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. $$ e^{y} y^{\prime}+\frac{t}{y+1}=\frac{2}{y+1}, \quad y(1)=2 $$

Short answer

Based on the given information and the steps followed, provide a short answer related to the initial value problem: The implicit solution to the given initial value problem involving the differential equation $e^{y}y'+\frac{t}{y+1}=\frac{2}{y+1}$ with initial condition $y(1)=2$, is: $$ (\ln{|y+1|})e^{-t}=-\frac{t^2}{2}+2t+\frac{3}{2}-(\ln{3})e^{-1}. $$ Due to the complexity of the equation, an explicit solution for $y(t)$ and the interval of existence couldn't be found in this case. For some problems like this one, further methods may be required to determine those values, such as the Picard-Lindelöf theorem or others related to existence and uniqueness.

Step-by-step solution

Separate the Variables

Rearrange the equation to make it suitable for integration: $$ e^{y} y^{\prime} = \frac{2-t}{y+1} $$ Next, separate the variables: $$ e^{-y} \frac{dy}{dt} = \frac{2-t}{y+1} $$ Divide both sides by y+1: $$ \frac{e^{-y}}{y+1} dy = (2-t) dt $$ Now we are ready to integrate each side.

Integrate Both Sides

Integrate both sides of the equation with respect to their respective variables: $$ \int \frac{e^{-y} }{y+1} dy = \int (2-t) dt $$ For the left side of the equation, perform integration by substitution, let u = y + 1: $$ \int \frac{e^{-u}}{u} du = \int (2-t) dt $$ The left side of the equation now represents the integration of the function e^{-u} / u, which is Ln|u|(-e^{t}) + C : $$ \int \frac{e^{-u}}{u} du = (\ln{|u|})e^{-t}+ C_1 $$ Substitute back u = y + 1 : $$ (\ln{|y+1|})e^{-t} = -\frac{t^2}{2}+2t + C_2 $$ This is the implicit solution to the differential equation.

Apply Initial Condition

Apply the initial condition y(1) = 2 to solve for the constant C_2: $$ (\ln{|2+1|})e^{-1} = -\frac{1^2}{2}+2(1) + C_2 $$ Calculating values and solving for C_2: $$ (\ln{3})e^{-1} = -\frac{1}{2}+2 + C_2 \Rightarrow C_2 = \frac{3}{2} - (\ln{3})e^{-1} $$ Substitute the constant C_2 in the implicit solution: $$ (\ln{|y+1|})e^{-t} = -\frac{t^2}{2}+2t + \frac{3}{2} - (\ln{3})e^{-1} $$

Solve for y(t)

In this particular case, it's difficult to find an explicit solution for y(t), so we can't provide one. However, we have the implicit solution: $$ (\ln{|y+1|})e^{-t} = -\frac{t^2}{2}+2t + \frac{3}{2} - (\ln{3})e^{-1} $$

Determine the Interval of Existence

Since an explicit solution couldn't be found for y(t), we are unable to find the interval of existence. In some cases, we need other methods to determine the interval of existence, like, for example, the Picard-Lindelöf theorem (or existence and uniqueness theorem) for ordinary differential equations.

Key concepts

Implicit Solution

When dealing with differential equations, especially with initial value problems, we often first encounter what is called an implicit solution. This solution is termed 'implicit' because the dependent variable, in this case, \(y\), cannot be isolated on one side of the equation. Thus, it remains "embedded" within the equation itself.

For example, in our problem, after separating variables and performing integration, we derive the implicit solution: \[(\ln{|y+1|})e^{-t} = -\frac{t^2}{2}+2t + \frac{3}{2} - (\ln{3})e^{-1}\]This equation relates \(y\) to the independent variable \(t\) in a non-direct way.

Implicit solutions are often the first step in solving more complex differential equations, as they can be manipulated or simplified to obtain further insights. Conversely, these solutions might not always lead to direct answers and thus, an implicit solution may be the endpoint for particularly complex problems.

Explicit Solution

An explicit solution provides a direct relationship between the dependent and independent variables. In other words, \(y\) is expressed entirely as a function of \(t\). Explicit solutions are usually more practical as they allow us to easily calculate the dependent variable for any given value of the independent variable.

However, in some problems, like the one presented, finding such a solution is not always possible due to complexities involved in isolating \(y\).

Though in many cases, converting an implicit solution to an explicit one involves algebraic manipulation or numerical methods, if the implicit form is not easily simplified to express \(y\) solely in terms of \(t\), it's not feasible to find an explicit solution. In this exercise, the complexity of isolating \(y\) means that the implicit form stands as the best representation of the solution.

Interval of Existence

The interval of existence refers to the range of the independent variable, usually \(t\), over which the solution to a differential equation is valid. This concept ensures that the solution remains continuous and does not encounter conditions that would lead to undefined behavior, such as division by zero or logarithm of a negative number.

In our exercise, since an explicit solution for \(y(t)\) could not be specified, determining the interval of existence becomes more challenging.

Ordinarily, the interval of existence is identified after obtaining an explicit solution, assessed by considering points where the solution might not be defined. Furthermore, the interval could be inferred using theorems like the Picard-Lindelöf theorem, which helps establish conditions for the existence and uniqueness of solutions.

Integration by Substitution

Integration by substitution is a technique to simplify the process of integration by making a clever substitution that transforms the integral into a more manageable form. This method proves especially useful when direct integration is complicated by the algebraic form of the function.

• Choose a substitution that simplifies a component of the integral.
• Replace the chosen part with a new variable which simplifies the process of integration.
• Perform the integral using the new variable.
• Substitute back the original variable to obtain the final result.

In the problem given, the substitution \(u = y + 1\) was utilized, transforming the integral in a pleasant form for solving, which initially was \(\int \frac{e^{-y}}{y+1} dy\). With substitution, it became \(\int \frac{e^{-u}}{u} du\).

The new integral facilitated easier calculation, ultimately contributing to solving for the implicit form of the solution provided.