Let y=y(x) be the solution of the differential equation. (x2+1)2dydx+2x(x2+1)y=1 such that y(0)=0 , then y(1) is equal to

dydx+(2x1+x2)y=1(1+x2)2I.F = e∫2x1+x2dx=elog(1+x2)=1+x2 ⇒∴(1+x2)y=∫11+x2dx ∴y(1+x2)=Tan−1x+c

Let y=y(x) be the solution of the differential equation. (x2+1)2dydx+2x(x2+1)y=1 such that y(0)=0 , then y(1) is equal to

dydx+(2x1+x2)y=1(1+x2)2I.F = e∫2x1+x2dx=elog(1+x2)=1+x2 ⇒∴(1+x2)y=∫11+x2dx ∴y(1+x2)=Tan−1x+c

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Let $y = y (x)$ be the solution of the differential equation. ${(x^{2} + 1)}^{2} \frac{d y}{d x} + 2 x (x^{2} + 1) y = 1$ such that $y (0) = 0$ , then $y (1)$ is equal to

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$\frac{3 π}{8}$

$\frac{π}{4}$

$\frac{π}{8}$

$\frac{π}{2}$

answer is A.

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Detailed Solution

$\frac{d y}{d x} + (\frac{2 x}{1 + x^{2}}) y = \frac{1}{{(1 + x^{2})}^{2}}$
I.F = $e^{\int \frac{2 x}{1 + x^{2}} d x} = e^{\log (1 + x^{2})} = 1 + x^{2}$
$\Rightarrow ∴ (1 + x^{2}) y = \int \frac{1}{1 + x^{2}} d x$
$∴ y (1 + x^{2}) = T a n^{- 1} x + c$