Question

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

• A. 1
• B. 2
• C. $\sqrt{2}$
• D. $\frac{1}{2}$

Answer 1

Answer: (A)

1

Answer 2

The given equation is:

$\int_{0}^{x} \sqrt{1 - (y'(t))^2} \, dt = \int_{0}^{x} y(t) \, dt.$

Differentiating both sides with respect to $x$:

$\sqrt{1 - (y'(x))^2} = y(x).$

Squaring both sides:

$1 - (y'(x))^2 = y(x)^2.$

Rearranging terms:

$(y'(x))^2 + y(x)^2 = 1.$

Differentiating this equation with respect to $x$:

$2y'(x)y''(x) + 2y(x)y'(x) = 0.$

Simplify:

$y'(x)(y''(x) + y(x)) = 0.$

Since $y'(x) \neq 0$ in general, it must be that:

$y''(x) + y(x) = 0.$

Now consider the term $y'' + y + 1$:

$y''(x) + y(x) + 1 = 0 + 1 = 1.$

Thus, at $x = 2$:

$1.$

Answer: (1) 1