Question

Let $\int_0^x \sqrt{1-(y'(t))^2} dt = \int_0^x y(t) dt, 0 \leq x \leq 3, y \geq 0, y(0)=0$. Then at $x=2, y''+y+1$ is equal to

Answer 1

1

Answer 2

Differentiating both sides of the given integral equation with respect to $x$ using the Fundamental Theorem of Calculus, we get: $\sqrt{1-(y'(x))^2} = y(x)$ Since $y(x) \geq 0$, we can square both sides: $1-(y'(x))^2 = (y(x))^2$ Rearranging the terms, we obtain the differential equation: $(y'(x))^2 + (y(x))^2 = 1$ Differentiating this equation with respect to $x$: $2y'(x)y''(x) + 2y(x)y'(x) = 0$ $2y'(x)(y''(x) + y(x)) = 0$ This implies that for any $x$, either $y'(x) = 0$ or $y''(x) + y(x) = 0$.

If $y'(x) = 0$, then from $(y'(x))^2 + (y(x))^2 = 1$, we have $0^2 + (y(x))^2 = 1$, which means $y(x) = \pm 1$. Since we are given $y \geq 0$, $y(x) = 1$. In this case, $y''(x) = 0$. So, $y''(x) + y(x) + 1 = 0 + 1 + 1 = 2$.

If $y''(x) + y(x) = 0$, then the expression $y''(x) + y(x) + 1$ becomes $0 + 1 = 1$.

To determine which case applies, we need to find the specific solution $y(x)$. From $\sqrt{1-(y'(x))^2} = y(x)$, we have $\frac{dy}{dx} = \pm \sqrt{1-y^2}$. If $\frac{dy}{dx} = \sqrt{1-y^2}$, separating variables gives $\frac{dy}{\sqrt{1-y^2}} = dx$. Integrating yields $\arcsin(y) = x + C$. Thus, $y(x) = \sin(x+C)$. Using $y(0)=0$, we get $C=n\pi$. So $y(x) = \sin(x+n\pi)$. For $y \geq 0$ on $[0, 3]$, we must have $y(x) = \sin(x)$ (when $n$ is even). If $\frac{dy}{dx} = -\sqrt{1-y^2}$, separating variables gives $\frac{dy}{\sqrt{1-y^2}} = -dx$. Integrating yields $\arcsin(y) = -x + C$. Thus, $y(x) = \sin(C-x)$. Using $y(0)=0$, we get $C=n\pi$. So $y(x) = \sin(n\pi-x)$. For $y \geq 0$ on $[0, 3]$, we must have $y(x) = \sin(x)$ (when $n$ is odd).

So, the unique solution is $y(x) = \sin(x)$. For $y(x) = \sin(x)$, we have $y'(x) = \cos(x)$ and $y''(x) = -\sin(x)$. The expression $y''(x) + y(x) + 1$ becomes $-\sin(x) + \sin(x) + 1 = 1$. This is consistent with the case where $y''(x)+y(x)=0$. The case $y'(x)=0$ occurs when $\cos(x)=0$, i.e., $x=\frac{\pi}{2}$. At this point, $y(\frac{\pi}{2}) = \sin(\frac{\pi}{2}) = 1$. Here $y''(x)+y(x)+1 = -1+1+1=1$. Thus, for $y(x)=\sin(x)$, the expression $y''+y+1$ is always equal to 1. Therefore, at $x=2$, $y''(2)+y(2)+1 = 1$.