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Let $$\int_0^x \sqrt{1 - (y'(t))^2} \, dt = \int_0^x y(t) \, dt, \, 0 \le x \le 3, \, y \ge 0, \, y(0) = 0$$. Then at $$x = 2$$, $$y'' + y + 1$$ is equal to
Differentiate both sides of
$$\int_0^x\sqrt{1-(y'(t))^2}dt=\int_0^xy(t)dt$$
$$\Rightarrow\sqrt{1-(y')^2}=y$$
Square:
$$1-(y')^2=y^2;\Rightarrow;(y')^2=1-y^2$$
$$Since(y\ge0)and(y(0)=0),takepositiveroot:$$
$$y'=\sqrt{1-y^2}$$
Separate:
$$\frac{dy}{\sqrt{1-y^2}}=dx$$
$$\Rightarrow\sin^{-1}(y)=x$$
$$\Rightarrow y=\sin x$$
$$y''=-\sin x$$
$$ y''+y+1=(-\sin x)+(\sin x)+1=1$$
At (x=2): 1
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