Let y = y(x) be the solution of the differential equation d y d x = 2 x ( x + y ) 3 - x ( x + y ) - 1 , y ( 0 ) = 1 . Then, ( 1 2 + y ( 1 2 ) ) 2 equals: [2024]

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Solution

Q.
Let y = y(x) be the solution of the differential equation $\frac{d y}{d x} = 2 x {(x + y)}^{3} - x (x + y) - 1, y (0) = 1$ . Then, $(\frac{1}{\sqrt{2}} + y {(\frac{1}{\sqrt{2}}))}^{2}$ equals: [2024]

1 $\frac{1}{2 - \sqrt{e}}$

2 $\frac{4}{4 + \sqrt{e}}$

3 $\frac{2}{1 + \sqrt{e}}$

4 $\frac{3}{1 - \sqrt{e}}$

Ans.
(1)

We have, $\frac{d y}{d x} = 2 x {(x + y)}^{3} - x (x + y) - 1, y (0) = 1$

Let $x + y = t \Rightarrow 1 + \frac{d y}{d x} = \frac{d t}{d x}$

Now, $\frac{d t}{d x} - 1 = 2 x t^{3} - x t - 1$

$\Rightarrow \frac{d t}{d x} = 2 x t^{3} - x t \Rightarrow \frac{1}{t^{3}} \frac{d t}{d x} + \frac{x}{t^{2}} - 2 x = 0$

Put $\frac{1}{t^{2}} = u \Rightarrow \frac{- 2}{t^{3}} \frac{d t}{d x} = \frac{d u}{d x}$ we get

$\frac{- 1}{2} \frac{d u}{d x} + x u = 2 x \Rightarrow \frac{d u}{d x} - 2 x u = - 4 x$

$∴ IF = e^{\int - 2 x d x = e^{- x^{2}}}$

Required solution = $u \cdot e^{- x^{2}} = \int e^{- x^{2}} \cdot (- 4 x) d x$

$\Rightarrow \frac{e^{- x^{2}}}{t^{2}} = \int e^{- x^{2}} \cdot (- 4 x) d x$

$\Rightarrow \frac{e^{- x^{2}}}{{(x + y)}^{2}} = 2 e^{- x^{2}} + C \Rightarrow \frac{1}{{(x + y)}^{2}} = 2 + C e^{x^{2}}$

At x = 0, y = 1, we get $1 = 2 + C \Rightarrow C = - 1$

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

At $x = \frac{1}{\sqrt{2}}, {(y + \frac{1}{\sqrt{2}})}^{2} = \frac{1}{2 - \sqrt{e}}$

$∴ (y {(\frac{1}{\sqrt{2}}) + \frac{1}{\sqrt{2}})}^{2} = (\frac{1}{2 - \sqrt{e}})$ .

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