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. $$ (\ln y) y^{\prime}+t=1, \quad y(3)=e $$

Short answer

Answer: The implicit solution of the given initial value problem is: $$y(t) = e + \int_{3}^{t} \frac{1-t'}{\ln y(t')} dt'$$. Since we could not find an explicit solution for y(t), no t-interval of existence can be determined at this time.

Step-by-step solution

Identify the type of differential equation

The given equation is \((\ln y) y^{\prime}+t=1\), which is a first-order, inhomogeneous, nonlinear ordinary differential equation (ODE). The prime symbol represents the derivative with respect to t.

Rewrite the equation

We rewrite the given equation: $$(\ln y) y^{\prime}+t=1$$ as $$(\ln y) \frac{dy}{dt}+t=1$$.

Separate variables

To separate variables, we want y and dy on one side, and t and dt on the other side of the equation. Divide both sides by \((\ln y)\) and isolate dy as follows: $$ \frac{dy}{dt}=\frac{1-t}{\ln y}$$. Now, multiply both sides by dt: $$ dy=\frac{1-t}{\ln y} dt $$.

Integrate both sides

Integrate both sides of the equation with respect to their respective variables: $$\int dy = \int \frac{1-t}{\ln y} dt$$. After integration, we get: $$y = \int \frac{1-t}{\ln y} dt$$.

Apply initial condition

We were given that when \(t=3\), \(y=e\). Substitute these values into the integrated equation: $$e = \int_{3}^{t} \frac{1-t'}{\ln e} dt'$$.

Solve for y(t) explicitly if possible

Without further information or restrictions on the function y(t), it is difficult to solve the integral explicitly for y(t). Therefore, we can only write an implicit solution of the initial value problem as: $$y(t) = e + \int_{3}^{t} \frac{1-t'}{\ln y(t')} dt'$$.

Key concepts

Implicit Solution

An implicit solution to a differential equation is one where the solution is given in terms of both the dependent and independent variables without necessarily solving for one variable explicitly. In other words, you might have the solution in a form that is not solved for the dependent variable, like \( F(t, y) = 0 \).
In the context of our exercise, we ended up with an implicit solution:

• \( y(t) = e + \int_{3}^{t} \frac{1-t'}{\ln y(t')} dt' \)

This means that the solution expresses \( y \) in terms of both \( t \) and the integral, but we have not separated \( y \) completely from the equation. Implicit solutions are often useful even if they aren't explicit, especially when it's tough to express the solution in a closed or straightforward form.
This captures the relationship between the variables according to the given differential equation and respects the initial condition \( y(3) = e \). Implicit solutions are crucial because not every differential equation can be solved explicitly, yet they still provide valuable insights into the behavior of the solution.

Explicit Solution

An explicit solution of a differential equation is one where you can solve for the dependent variable, usually denoted as \( y \), solely in terms of the independent variable, \( t \). This means you rewrite the equation to isolate \( y \) on one side:

• For example, \( y = f(t) \).

In our current exercise, after applying integration and considering the initial condition \( y(3) = e \), it turned out that deriving an explicit solution, \( y \) expressed completely in terms of \( t \), was extremely challenging.
This is often the case with complex nonlinear ordinary differential equations like the one we are dealing with.
Though obtaining an implicit solution can be achieved, an explicit expression for \( y(t) \) makes it easier to directly compute values or analyze the solution's behavior for specific \( t \)-intervals.
However, due to the complicated nature of the equation especially without further simplification possibilities, our explicit solution remained elusive, illustrating a real-world challenge often encountered in advanced calculus problems.

Initial Value Problem

An initial value problem (IVP) is a type of differential equation paired with a condition that specifies the value of the unknown function at a particular point. This initial condition helps in determining a unique solution to the differential equation.
The exercise presented is an initial value problem because it provides the condition \( y(3) = e \).

• This condition helps to uniquely identify the solution out of the possibly many solutions an ordinary differential equation might have.
• Initial conditions are crucial in finding specific solutions that satisfy both the differential equation and the given constraints.

Solving an IVP often involves integrating the differential equation and then using the initial conditions to solve for any integration constants.
Initial value problems are prevalent in many fields such as physics, engineering, and economics, where they help model real-world scenarios with known starting conditions.

First-Order Differential Equation

A first-order differential equation is a type of ordinary differential equation that involves derivatives of the first order, but no higher-order derivatives are involved in the equation.
Typically, these equations can be written in the form:

• \( y' = f(t, y) \)

The exercise equation, \((\ln y) y' + t = 1\), is identified as a first-order differential equation. In this equation, \( y' \) (the first derivative of \( y \) with respect to \( t \)) appears without any second or higher-order derivatives.

• First-order equations can be linear or nonlinear, homogeneous or inhomogeneous.
• The given equation is nonlinear due to the presence of \( \ln y \).

Understanding the type of differential equation is crucial in determining the methods or techniques that can be applied to solve it, such as separation of variables, integrating factors, or other techniques depending on its form.
This foundational understanding serves as the starting point for solving the original problem.