Let [_0^x √1-(y^(t))^2 d t=_0^x y(t) d t, 0 x 3, y 0, y(0)=0].

Question

Let $$\int_\limits0^x \sqrt{1-\left(y^{\prime}(t)\right)^2} d t=\int_0^x y(t) d t, 0 \leq x \leq 3, y \geq 0, y(0)=0$$. Then at $$x=2, y^{\prime \prime}+y+1$$ is equal to

Accepted Answer

$$\int_\limits0^x \sqrt{1-\left(y^{\prime}(t)\right)^2} d t=\int_\limits0^x y(t) d t$$

Differentiating both side

$$\begin{aligned}
& \sqrt{1-\left(y^{\prime}(x)\right)^2}=y(x) \
& \left(\frac{d y}{d x}\right)^2+y^2=1 \
& y^{\prime 2}+y^2=1 \
& 2 y^{\prime} y^{\prime \prime}+2 y y^{\prime}=0 \
& y^{\prime \prime}+y=0
\end{aligned}$$

$$\therefore$$

Solutions · 1