Let y = y(x) be the solution of the differential equation ( 1 + x 2 ) d y d x + y =e tan - 1 x , y(1) = 0. Then y(0) is [2024]

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Solution

Q.
Let y = y(x) be the solution of the differential equation $(1 + x^{2}) \frac{d y}{d x} + y = e^{\tan^{- 1} x}$ , y(1) = 0. Then y(0) is [2024]

1 $\frac{1}{2} (e^{π / 2} - 1)$

2 $\frac{1}{2} (1 - e^{π / 2})$

3 $\frac{1}{4} (e^{π / 2} - 1)$

4 $\frac{1}{4} (1 - e^{π / 2})$

Ans.
(2)

We Have, $(1 + x^{2}) \frac{d y}{d x} + y = e^{\tan^{- 1} x}$

$\Rightarrow \frac{d y}{d x} + \frac{y}{1 + x^{2}} = \frac{e^{\tan^{- 1} x}}{1 + x^{2}}$

$IF = e^{\int \frac{1}{1 + x^{2}} d x} = e^{\tan^{- 1} x}$

$∴ y e^{\tan^{- 1} x} = \int \frac{e^{\tan^{- 1} x}}{1 + x^{2}} \cdot e^{\tan^{- 1} x} d x + c$

Let $\tan^{- 1} x = t$

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

$\Rightarrow y e^{t} = \frac{e^{2 t}}{2} + c \Rightarrow y e^{\tan^{- 1} x} = \frac{e^{2 \tan^{- 1} x}}{2} + c$

Since y(1) = 0

$\Rightarrow 0 e^{π / 4} = \frac{e^{π / 2}}{2} + c \Rightarrow c = \frac{- e^{π / 2}}{2}$

$∴ y = \frac{e^{\tan^{- 1} x}}{2} - \frac{e^{π / 2}}{2 e^{\tan^{- 1} x}}$

$\Rightarrow y (0) = \frac{1}{2} - \frac{e^{π / 2}}{2} = \frac{1 - e^{π / 2}}{2}$

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