Let $\int_0^x \sqrt{1-\left(y^{\prime}(t)\right)^2} d t=\int_0^x y(t) d t, 0 \leq x \leq 3, y \geq 0, y(0)=0$. Then at $x=2, y^{\prime \prime}+y+1$ is equal to

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Aicharya · Accepted Answer

$\begin{aligned} & \sqrt{1-\left(y^{\prime}(x)ight)^2}=y(x) \ & 1-\left(\frac{d y}{d x}ight)^2=y^2 \ & \left(\frac{d y}{d x}ight)^2=1-y^2\end{aligned}$ $\begin{aligned} & \frac{d y}{\sqrt{1-y^2}}=d x 	ext { OR } \frac{d y}{\sqrt{1-y^2}}=-d x \ & \Rightarrow \sin ^{-1} y=x+c, \sin ^{-1} y=-x+c \ & x=0, y=0 \Rightarrow c=0 \ & \sin ^{-1} y=x, 	ext { as } y \geq 0 \ & \sin x=y \ & \Rightarrow \frac{d y}{d x}=\cos x \ & \frac{d^2 y}{d x^2}=-\sin x \ & \Rightarrow-\sin x+\sin x+1=1\end{aligned}$

Let $\int_0^x \sqrt{1-\left(y^{\prime}(t)\right)^2} d t=\int_0^x y(t) d t, 0 \leq x \leq 3, y \geq 0,…

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