Problem 28 - Differential Equations and Linear Algebra

Question

Solve the given initial-value problem. $$y^{\prime}-2 y=u_{2}(t) e^{t-2}, \quad y(0)=2$$.

Step-by-Step Solution

Answer

The solution to the given initial-value problem is \(y(t) = e^{2t}(\int u_{2}(t)e^{-t} dt + 2)\).

Step 1: Identify the given differential equation

The given initial-value problem is a linear first-order differential equation: $$y^{\prime}-2 y=u_{2}(t) e^{t-2}, \quad y(0)=2$$

Step 2: Find the integrating factor

Observe that the given linear differential equation is of the form: $$y^{\prime}-2 y = g(t)$$ The integrating factor is given by: $$\mu(t) = e^{\int -2 dt} = e^{-2t}$$

Step 3: Multiply the differential equation by the integrating factor

Multiply the given differential equation by \(\mu(t)\): $$e^{-2t}(y^{\prime}-2 y) = u_{2}(t) e^{t-2}e^{-2t}$$ Simplifying, we get: $$y^{\prime}e^{-2t} - 2ye^{-2t} = u_{2}(t)e^{-t}$$

Step 4: Integrate both sides of the equation

The left-hand side is an exact derivative: $$\frac{d}{dt}(ye^{-2t})= u_{2}(t)e^{-t}$$ Now integrate both sides with respect to \(t\): $$\int \frac{d}{dt}(ye^{-2t})dt = \int u_{2}(t)e^{-t} dt$$ $$ye^{-2t}= \int u_{2}(t)e^{-t} dt + C$$

Step 5: Solve for y(t) and apply the initial condition

Multiply both sides by \(e^{2t}\) to solve for \(y(t)\): $$y(t) = e^{2t}(\int u_{2}(t)e^{-t} dt + C)$$ Now, apply the initial condition \(y(0) = 2\): $$2 = e^{0}(\int u_{2}(0)e^{0} dt + C)$$ So, \(C = 2\).

Step 6: Write the final solution

The solution to the given initial-value problem is: $$y(t) = e^{2t}(\int u_{2}(t)e^{-t} dt + 2)$$

Key Concepts

Linear First-Order Differential Equation

A linear first-order differential equation is a type of equation that can be expressed in a certain standard form. In general, it appears as:

• \( \frac{dy}{dt} + P(t)y = Q(t) \)

The term "linear" signifies that the dependent variable \( y \) and its derivative \( \frac{dy}{dt} \) appear only to the first power, not squared or cubed, for example. The coefficients \( P(t) \) and \( Q(t) \) may still be functions of \( t \). These equations are common in scenarios where change is proportional to the current state, such as growth and decay models.
In our example, the given linear first-order differential equation is:

• \( y^{\prime} - 2y = u_{2}(t) e^{t-2} \) with initial condition \( y(0) = 2 \)

We'll see that this form allows us to proceed with solving the problem using an effective method called an integrating factor.

Integrating Factor

An integrating factor is a clever technique used to solve linear first-order differential equations. This method involves multiplying the differential equation by an "integrating factor" that simplifies the left side of the equation, making it an exact derivative.
The integrating factor \( \mu(t) \) is determined as follows:

• \( \mu(t) = e^{\int P(t) \, dt} \)

For our differential equation \( y^{\prime} - 2y = u_{2}(t) e^{t-2} \), \( P(t) = -2 \). Hence, the integrating factor is calculated:

• \( \mu(t) = e^{-2t} \)

This choice turns the left-hand side of the equation into an exact derivative, forming a key step in solving the equation by integration.

Exact Derivative

When solving a linear differential equation using an integrating factor, a key observation is that the modified equation becomes an exact derivative. This means the equation looks like the derivative of a single function.

• For example, \( \frac{d}{dt}(ye^{-2t}) \)

The concept of exactness means that when differentiated, this expression yields what was initially on the left side of the manipulated equation. Therefore, finding this exact form lets us simplify the task into basic integration.
In the step-by-step solution, the left-hand side of our equation becomes:

• \( \frac{d}{dt}(ye^{-2t}) = u_{2}(t)e^{-t} \)

This shows how integral calculus is used practically to solve the differential equation via simple antiderivatives.

Initial Condition

An initial condition is a crucial part of solving an initial-value problem. It specifies the value of the unknown function at a particular point, which is vital for determining the unique solution of a differential equation.

• In our example, the initial condition provided is \( y(0) = 2 \).

This means at time \( t = 0 \), the function \( y \) is equal to 2. After integrating, we obtain a solution that involves a constant \( C \).
To find \( C \), we apply the initial condition:

• Substitute: \( 2 = e^{0}(\int u_{2}(0)e^{0} dt + C) \)
• Simplifying gives \( C = 2 \).

This crucial step ensures that our solution satisfies not only the differential equation but also the given condition at \( t = 0 \).