Problem 36 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that the differential equation $$ \frac{d y}{d t}=a y+b, y(0)=y_{0} $$ has solution $$ y=\left(y_{0}+\frac{b}{a}\right) e^{a t}-\frac{b}{a} $$ Assume that \(a \neq 0\).

Step-by-Step Solution

Answer

The differential equation solution is verified as \( y(t) = \left(y_0 + \frac{b}{a}\right) e^{at} - \frac{b}{a} \).

Step 1: Identify the Type of Differential Equation

The given differential equation \( \frac{d y}{d t} = a y + b \) is a first-order linear differential equation. It has the standard form \( \frac{d y}{d t} + P(t)y = Q(t) \) where \( P(t) = -a \) and \( Q(t) = -b \).

Step 2: Find the Integrating Factor

To solve the differential equation, we first need an integrating factor. The integrating factor, \( \mu(t) \), is given by \( e^{\int P(t)\, dt} = e^{-at} \).

Step 3: Multiply by the Integrating Factor

Multiply both sides of the differential equation by the integrating factor: \[ e^{-at} \frac{d y}{d t} + e^{-at} (-a)y = -b e^{-at} \] This allows us to write the left-hand side as a derivative of a product: \[ \frac{d}{dt}(y e^{-at}) = -b e^{-at} \]

Step 4: Integrate Both Sides

Integrate both sides with respect to \( t \):\[ \int \frac{d}{dt}(y e^{-at}) \, dt = \int -b e^{-at} \, dt \]This simplifies to:\[ y e^{-at} = \frac{b}{a} e^{-at} + C \] where \( C \) is the integration constant.

Step 5: Solve for y(t)

To find \( y(t) \), multiply through by \( e^{at} \):\[ y = \frac{b}{a} + Ce^{at} \] Now solve for \( C \) using the initial condition \( y(0) = y_0 \):\[ y_0 = \frac{b}{a} + C \] This gives: \( C = y_0 - \frac{b}{a} \)

Step 6: Substitute Back to Get the Final Solution

Substitute the value of \( C \) back into the equation:\[ y(t) = \frac{b}{a} + \left(y_0 - \frac{b}{a}\right)e^{at} \] Simplify the solution:\[ y(t) = \left(y_0 + \frac{b}{a}\right)e^{at} - \frac{b}{a} \] This is the required solution. Therefore, the differential equation is verified.

Key Concepts

Integrating Factor

The integrating factor is a crucial tool for solving first-order linear differential equations. It's what allows us to convert a difficult differential equation into an easier one that can be directly integrated. Think of it as a multiplier that simplifies the problem.

For a standard first-order linear differential equation of the form \( \frac{dy}{dt} + P(t)y = Q(t) \), the integrating factor, \( \mu(t) \), is given by:

• \( \mu(t) = e^{\int P(t) \, dt} \)

In our case, \( P(t) = -a \). So, the integrating factor becomes \( \mu(t) = e^{-at} \).

Once you have the integrating factor, you multiply it by the entire differential equation. This makes the left side a derivative of a product, which is much easier to integrate. This is the beauty of utilizing an integrating factor: it systematically separates variables, allowing straightforward integration.

Initial Value Problem

An initial value problem determines a specific solution from a family of possible solutions to a differential equation. It does so by providing a start point, known as the initial condition.

In this equation, the initial value is given as \( y(0) = y_0 \). This tells us the value of \( y \) when \( t = 0 \).

Using initial conditions, we find the constant of integration \( C \) that arises when solving the differential equation. For our problem, after integrating, we reached the expression:

• \( y = \frac{b}{a} + Ce^{at} \)
• Substituting the initial value, \( y_0 = \frac{b}{a} + C \) gives \( C = y_0 - \frac{b}{a} \).

This step is critical because it allows us to find the unique solution that satisfies both the differential equation and the given initial condition. Thus, initial value problems connect differential equations to real-world initial conditions, yielding practical and relevant solutions.

Exponential Function

The exponential function often appears in solutions to differential equations, thanks to its unique property of being its own derivative. This feature makes it especially valuable when dealing with damping, growth problems, or any process that changes at a rate proportional to its current state.

In the solution to the differential equation, the term \( e^{at} \) arises naturally. When we multiply the integrated result by the exponential \( e^{at} \), we effectively "undo" the effect of the integrating factor, isolating \( y \).

Let's note some key properties of the exponential function involved:

• \( e^{at} \) means the function grows or decays at a rate determined by \( a \).
• If \( a \) is positive, the function grows exponentially. If negative, it decays exponentially.

In this specific solution, \( y(t) = \left(y_0 + \frac{b}{a}\right)e^{at} - \frac{b}{a} \), it is the exponential function \( e^{at} \) that scales the solution over time, reflecting how the initial condition and rate of change interact dynamically.