Let 0 x 1 - ( y ' ( t ) ) 2 d t = 0 x y ( t ) d t ,0 ? x ? 3 , y ? 0 , y(0) = 0. Then at x = 2, y" + y + 1 is equal to [2024]

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Solution

Q.
Let $\int_{0}^{x} \sqrt{1 - (y' {(t))}^{2}} d t = \int_{0}^{x} y (t) d t$ , $0 \leq x \leq 3, y \geq 0$ , y(0) = 0. Then at x = 2, y" + y + 1 is equal to [2024]

1 2

2 $\frac{1}{2}$

3 $\sqrt{2}$

4 1

Ans.
(4)

We have $\int_{0}^{x} \sqrt{1 - (y' {(t))}^{2}} d t = \int_{0}^{x} y (t) d t$

On differentiating both sides, we get

$\sqrt{1 - (y' {(x))}^{2}} = y (x)$

$\Rightarrow 1 - {(y')}^{2} = y^{2} \Rightarrow y^{2} + {(y')}^{2} = 1$

$\Rightarrow 2 y y' + 2 y " = 0 \Rightarrow y y' + y " = 0$

So, at x = 2, y" + y + 1 = 0 + 1 = 1.

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