Let _0^x√1-(y^'(t))^2dt= _x^0y(t)dt,0 x 3,y 0,y(0)=0.

Question

Let $\int _{0}^{x}\sqrt{1-{({y}^{'}(t))}^{2}}dt={\int }_{x}^{0}y(t)dt,0\le x\le 3,y\ge 0,y(0)=0$. Then at $x=2,{y}^{''}+y+1$ is equal to

Accepted Answer

$\int _{0}^{x}\sqrt{1-{({y}^{'}(t))}^{2}}dt=\int _{0}^{x}y(t)dt$

Differentiating both side

$\begin{gathered} \sqrt{1-{({y}^{'}(x))}^{2}}=y(x) \ {(\frac{dy}{dx})}^{2}+{y}^{2}=1 \ {y}^{'2}+{y}^{2}=1 \ 2{y}^{'}{y}^{''}+2y{y}^{'}=0 \ {y}^{''}+y=0 \end{gathered}$

$	herefore $

Solutions · 1