Let y = y(x) be the solution of the differential equation ; y(0) = 0. The the area enclosed by the curve and the line y – x = 4 is __________. [2024]

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Solution

Q.
Let y = y(x) be the solution of the differential equation $\frac{d y}{d x} + \frac{2 x}{{(1 + x^{2})}^{2}} y = x e^{\frac{1}{(1 + x^{2})}}$ ; y(0) = 0. Then the area enclosed by the curve $f (x) = y (x) e^{- \frac{1}{(1 + x^{2})}}$ and the line y – x = 4 is __________. [2024]

Ans.
(18)

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

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

$∴$ General solution of given differential equation is,

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

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

Since y(0) = 0

$\Rightarrow 0 = 0 + c$

$\Rightarrow c = 0$

$∴ y (x) = \frac{x^{2}}{2} e^{\frac{1}{1 + x^{2}}}$

$\Rightarrow f (x) = \frac{x^{2}}{2}$

So, intersection of curve $y = f (x) = \frac{x^{2}}{2}$ and y = x + 4 is given by $\frac{x^{2}}{2} = x + 4$

$\Rightarrow x^{2} - 2 x - 8 = 0 \Rightarrow (x - 4) (x + 2) = 0$

$\Rightarrow x = 4, - 2 \Rightarrow y = 8, 2$

$∴$ Required area = $\int_{- 2}^{4} ((x + 4) - \frac{x^{2}}{2}) d x$

= ${[\frac{x^{2}}{2} + 4 x - \frac{x^{3}}{6}]}_{- 2}^{4} = [8 + 16 - \frac{32}{3} - 2 + 8 - \frac{8}{6}]$

= 30 – 12 = 18.

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