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

The Correct Option is A

To solve the given problem, we need to understand that the problem involves an integral equation and a differential equation. We will proceed by analyzing the conditions and deriving the necessary expressions.

The given equation is:

\[\int_{0}^{x} \sqrt{1 - (y'(t))^2} \, dt = \int_{0}^{x} y(t) \, dt, \quad 0 \leq x \leq 3, \, y \geq 0, \, y(0) = 0.\]

We differentiate both sides with respect to \( x \) to get:

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

Squaring both sides, we get:

\[1 - (y'(x))^2 = y(x)^2.\]

This can be rearranged to:

\[(y'(x))^2 = 1 - y(x)^2.\]

Taking the second derivative with respect to \( x \), we differentiate both sides:

\[2y'(x)y''(x) = -2y(x)y'(x).\]

Assuming \( y'(x) \neq 0 \), we can divide by \( 2y'(x) \):

\[y''(x) = -y(x).\]

Substituting into the expression \( y'' + y + 1 \) gives us:

\[y''(2) + y(2) + 1 = -y(2) + y(2) + 1 = 1.\]

Thus, at \( x = 2 \), the value of \( y'' + y + 1 \) is 1.

Correct choice: 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