Question

(a) Obtain an implicit solution and, if possible, an explicit solution of the initial value problem. (b) If you can find an explicit solution of the problem, determine the \(t\)-interval of existence. $$ t^{2} y^{\prime}+\sec y=0, \quad y(-1)=0 $$

Short answer

Based on the solution process, provide a short answer to the problem: We obtained an implicit solution to the given IVP as \(\sin y = \frac{1}{t} + 1\). An explicit solution could not be found due to the domain constraints of the arcsine function. Consequently, the t-interval of existence could not be determined.

Step-by-step solution

Rewrite the given equation

The given equation is a first-order differential equation as follows: \(t^2 y' + \sec y = 0\). To make the equation more suitable for solving, first rewrite the equation as \(y' = -\frac{\sec y}{t^2}\).

Separate variables and integrate

Now, express the equation in terms of \(dy\) and \(dt\): $$\frac{dy}{\sec y} = -\frac{dt}{t^2}.$$ To separate the variables, multiply both sides by the corresponding denominator: $$\cos y dy = -t^{-2} dt.$$ Integrate both sides with respect to \(y\) and \(t\) respectively: $$\int \cos y dy = -\int t^{-2} dt.$$ The integral of \(\cos y\) with respect to \(y\) is \(\sin y \) and the integral of \(t^{-2}\) with respect to \(t\) is \(\frac{t^{-1}}{-1}\). So, we have: $$\sin y = \frac{1}{t}+ C.$$

Apply the initial condition

Next, apply the initial condition \(y(-1) = 0\) to the implicit solution to obtain the value of the constant \(C\). Using the initial condition, we get: $$\sin (0) = \frac{1}{-1} + C.$$ Since \(\sin(0) = 0\), the above equation becomes: $$ 0 = -1 + C.$$ Hence, \(C = 1\). Thus, the implicit solution is: $$\sin y = \frac{1}{t} + 1.$$

Attempt to find an explicit solution

An explicit solution attempts to isolate \(y\) in the equation. However, in this case, we would need to find the inverse sine of the equation: $$y = \arcsin\left(\frac{1}{t} + 1\right).$$ The domain of the arcsine function is \([-1,1]\), which means that the expression inside the parentheses, \(\frac{1}{t} + 1\), must be within that range. The problem with this is that there is no guarantee that the expression stays within this range for all \(t\), so we cannot proceed to find an explicit solution.

Determine the t-interval of existence (if possible)

Since we could not find an explicit solution, we cannot determine the t-interval of existence for this problem. To summarize the solution: 1. We rewrote the given equation into a more suitable form: \(y' = -\frac{\sec y}{t^2}\). 2. We separated variables and integrated both sides. 3. We applied the initial condition \(y(-1) = 0\) to find the constant \(C\). 4. We attempted to find an explicit solution, but failed due to the domain of the arcsine function. 5. We could not determine the t-interval of existence since an explicit solution could not be found.

Key concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. They are vital in various scientific disciplines for modeling how quantities change over time or space. The order of a differential equation is determined by the highest derivative it contains. For instance, a first-order differential equation includes only the first derivative of the function (denoted by \(y'\) or \(\frac{dy}{dt}\)), such as the example \(t^2 y' + \sec y = 0\).
In the given exercise, we're dealing with a first-order differential equation. Solving such equations involves finding the function \(y(t)\) that satisfies the equation for a given initial condition. The initial condition, like \(y(-1)=0\), specifies the value of the function at a particular point, which is necessary for finding a unique solution.

Separation of Variables

The technique of separation of variables is used to solve certain types of differential equations. This method involves rearranging the equation so that each variable and its differential are on separate sides of the equation, allowing us to integrate them independently. For example:
$$\frac{dy}{\sec y} = -\frac{dt}{t^2}.$$
By separating the \(dy\) and \(dt\) terms to different sides, and each associated with only one variable \(y\) or \(t\), we can utilize integral calculus to find functions that satisfy each discrete part of the equation, such as:
$$\int \cos y \, dy = -\int t^{-2} \, dt.$$
Successfully separating variables and integrating is a crucial step in finding the implicit solution to a differential equation.

Implicit and Explicit Solutions

Implicit solutions of differential equations are equations where the dependent variable (often \(y\)) and the independent variable (commonly \(t\)) are not separated; the solution gives a relationship between \(y\) and \(t\) without explicitly solving for \(y\) alone. An explicit solution, on the other hand, is when the dependent variable is written explicitly as a function of the independent variable, such as \(y = f(t)\).
The implicit solution of the initial value problem given in the example can be shown as \(\sin y = \frac{1}{t} + 1\). We attempt to express \(y\) explicitly as \(y = \arcsin\left(\frac{1}{t} + 1\right)\), but this requires the \(\arcsin\) function to be defined for the resulting values, which isn't always the case. Hence, sometimes we may not be able to find an explicit solution.

Interval of Existence

The interval of existence refers to the set of values of the independent variable (usually \(t\)) for which a solution to a differential equation is defined. For explicit solutions, this is tied to the domain of the resulting function. Due to the domain of the \(\arcsin\) function being \( [-1,1]\), a strict condition is imposed on the explicit form of the solution—\(\frac{1}{t} + 1\) must be within this range.
As we could not find an explicit solution for the given exercise, the interval of existence could not be determined. However, the concept is important for understanding the limitations of solutions, as they often only apply for certain ranges of the independent variable. For instance, if an explicit solution had been attainable, then saying it is valid for \( t > a \) and \( t < b \) where \(a\) and \(b\) are specific numbers, would describe the interval of existence for that solution.