Let ∫ √1- (y' (t))^2 dt ; t ∈[0, x] = ∫ y(t)dt, ; t ∈[0, x] 0 ≤ x ≤ 3, y ≥ 0, y (o) = 0. then at x = 2, y" + y + 1 is equal to :

Question

Let \(\int\limits_{0}^{\mathrm{x}} \sqrt{1-\left(\mathrm{y}^{\prime}(\mathrm{t})\right)^{2}} \ \mathrm{dt}=\int\limits_{0}^{\mathrm{x}} \mathrm{y}(\mathrm{t}) \mathrm{dt}, 0 \leq \mathrm{x} \leq 3, \mathrm{y} \geq 0\),

\(y(0)=0\). Then at \(x=2, \ y^{\prime \prime}+y+1\) is equal to : 

(1) 1

(2) 2

(3) \( \sqrt{2}\)

(4) \(1 / 2\) 

Answer 1

Correct option is : (1) 1 

\(\sqrt{1-\left(y^{\prime}(x)\right)^{2}}=y(x)\)

\(1-\left(\frac{d y}{d x}\right)^{2}=y^{2}\)

\(\left(\frac{d y}{d x}\right)^{2}=1-y^{2}\) 

\(\frac{d y}{\sqrt{1-y^{2}}}=d x\) OR \(\frac{d y}{\sqrt{1-y^{2}}}=-d x\)

\(\Rightarrow \sin ^{-1} y=x+c,\ \sin ^{-1} y=-x+c\)

\(\mathrm{x}=0, \ \mathrm{y}=0 \Rightarrow \mathrm{c}=0\)

\(\sin ^{-1} y=x, \ \text{as} \ y \geq 0\)

\(\sin x=y\)

\(\Rightarrow \frac{d y}{d x}=\cos x\)

\(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\sin \mathrm{x}\)

\(\Rightarrow-\sin x+\sin x+1=1\)