Question

Let (-c,c) be the largest open interval in \(\R\) (where c is either a positive real number or c = ∞) on which the solution y(x) of the differential equation \(\frac{dy}{dx}=x^2+y^2+1\) with initial condition y(0) = 0 exists and is unique. Then which of the following is/are true?

• A. y(x) is an odd function on (-c, c).
• B. y(x) is an even function on (-c, c).
• C. (y(x))² has a local minimum at 0
• D. (y(x))² has a local maximum at 0

Answer 1

The Correct Option is A, C

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

\(\frac{dy}{dx} = x^2 + y^2 + 1\) 

with the initial condition \(y(0) = 0\). Let's step through the analysis:

1. Existence and Uniqueness: The function \(f(x, y) = x^2 + y^2 + 1\) is continuous and has continuous first partial derivatives with respect to \(y\) in the entire plane \(\mathbb{R} \times \mathbb{R}\). By the Picard-Lindelöf theorem (existence and uniqueness theorem for ordinary differential equations), there exists a unique solution to the initial value problem around \(x = 0\). As there are no restrictions on \(x\) and \(y\) that would violate the conditions of the theorem for \((x, y)\) given the initial condition, the solution exists and is unique for all \(x \in (-c, c)\) with \(c = \infty\).
2. Checking if \(y(x)\) is an odd or even function: A function \(y(x)\) is odd if \(y(-x) = -y(x)\) holds for all \(x\) in the domain. Substituting into the differential equation, if \(y(x)\) satisfies \(\frac{dy}{dx} = x^2 + y^2 + 1\), replace \(x\) by \(-x\) and \(y(-x)\) by \(-y(x)\), we get:

\(\frac{d(-y)}{d(-x)} = (-x)^2 + (-y)^2 + 1 = x^2 + y^2 + 1\)

1. Since the form remains the same, if \(y(x)\) is a solution, \(-y(x)\) is also a correct form inside the domain, thus verifying it is indeed an odd function.
2. Analyzing \((y(x))^2\) at \(x = 0\):
   • Since \(y(x)\) is an odd function, \((y(x))^2\) is even, meaning \((y(x))^2\) at \(x = 0\) is a local extremum.
   • Evaluating the derivative \(\frac{d}{dx}((y(x))^2)\) at \(x = 0\), it simplifies to \(0\), so we check the second derivative \(\frac{d^2}{dx^2}((y(x))^2)\) to determine if it's a minimum or maximum.
   • From the differential equation, \(y'(x)\) and hence \(\frac{d}{dx}((y(x))^2)\) involve squares that increase away from \((y(0))^2 = 0\), indicating \((y(x))^2 > 0\) when \(x \neq 0\).
   • Therefore, \((y(x))^2\) has a local minimum at \(x = 0\).

Conclusion: Based on the above analysis:

• \(y(x)\) is an odd function on \((-c, c)\).
• \( (y(x))^2 \) has a local minimum at 0.