Let y = y(x) be the solution curve of the differential equation ( 1 + x 2 ) ? d y + ( y - tan - 1 x ) d x = 0 , y ( 0 ) = 1 . Then the value of y(1) is: [2026]

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Q.
Let $y = y (x)$ be the solution curve of the differential equation $(1 + x^{2}) d y + (y - \tan^{- 1} x) d x = 0, y (0) = 1 .$ Then the value of $y (1)$ is: [2026]

1 $\frac{4}{e^{π / 4}} - \frac{π}{2} - 1$

2 $\frac{2}{e^{π / 4}} + \frac{π}{4} - 1$

3 $\frac{4}{e^{π / 4}} + \frac{π}{2} - 1$

4 $\frac{2}{e^{π / 4}} - \frac{π}{4} - 1$

Ans.
(2)

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

$I.F. = e^{\tan^{- 1} x}$

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

$y \times e^{\tan^{- 1} x} = \tan^{- 1} x (e^{\tan^{- 1} x}) - e^{\tan^{- 1} x} + c$

$y (0) = 1 \Rightarrow c = 2$

$y (1) = \frac{2}{e^{π / 4}} + \frac{π}{4} - 1$

Solution

Q.
Let $y = y (x)$ be the solution curve of the differential equation $(1 + x^{2}) d y + (y - \tan^{- 1} x) d x = 0, y (0) = 1 .$ Then the value of $y (1)$ is: [2026]

1 $\frac{4}{e^{π / 4}} - \frac{π}{2} - 1$

2 $\frac{2}{e^{π / 4}} + \frac{π}{4} - 1$

3 $\frac{4}{e^{π / 4}} + \frac{π}{2} - 1$

4 $\frac{2}{e^{π / 4}} - \frac{π}{4} - 1$

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Solution

Q. Let y=y(x) be the solution curve of the differential equation (1+x2)dy+(y-tan-1x)dx=0, y(0)=1. Then the value of y(1) is: [2026]

1 4eπ/4-π2-1

2 2eπ/4+π4-1

3 4eπ/4+π2-1

4 2eπ/4-π4-1

Ans. (2) dydx+yx2+1=tan-1xx2+1 I.F.=etan-1x y×etan-1x=∫etan-1x·tan-1x1+x2dx y×etan-1x=tan-1x(etan-1x)-etan-1x+c y(0)=1 ⇒ c=2 y(1)=2eπ/4+π4-1

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Q.
Let $y = y (x)$ be the solution curve of the differential equation $(1 + x^{2}) d y + (y - \tan^{- 1} x) d x = 0, y (0) = 1 .$ Then the value of $y (1)$ is: [2026]

1 $\frac{4}{e^{π / 4}} - \frac{π}{2} - 1$

2 $\frac{2}{e^{π / 4}} + \frac{π}{4} - 1$

3 $\frac{4}{e^{π / 4}} + \frac{π}{2} - 1$

4 $\frac{2}{e^{π / 4}} - \frac{π}{4} - 1$