Question

Consider the initial value problem $$ y^{\prime}=\sqrt{1-y^{2}}, \quad y(0)=0 . $$ (a) Show that \(y=\sin t\) is an explicit solution on the \(t\)-interval \(-\pi / 2 \leq t \leq \pi / 2\). (b) Show that \(y=\sin t\) is not a solution on either of the intervals \(-3 \pi / 2t^{*}\) ? (d) Show that the solution of the initial value problem is given by $$ y=\left\\{\begin{array}{lrl} -1, & -\infty &

Short answer

No, y=sin(t) is not a solution on the interval -3π/2 < t < -π/2, because the derivative y'=cos(t) is negative in this interval while the equation y'=√(1-y²) gives a non-negative value.

Step-by-step solution

(a) Verify the explicit solution on interval -π/2 ≤ t ≤ π/2.

To show that y=sin(t) is an explicit solution of the differential equation y'=√(1-y²), we need to differentiate sin(t) with respect to t and see whether the given equation is satisfied in the interval -π/2 ≤ t ≤ π/2. Let's do that: $$ y' = \frac{d}{dt} \sin{t} = \cos{t} $$ Now we get y'=cos(t). Next, we need to substitute y=sin(t) into the given equation and see if both sides are equal: $$ y' = \sqrt{1-y^2} \Rightarrow \cos{t} = \sqrt{1-\sin^2{t}} $$ Since we already know that cos²(t) + sin²(t) = 1, we can see that both cos(t) and √(1-sin²(t)) are the same, and therefore the given solution is an explicit solution on the interval -π/2 ≤ t ≤ π/2.

(b) Verify the explicit solution on the intervals -3π/2 < t < -π/2 and π/2 < t < 3π/2.

Since we already established that y=sin(t) and y'=cos(t) in part (a), we just need to check whether the given equation y'=√(1-y²) is satisfied on the intervals -3π/2 < t < -π/2 and π/2 < t < 3π/2. However, we know that in these two intervals, cos(t) is negative. Meanwhile, the right side of the given equation, √(1-y²), is non-negative. Therefore, y'=cos(t) cannot be equal to √(1-y²) on these intervals. Thus, y=sin(t) is not a solution on the intervals -3π/2 < t < -π/2 and π/2 < t < 3π/2.

(c) Find the equilibrium solutions and analyze the graph of the function at equilibrium points.

We need to find values of y when the function y'=√(1-y²) becomes zero. The function will be zero when the quantity under the square root (1-y²) is zero. Therefore, $$ 1 - y^2 = 0 \Rightarrow y^2 = 1 \Rightarrow y = \pm 1 $$ Thus, the equilibrium solutions are y=1 and y=-1. Suppose, the solution y(t) reaches an equilibrium value at t=t*. Then, for t>t*, the graph of y(t) will be a horizontal line at y=1 or y=-1, because the derivative (i.e., the rate of change) will be zero at the equilibrium points.

(d) Show the given piecewise function as the solution to the initial value problem.

We are given the following piecewise function as the solution to the initial value problem: $$ y=\left\{\begin{array}{lrl} -1, & -\infty & <t<-\pi / 2 \\ \sin t, & -\pi / 2 & \leq t \leq \pi / 2 \\ 1, & \pi / 2 & <t<\infty . \end{array}\right. $$ We have already established that y=sin(t) is a solution for the interval -π/2 ≤ t ≤ π/2 in part (a). We know that the function has equilibrium solutions at y=1 and y=-1 and that the graph is a horizontal line at these values when the derivative is zero. In part (b), we have already shown that y=sin(t) is not a solution in the other intervals, i.e., -∞ < t<-π/2 and π/2 < t < ∞. Since y=-1 and y=1 are the equilibrium solutions and lie in those intervals, we now have a complete piecewise solution for the initial value problem: $$ y=\left\{\begin{array}{lrl} -1, & -\infty & <t<-\pi / 2 \\ \sin t, & -\pi / 2 & \leq t \leq \pi / 2 \\ 1, & \pi / 2 & <t<\infty . \end{array}\right. $$

Key concepts

Equilibrium Solutions

An equilibrium solution in the context of differential equations refers to a constant solution where the rate of change is zero. For the differential equation \( y' = \sqrt{1-y^2} \), equilibrium occurs when the derivative \( y' \) is equal to zero. This means:

• \( \sqrt{1-y^2} = 0 \)
• \( 1-y^2 = 0 \)
• \( y^2 = 1 \)
• \( y = \pm 1 \)

Thus, \( y = 1 \) and \( y = -1 \) are equilibrium solutions. When a function reaches these values, it stays there because further changes in \( y \) will result in no change in the derivative, indicating the graph of the function flattens out. At these equilibrium points, the function becomes a horizontal line, showing that these are stable states where the system does not change.

Piecewise Function

A piecewise function is a function that is defined by different expressions depending on the value of the independent variable. In our initial value problem, the solution to the differential equation is given by:\[y=\left\{\begin{array}{lll}-1, & -\infty < t < -\pi/2 \\sin t, & -\pi/2 \leq t \leq \pi/2 \1, & \pi/2 < t < \infty \\end{array}\right.\]This piecewise function combines three parts:

• For \( t \) values less than \(-\pi/2\), \( y \) remains at the equilibrium solution \(-1\).
• For \( t \) within \([-\pi/2, \pi/2]\), \( y \) follows a \( \sin t \) pattern, representing the dynamic part of the function.
• For \( t \) greater than \( \pi/2 \), \( y \) is at the other equilibrium solution \(1\).

Piecewise functions are useful for modeling situations where a behavior changes at specific points or intervals, capturing transitions in simple terms.

Explicit Solution

An explicit solution to a differential equation expresses the dependent variable directly in terms of the independent variable. In the initial value problem \( y' = \sqrt{1-y^2} \), the explicit solution \( y = \sin t \) is verified over the interval \( -\pi/2 \leq t \leq \pi/2 \). This means you can clearly define \( y \) using cosine, as shown:- Differentiate \( y = \sin t \) to get \( y' = \cos t \).- Substitute \( y = \sin t \) into the equation, verifying that \( \cos t = \sqrt{1-\sin^2 t} \).This identity is proven using the Pythagorean identity \( \cos^2 t + \sin^2 t = 1 \). As a result, the solution satisfies the differential equation explicitly, showing a direct expression for \( y \) over the given interval, while explaining why it doesn't work beyond it, due to negative cosine values that do not satisfy the original equation.