Problem 26 - Advanced Engineering Mathematics

Question

In Problems 23-28, 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: \( y = \frac{1}{2} + 2e^{-2t} \); implicit form: \( e^{2t}y = \frac{1}{2}e^{2t} + 2 \).

Step 1: Separate Variables

Rewrite the given first-order differential equation \( \frac{dy}{dt} + 2y = 1 \) in a format that allows separation of variables. To do this, we realize that this is a linear differential equation and we can instead use the integrating factor method rather than separating the variables directly.

Step 2: Find the Integrating Factor

The standard form for a linear first-order differential equation is \( \frac{dy}{dt} + P(t) y = Q(t) \). Here \( P(t) = 2 \) and \( Q(t) = 1 \). The integrating factor is calculated as \( \mu(t) = e^{\int P(t) \, dt} = e^{2t} \).

Step 3: Multiply by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( e^{2t} \): \[ e^{2t} \frac{dy}{dt} + 2e^{2t} y = e^{2t} \] This can be rewritten using the product rule: \[ \frac{d}{dt}(e^{2t} y) = e^{2t} \]

Step 4: Integrate Both Sides

Integrate both sides with respect to \( t \):\[ \int \frac{d}{dt}(e^{2t} y) \ dt = \int e^{2t} \ dt \] The left side becomes \( e^{2t} y \), and the right side integrates to \( \frac{1}{2} e^{2t} + C \).

Step 5: Solve for y Explicitly

Solve for \( y \) by dividing by \( e^{2t} \): \[ y = \frac{1}{2} + Ce^{-2t} \] This is the explicit solution of the differential equation.

Step 6: Apply Initial Condition

Apply the initial condition \( y(0) = \frac{5}{2} \): \[ \frac{5}{2} = \frac{1}{2} + Ce^{0} \] Solving for \( C \), we find \( C = 2 \).

Step 7: Write Final Solutions

The specific explicit solution is \( y = \frac{1}{2} + 2 e^{-2t} \). The implicit solution is the equation in terms of \( y \) that does not isolate \( y \), which could be expressed as \( e^{2t} y = \frac{1}{2} e^{2t} + 2 \).

Key Concepts

Integrating Factor

An integrating factor is a useful mathematical tool that simplifies solving linear first-order differential equations. Essentially, an integrating factor transforms a differential equation into one that is easily integrable.
To find the integrating factor, first express your linear differential equation in standard form which looks like this:

• \( \frac{dy}{dt} + P(t)y = Q(t) \)

For the given problem, \( P(t) = 2 \) and \( Q(t) = 1 \).
The integrating factor \( \mu(t) \) is obtained by taking the exponential of the integral of \( P(t) \):

• \( \mu(t) = e^{\int P(t) \, dt} = e^{2t} \)

This integrating factor, \( e^{2t} \), is then used to multiply every term in the equation. The goal is to rewrite the left side in a form that is the derivative of a product of two functions. This manipulation makes the equation easy to integrate as seen in the original solution plan.

Implicit Solution

An implicit solution represents a solution to the differential equation where terms involve the dependent variable (like \( y \)) in a non-isolated form. In other words, \( y \) isn't written just in terms of other known quantities.
For the equation \( \frac{dy}{dt} + 2y = 1 \), after multiplying by the integrating factor \( e^{2t} \), the left-hand side becomes a derivative:

• \( \frac{d}{dt}(e^{2t} y) = e^{2t} \)

Once integrated, this becomes:

• \( e^{2t} y = \frac{1}{2} e^{2t} + C \)

This is an implicit solution. Notice \( y \) is not explicitly expressed alone on one side of the equation; instead, it's still part of a relationship involving exponential functions. Implicit solutions are sometimes preferred, particularly when they make further analysis or computation more straightforward.

Explicit Solution

An explicit solution clearly expresses the dependent variable, \( y \), as a function of the independent variable, \( t \), without any ambiguity.
Take the implicit solution \( e^{2t} y = \frac{1}{2} e^{2t} + C \) and solve it for \( y \):

• \( y = \frac{1}{2} + Ce^{-2t} \)

This formulation is an explicit solution since \( y \) is isolated on one side of the equation.
To find the specific explicit solution for the initial-value problem, we use the initial condition \( y(0) = \frac{5}{2} \). By substituting this into the explicit form and solving for \( C \), we find:

• \( C = 2 \)

Thus, the explicit solution is \( y = \frac{1}{2} + 2e^{-2t} \). Explicit solutions are often more useful for understanding the behavior of the solution over time.

Initial-value Problem

An initial-value problem is a type of differential equation along with a specified value for the unknown function at a given point, which sets the stage for finding a unique solution.
In this case, we begin with the equation:

• \( \frac{dy}{dt} + 2y = 1 \)

and the specified initial condition:

• \( y(0) = \frac{5}{2} \)

The initial condition provides a precise starting point to determine the constant \( C \) in the general solution. This condition distinguishes the particular solution from the family of possible solutions.
Using the initial condition in the explicit solution \( y = \frac{1}{2} + Ce^{-2t} \), we substituted \( t=0 \) and \( y=\frac{5}{2} \) to solve for \( C \). Finding \( C = 2 \) gives us the particular solution that meets the specific requirements of the problem. Initial-value problems are important in many scientific and engineering applications where initial conditions are known.