Problem 26 - Advanced Engineering Mathematics

Question

In Problems, find an implicit and an explicit solution of the given initial- value problem. \(\frac{d y}{d t}+2 y=1, \quad y(0)=\frac{5}{2}\)

Step-by-Step Solution

Answer

Explicit solution: \( y(t) = \frac{1}{2} + 2e^{-2t} \); Implicit solution: \( (y - \frac{1}{2})e^{2t}=2 \).

Step 1: Identify and Rearrange the Differential Equation

The given differential equation is \( \frac{dy}{dt} + 2y = 1 \). This is a first-order linear ordinary differential equation (ODE). First, rearrange it to the standard form: \( \frac{dy}{dt} = 1 - 2y \).

Step 2: Use an Integrating Factor

To solve this linear ODE, we first determine the integrating factor, \( \mu(t) \), given by \( \mu(t) = e^{\int 2 \, dt} = e^{2t} \).

Step 3: Multiply Through by the Integrating Factor

Multiply the entire differential equation \( \frac{dy}{dt} + 2y = 1 \) by the integrating factor \( e^{2t} \) to obtain \( e^{2t} \frac{dy}{dt} + 2e^{2t} y = e^{2t} \).

Step 4: Recognize as a Derivative

Notice that the left side of the equation \( e^{2t} \frac{dy}{dt} + 2e^{2t} y \) is the derivative of \( y(t) e^{2t} \). Therefore, the equation becomes \( \frac{d}{dt}(y(t) e^{2t}) = e^{2t} \).

Step 5: Integrate Both Sides

Integrate both sides with respect to \( t \) to obtain \( y(t) e^{2t} = \int e^{2t} \, dt + C \), where \( C \) is the constant of integration. The right side integrates to \( \frac{1}{2} e^{2t} + C \).

Step 6: Solve for the Explicit Solution

To find \( y(t) \), divide both sides by \( e^{2t} \): \( y(t) = \frac{1}{2} + Ce^{-2t} \). This is the explicit solution.

Step 7: Use Initial Condition to Find the Constant

Apply the initial condition \( y(0) = \frac{5}{2} \). Substitute \( t = 0 \) and \( y(0) = \frac{5}{2} \) into the explicit solution: \( \frac{5}{2} = \frac{1}{2} + C \). Solve for \( C \): \( C = 2 \).

Step 8: Write the Particular Solution

Using \( C = 2 \), the explicit solution becomes \( y(t) = \frac{1}{2} + 2e^{-2t} \).

Step 9: Identify the Implicit Solution

The implicit solution is derived from the equation we obtained before isolating \( y(t) \): \( (y - \frac{1}{2})e^{2t} = 2 \).

Key Concepts

Initial Value Problem

An Initial Value Problem consists of a differential equation along with a specified value at a given point. In our exercise, the differential equation is \( \frac{dy}{dt} + 2y = 1 \), and the initial condition provided is \( y(0) = \frac{5}{2} \).
This means we are not only trying to find a function \( y(t) \) that satisfies the differential equation, but also adheres to the condition \( y(t) = \frac{5}{2} \) when \( t = 0 \).

• It narrows down the possibilities to a unique solution that both satisfies the differential equation and passes through a given point.
• Our task is to solve the differential equation and adjust the function so it matches the initial value given.

The initial value assures that any constants derived during the integration process are specifically defined, creating a precise solution to the problem.

Integrating Factor Method

The Integrating Factor Method is a technique to solve linear ordinary differential equations. The goal is to multiply the entire differential equation by the integrating factor, transforming it into a form that is easier to integrate.
For the equation \( \frac{dy}{dt} + 2y = 1 \), the integrating factor \( \mu(t) \) is determined by:
\[ \mu(t) = e^{\int 2 \, dt} = e^{2t} \]

• This factor helps in restructuring the differential equation into a complete derivative form.
• It transforms the left-hand side into the derivative of a product, simplifying integration.

Once both sides are multiplied by \( \mu(t) \), the left side becomes \( \frac{d}{dt}(y(t) e^{2t}) \), which is straightforward to integrate.

Linear Ordinary Differential Equation

A Linear Ordinary Differential Equation (or linear ODE) involves derivatives of a function and no products of the function or its derivatives. Our example, \( \frac{dy}{dt} + 2y = 1 \), perfectly represents this type.

• It is linear because the expression involves terms with \( y \) and \( \frac{dy}{dt} \) raised only to the first power.
• It is ordinary because it contains derivatives with respect to just one variable (\( t \)).

This characteristic allows us to use specific strategies like the Integrating Factor Method for solving it.

Explicit Solution

An explicit solution of a differential equation is one where the dependent variable (in our case \( y(t) \)) is isolated on one side of the equation. We derive such solutions to make predicting specific values easier.
From the solution process, our explicit solution is:
\[ y(t) = \frac{1}{2} + 2e^{-2t} \]

• This represents \( y \) directly in terms of \( t \).
• It allows immediate calculation of \( y \) at any given \( t \) by simply plugging in the value.

This type of solution is highly useful as it clearly describes the behavior of \( y(t) \) over time.

Implicit Solution

An implicit solution expresses the relationship between \( y \) and \( t \) without necessarily isolating \( y \). Instead, it might involve a combination or constraint containing both variables.
During the solution process, we derived an implicit form:
\[ (y - \frac{1}{2})e^{2t} = 2 \]

• Such expressions bind all variables in a unified statement but do not solve directly for \( y \).
• Implicit solutions can be transformed into explicit ones but are useful as they sometimes capture more complex relationships or are easier to derive initially.

These solutions are valuable for understanding relational aspects of variables and are typically used when explicit formulation is complex or unnecessary.