - 1) + C$

$\Rightarrow \frac{- π}{4} = - \frac{π}{4} + C \Rightarrow C = 0$

So, $y (x) = \tan^{- 1} (x) + \tan^{- 1} (1 + x)$

Now, $y (0) = \tan^{- 1} (0) + \tan^{- 1} (1 + 0) = \frac{π}{4}$ .

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