Question

On the open interval \((-c, c)\), where \( c \) is a positive real number, \( y(x) \) is an infinitely differentiable solution of the differential equation\[ \frac{dy}{dx} = y^2 - 1 + \cos x, \] with the initial condition \( y(0) = 0 \). Then which one of the following is correct?

• A. \( y(x) \) has a local maximum at the origin.
• B. \( y(x) \) has a local minimum at the origin.
• C. \( y(x) \) is strictly increasing on the open interval \((- \delta, \delta)\) for some positive real number \(\delta\).
• D. \( y(x) \) is strictly decreasing on the open interval \((- \delta, \delta)\) for some positive real number \(\delta\).

Answer 1

The Correct Option is D

To solve this problem, we need to analyze the behavior of the differential equation given:

\[\frac{dy}{dx} = y^2 - 1 + \cos x,\]

with the initial condition \(y(0) = 0\). Our goal is to determine the nature of \(y(x)\) around the origin.

1. **Evaluate at the origin**: Begin by evaluating the expression at the initial point \(x = 0\) with the initial condition \(y(0) = 0\):

\[\frac{dy}{dx}\bigg|_{x=0} = (0)^2 - 1 + \cos(0) = -1 + 1 = 0.\]

This indicates that the derivative is zero at \(x = 0\). However, this does not yet tell us whether the point is a minimum, maximum, or even a saddle point. Further analysis is needed.

2. **Second derivative test**: Evaluate the second derivative at the origin to determine the concavity of the function:

From the equation \(d\left(\frac{dy}{dx}\right)/dx = \frac{d}{dx}(y^2 - 1 + \cos x)\), we find:

\[\frac{d^2y}{dx^2} = 2y\frac{dy}{dx} - \sin x.\]

Substitute \(x = 0\) and \(y = 0\):

\[\frac{d^2y}{dx^2}\bigg|_{x=0} = 2 \cdot 0 \cdot 0 - \sin(0) = 0.\]

This result is inconclusive for determining whether \(y(x)\) has a local min or max at this point, so we investigate further.

3. **Behavior around the origin**: Check if \(y(x)\) is increasing or decreasing around \(x = 0\). Since the first derivative is zero, use the second derivative at points near the origin:

For small \(x\) around zero, since \(\cos x \approx 1\) when \(x\) is small, the equation approximates:

\[\frac{dy}{dx} \approx y^2 - 1 + 1 = y^2.\]

Since the term \(y^2\) is non-negative, and given that \(y(0) = 0\), as \(x\) moves away from zero, we need further detail on the behavior, but largely it implies if \(y(x)\) remains under zero (which is confirmed by examining very close behavior of its second change considering the non increase activity), it will tend towards drop since increase is conditional. Hence, it intuitively leads to conclude a likely lowering/strict decreasing path immediately around this given start:

Therefore, provided \(y(x)\) cannot suddenly jump positive initially and exhibit reduction tendencies, the strict decreasing nature on a small interval surrounding zero is logically convincing.

From the equations evaluated and given choices, therefore, the correct choice is:

\( y(x) \) is strictly decreasing on the open interval \((- \delta, \delta)\) for some positive real number \(\delta\).