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. $$ \frac{d y}{d t}=1+y^{2}, \quad y(\pi / 4)=-1 $$

Short answer

Differential Equation: \(\frac{dy}{dt} = 1+y^2\) Initial Condition: \(y(\frac{\pi}{4}) = -1\) Answer: The explicit solution for the given IVP is \(y(t) = \tan(t - \frac{\pi}{2})\). The interval of existence for this solution is \(-\frac{\pi}{2} < t < \frac{\pi}{2}\).

Step-by-step solution

Integrate the Differential Equation

First, let's rewrite the given differential equation as a separable equation, and then integrate both sides: $$ \frac{dy}{dt} = 1 + y^2 \implies \frac{1}{1+y^2} dy = dt $$ Now, integrate both sides with respect to their variables: $$ \int \frac{1}{1+y^2} dy = \int dt $$ The integral results in: $$ \arctan(y) = t + C $$ where \(C\) is the constant of integration.

Apply the Initial Condition

Now we have to apply the initial condition \(y(\pi/4) = -1\) to find the constant \(C\). Plug the initial condition into the equation: $$ \arctan(-1) = \pi/4 + C $$ Since \(\arctan(-1) = -\pi/4\), we get: $$ -\pi/4 = \pi/4 + C \implies C = -\pi/2 $$ Now, our implicit equation becomes: $$ \arctan(y) = t - \frac{\pi}{2} $$

Find the Explicit Solution (If Possible)

To find the explicit solution if possible, we need to isolate the \(y\) on one side: $$ y = \tan(t - \frac{\pi}{2}) $$ Thus, we have found an explicit solution: $$ y(t) = \tan(t - \frac{\pi}{2}) $$

Determine the t-interval of Existence

Since our explicit solution involves a tangent function, there is a vertical asymptote where the function is undefined. The tangent function has vertical asymptotes at \((2n+1) \frac{\pi}{2}\), where n is an integer. Using the initial condition, we have \(t = \frac{\pi}{4}\), which is between the vertical asymptotes for \(n = 0\) and \(n = -1\), which are at \(t = \frac{\pi}{2}\) and \(t = -\frac{\pi}{2}\). Therefore, the t-interval of existence for the explicit solution is: $$ -\frac{\pi}{2} < t < \frac{\pi}{2} $$

Key concepts

Implicit Solution

When solving differential equations, an implicit solution is one where the function we're looking for, typically denoted as y(t), is not by itself on one side of the equation. Rather, y(t) is intertwined with the variable t in a way that cannot be simplified to express y explicitly as a function of t. In our example, after integrating both sides of the separable equation and applying the initial condition, we arrive at the equation \( \arctan(y) = t - \frac{\pi}{2} \). This is our implicit solution because y is not isolated; it's contained within the arctangent function coupled with t. Solving some differential equations may only yield implicit solutions, especially when it's difficult, or sometimes impossible, to express y in terms of t directly.

Explicit Solution

An explicit solution to a differential equation is one where y(t), the function we’re searching for, stands alone on one side of the equation and is expressed as a function of t. This is seen as the 'solved for' or 'isolated' form of y(t). In our textbook problem, to find an explicit solution, the arctan function is manipulated to isolate y, yielding \( y(t) = \tan(t - \frac{\pi}{2})\). Not all implicit solutions can be converted into an explicit form due to the complexity of the relationship between y and t. However, when possible, the explicit form is often more practical for understanding the behavior of the function and for calculating specific values.

Interval of Existence

The interval of existence refers to the range of t-values for which the solution to a differential equation is defined and behaves properly without encountering any discontinuities or singularities. For the explicit solution \( y(t) = \tan(t - \frac{\pi}{2}) \) provided by the exercise, we must consider the properties of the tangent function. Since the tangent function has vertical asymptotes at \( (2n+1) \frac{\pi}{2} \) where n is any integer, it is undefined at these points. By looking at the initial condition \( t = \frac{\pi}{4} \) and considering where the nearest asymptotes occur, we determine the interval of existence for our solution to be \( -\frac{\pi}{2} < t < \frac{\pi}{2} \), thus avoiding any discontinuities and ensuring the solution is valid within this interval.

Separable Differential Equations

Separable differential equations are a class of differential equations where the variables can be separated on opposite sides of the equation. Specifically, they can be written in the form \( g(y)dy = f(t)dt \), which allows us to integrate both sides with respect to their respective variables. This technique is particularly powerful because it reduces the problem of solving the differential equation to the problem of evaluating two integrals. In our textbook example, we rewrote the initial differential equation as \( \frac{1}{1+y^2} dy = dt \) by separating the y-dependent terms from the t-terms. This is a classic move in the playbook for solving separable differential equations and leads us to find the implicit solution by integrating both sides, paving the way towards finding the explicit solution as well.