Differential Equation Problem Analysis

We are given the ordinary differential equation (ODE) $y' = (1 - y^2)^{10} \cos y$ with the initial condition $y(0) = 0$. We need to determine the maximal interval of existence $J$ and the range $K$ of the solution $y(t)$.

Understanding the ODE

The given ODE is an autonomous first-order differential equation of the form $y' = f(y)$, where $f(y) = (1 - y^2)^{10} \cos y$.

For the existence and uniqueness of a solution in a neighborhood of the initial point $(t_0, y_0) = (0, 0)$, the function $f(y)$ and its partial derivative with respect to $y$, $\frac{\partial f}{\partial y}$, must be continuous in a region containing $y_0$.

• The function $f(y) = (1 - y^2)^{10} \cos y$ is a product of a polynomial and a trigonometric function, both of which are continuous for all real values of $y$. Thus, $f(y)$ is continuous for all $y \in \mathbb{R}$.
• The partial derivative is $\frac{\partial f}{\partial y} = \frac{d}{dy} [(1 - y^2)^{10} \cos y] = 10(1 - y^2)^9 (-2y) \cos y + (1 - y^2)^{10} (-\sin y) = -20y (1 - y^2)^9 \cos y - (1 - y^2)^{10} \sin y$. This expression is also continuous for all $y \in \mathbb{R}$.

Since $f(y)$ and $\frac{\partial f}{\partial y}$ are continuous for all $y \in \mathbb{R}$, by the Existence and Uniqueness Theorem, a unique solution exists for some interval around $t=0$.

Equilibrium Points and Range of Solution

Equilibrium points are constant solutions where $y' = 0$. This occurs when $f(y) = 0$.

$(1 - y^2)^{10} \cos y = 0$

This equation is satisfied if:

• $(1 - y^2)^{10} = 0 \implies 1 - y^2 = 0 \implies y^2 = 1 \implies y = 1$ or $y = -1$.
• $\cos y = 0 \implies y = \frac{\pi}{2} + n\pi$ for any integer $n$.

The constant solutions relevant to the initial condition $y(0)=0$ are $y(t) = 1$ and $y(t) = -1$. The initial condition is $y(0) = 0$. Since $-1 < 0 < 1$, and by the uniqueness theorem, the solution curve starting at $y(0)=0$ can never cross the constant solutions $y=1$ and $y=-1$. This means the solution $y(t)$ must always satisfy $-1 < y(t) < 1$ for all $t$ in its interval of existence.

Thus, the range $K$ of the solution must be a subset of the interval $(-1, 1)$.

Let's examine the sign of $y' = (1 - y^2)^{10} \cos y$ for $-1 < y < 1$. In this interval, $(1 - y^2)^{10} > 0$. The sign of $y'$ depends on $\cos y$. Since $-1 < 0 < 1$, and approximately $-1 \approx -\pi/3$ and $1 \approx \pi/3$, the interval $(-1, 1)$ is contained within $(-\pi/2, \pi/2)$. For $y \in (-1, 1)$, $\cos y > 0$. Therefore, for $y \in (-1, 1)$, $y' > 0$.

Starting at $y(0)=0$, $y'(0) = (1-0)^{10} \cos(0) = 1 > 0$. The solution starts increasing. Since $y(t)$ is increasing and bounded above by 1, it must approach a limit as $t \to \infty$. By properties of autonomous ODEs, this limit must be an equilibrium point. Since $y(t)$ starts at 0 and increases towards 1 without crossing it, we have $\lim_{t \to \infty} y(t) = 1$. However, it cannot reach 1 in finite time. Similarly, as $t \to -\infty$, since $y' > 0$ for $y \in (-1, 1)$, tracing back in time means $y(t)$ must decrease. Bounded below by -1, it must approach -1 as $t \to -\infty$. Thus, $\lim_{t \to -\infty} y(t) = -1$.

The range of the solution $y(t)$ is the set of all values $y(t)$ takes. As $t$ goes from $-\infty$ to $\infty$, $y(t)$ goes from approaching -1 to approaching 1, without ever reaching these values. Therefore, the range is $K = (-1, 1)$.

Maximal Interval of Existence

For an autonomous ODE $y' = f(y)$, if $f(y)$ and $\frac{\partial f}{\partial y}$ are continuous on an interval $(a, b)$, and the initial condition $y(t_0) = y_0$ is such that $y_0 \in (a, b)$, then the maximal interval of existence $J$ is $\mathbb{R}$ if the solution $y(t)$ remains within a closed bounded subinterval $[c, d] \subset (a, b)$ for all $t \in J$.

In this problem, $f(y)$ and $\frac{\partial f}{\partial y}$ are continuous on $\mathbb{R}$. We can choose $(a, b) = \mathbb{R}$. The initial condition is $y(0) = 0$. We found that the range of the solution is $K = (-1, 1)$. This range is contained within any closed bounded interval like $[-0.9, 0.9]$. This interval $[-0.9, 0.9]$ is a subset of $\mathbb{R}$.

Since the solution $y(t)$ is bounded (its range is $(-1, 1)$) and stays within a closed bounded interval (e.g., $[-0.9, 0.9]$) where $f$ and $\frac{\partial f}{\partial y}$ are continuous, the maximal interval of existence $J$ is $\mathbb{R}$.

Summary

Based on the analysis:

• The maximal interval of existence is $J = \mathbb{R}$.
• The range of the solution is $K = (-1, 1)$.

Let's review the given options:

• Option 1: $J = \mathbb{R}$ - This statement is true.
• Option 2: $K = (-1, 1)$ - This statement is true.
• Option 3: $J = (-1, 1)$ - This statement is false. The interval of existence is a time interval.
• Option 4: $K = [-1, 1]$ - This statement is false. The solution cannot reach the values -1 or 1.

The statements that are true are Option 1 and Option 2. The problem asks which statements are true, and the provided correct answer text indicates that both $J = \mathbb{R}$ and $K = (-1, 1)$ are correct.