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. $$ e^{t} y^{\prime}+(\cos y)^{2}=0, \quad y(0)=\pi / 4 $$

Short answer

Question: Solve the given initial value problem (IVP) and determine the interval of existence of the solution: $$ e^{t} y^{\prime}+(\cos y)^{2}=0, \quad y(0)=\pi / 4 $$ Answer: The implicit solution to the given initial value problem is: $$ y(t) = \arctan(-\int e^{-t} dt + 1) $$ The solution exists for all $t \in \mathbb{R}$ as there are no restrictions on the domain or singularities in the initial value problem.

Step-by-step solution

Separate variables

First, we need to separate variables. Rearrange the given equation to get: $$ y^{\prime} = - (\cos y)^{2} e^{-t} $$ Now, we'll separate the variables: $$ \frac{dy}{dt} = - (\cos y)^{2} e^{-t} $$ $$ \frac{dy}{(\cos y)^{2}} = -e^{-t} dt $$

Integrate both sides

Now, we'll integrate both sides with respect to the respective variables: $$ \int \frac{dy}{(\cos y)^{2}} = \int -e^{-t} dt $$ On the left side, we have a well-known integral, which is the integral of the secant function squared: $$ \tan y = -\int e^{-t} dt + C_1 $$

Solving for C_1 using the initial condition

Now, we need to use the initial condition \(y(0) = \pi/4\) to solve for the constant \(C_1\) $$ \tan (\pi / 4) = -\int_0^0 e^{-t} dt + C_1 $$ Since \(\tan (\pi / 4) = 1\), we have: $$ 1= C_1 $$ So we have the implicit solution: $$ \tan y = -\int e^{-t} dt + 1 $$

Finding the explicit solution

To find the explicit solution, we need to solve for \(y\): $$ y(t) = \arctan(-\int e^{-t} dt + 1) $$ However, the integral of \(-e^{-t}\) doesn't have a simple closed-form expression, so we can't simplify this further. We're left with the implicit solution: $$ y(t) = \arctan(-\int e^{-t} dt + 1) $$

Interval of existence

Since we can't find a simpler explicit solution, we can't determine the interval of existence in this case. However, we can say that the solution exists for all \(t \in \mathbb{R}\) as there are no restrictions on the domain or singularities in the initial value problem.

Key concepts

Implicit Solution

An implicit solution arises when the relationship between the dependent and independent variables in a differential equation isn't explicitly expressed as a function. Here, you don't solve directly for one variable in terms of another. Instead, the solution mantras the original variables in a related form.

• It can often be easier to obtain than explicit solutions.
• In some cases, it may not be possible to express the solution explicitly using standard functions.
• Implicit solutions retain all initial information from the equation, making manipulations straightforward.

For the given initial value problem, we derived:\[\tan y = -\int e^{-t} dt + 1\]This expression leaves \(y\) defined implicitly. While it's not neatly \(y = ...\), an implicit form can sometimes still offer valuable insights.
Examining implicit solutions thoroughly can reveal criteria and insights that might be cumbersome to observe otherwise.

Explicit Solution

An explicit solution is a more straightforward solution type, offering an exact expression for the dependent variable. This means you express the dependent variable explicitly as a function of the independent one, \(y = f(t)\).

• Explicit solutions help provide clear and direct results.
• They allow for easier evaluation of the function at given points.
• Sometimes, calculating an explicit solution may not be feasible or can lead to complexities.

In the initial value problem at hand, an explicit solution was sought:\[y(t) = \arctan(-\int e^{-t} dt + 1)\]However, the integral component here doesn't neatly simplify into an elementary function. Hence, a fully simplified explicit form isn't attainable. Yet, understanding this limitation is vital, as it highlights the nature and limitations of differentiable calculus and integral expressions.

Interval of Existence

The interval of existence refers to the range of independent variable values over which the solution to a differential equation remains valid. Determining this interval ensures understanding the practical applicability and constraints within which the solution operates.

• The interval depends on the differential equation and initial conditions.
• For some equations, singularities or peculiarities in the functions involved may limit existence to certain intervals.
• In other cases, solutions can extend indefinitely across the number line.

In our case, the solution exists for all \(t \in \mathbb{R}\). This broad interval of existence is due to the lack of restrictions in the original equation and the nature of exponential and trigonometric functions which generally pose no inherent boundary constraints. Understanding the interval of existence prevents misapplication and ensures solutions are used within their valid range.