Question

Consider the differential equation $$ d y / d t=-a y+b $$ where both \(a\) and \(b\) are positive numbers. $$ \begin{array}{l}{\text { (a) Solve the differential equation }} \\ {\text { (b) Sketch the solution for several different initial conditions. }} \\\ {\text { (c) Describe how the solutions change under each of the following conditions: }}\end{array} $$ $$ \begin{array}{l}{\text { i. } a \text { increases. }} \\ {\text { ii. } b \text { increases. }} \\ {\text { iii. Both } a \text { and } b \text { increase, but the ratio } b / a \text { remains the same. }}\end{array} $$

Short answer

Question: Solve the first-order linear differential equation, sketch the solution for several different initial conditions, and describe how the solutions change if the coefficients a, b, or their ratio b/a changes. Solution: The general solution to the differential equation is given by: $$ y(t) = -\frac{b}{a} + Ce^{at}. $$ The solution can be sketched for different initial conditions as follows: 1. If y(0) > -b/a, the solution starts above the equilibrium point and decays towards it. 2. If y(0) = -b/a, the solution remains at the equilibrium point for all t. 3. If y(0) < -b/a, the solution starts below the equilibrium point and increases towards it. As a increases, the solutions approach the equilibrium point more quickly. As b increases, the equilibrium point becomes more negative. If both a and b increase, but the ratio b/a remains the same, the equilibrium point remains unchanged, but the solutions approach it more quickly.

Step-by-step solution

(a) Solve the differential equation.

To solve the first-order linear differential equation$$ \frac{dy}{dt} = -ay+b, $$we first have to find an integrating factor. The integrating factor is given by$$ I(t) = \mathrm{e}^{\int -a dt} = \mathrm{e}^{-at}. $$We then multiply both sides of the differential equation by I(t) to get$$ \mathrm{e}^{-at} \frac{dy}{dt} + a\mathrm{e}^{-at}y = \mathrm{e}^{-at}b. $$The left side is now a perfect derivative:$$ \frac{d}{dt}(\mathrm{e}^{-at}y) = \mathrm{e}^{-at}b. $$Now, we can solve by integrating with respect to t on both sides:$$ \int \frac{d}{dt}(\mathrm{e}^{-at}y) dt = \int \mathrm{e}^{-at} b dt \\ \mathrm{e}^{-at}y = -\frac{b\mathrm{e}^{-at}}{a} + C \\ y(t) = -\frac{b}{a} + Ce^{at}, $$where C is a constant of integration.

(b) Sketch the solution for several different initial conditions.

Now that we have the general solution$$ y(t) = -\frac{b}{a} + Ce^{at}, $$we can sketch its graph for different initial conditions. The general shape of the solution depends on the sign of the constant C. 1. If the initial value y(0) > -b/a, then C > 0. The solution will start above the equilibrium point y = -b/a, and it will decrease towards but never reach the equilibrium point as t goes to infinity. 2. If y(0) = -b/a, then C = 0. In this case, the solution will start at the equilibrium point and remain there for all t. 3. If y(0) < -b/a, then C < 0. The solution will start below the equilibrium point and increase towards, but never reach, the equilibrium point as t goes to infinity.

(c) i. Describe how the solutions change as a increases.

If a increases, the exponential term in the solution$$ y(t) = -\frac{b}{a} + Ce^{at} $$will decay more rapidly. As a result, the solution will approach the equilibrium point more quickly. This can be seen from the graph, where the slope will be steeper for solutions with larger values of a.

(c) ii. Describe how the solutions change as b increases.

If b increases, the equilibrium point$$ -\frac{b}{a} $$will become more negative. Thus, the solutions will move downward in the vertical direction. The general behavior will remain the same, but the equilibrium point, which all solutions approach as t goes to infinity, will be lower.

(c) iii. Describe how the solutions change if both a and b increase, but the ratio b/a remains the same.

If a and b both increase while maintaining the same ratio b/a, the equilibrium point$$ -\frac{b}{a} $$will remain unchanged. However, the exponential term of the solution$$ y(t) = -\frac{b}{a} + Ce^{at} $$will decay faster as a increases, similar to part (c) i. Therefore, the solutions will approach the equilibrium point more quickly, but the equilibrium itself remains the same.

Key concepts

Integrating Factor

An integrating factor is a clever mathematical trick used to simplify solving certain types of differential equations. In our problem, the differential equation is given by \( \frac{dy}{dt} = -ay + b \), a first-order linear differential equation.
To solve it, we multiply both sides by an integrating factor to make the equation easier to solve. The integrating factor here is \( I(t) = e^{-at} \).

• **Purpose**: It makes the left-hand side of the equation a perfect derivative. This way, integrating becomes straightforward.
• **Calculation**: The integrating factor is derived by taking the exponential of the integral of the coefficient of \( y \), which is \(-a\) in this case.

Once you apply this factor and rearrange the terms, the differential equation becomes quite favorable for integration. Solving becomes more automatic, leading us closer to finding the general solution.

Equilibrium Point

The equilibrium point is essentially a steady state where the system does not change over time. In the context of our differential equation \( \frac{dy}{dt} = -ay + b \), the equilibrium point occurs when the rate of change, \( \frac{dy}{dt} \), is zero.

• To find this point, set \(-ay + b = 0\) and solve for \( y \), resulting in \( y = -\frac{b}{a} \).
• This gives us the value where the system stabilizes.

Understanding the equilibrium is crucial because it tells us the target value that solutions will tend toward over time. In sketches of solutions, you’ll notice that irrespective of the initial conditions, the solutions will trend toward this equilibrium, when \( t \) tends to infinity.

Initial Conditions

Initial conditions are vital in determining the specific path taken by a solution of a differential equation. It refers to the value of the solution at \( t = 0 \), commonly denoted as \( y(0) \).
In solving \( y(t) = -\frac{b}{a} + Ce^{at} \), \( C \) is determined using the initial condition.

• If \( y(0) > -\frac{b}{a} \), then \( C > 0 \), indicating that the solution starts above the equilibrium point and asymptotically approaches from above.
• If \( y(0) = -\frac{b}{a} \), then \( C = 0 \), and the solution remains exactly at the equilibrium point.
• If \( y(0) < -\frac{b}{a} \), then \( C < 0 \), showing it starts below and increases towards the equilibrium.

Initial conditions are your way of 'anchoring' the solution in real-world scenarios, making it crucial in practical applications.

Exponential Decay

Exponential decay refers to the decreasing nature of a value at a rate proportional to its current value. In the equation solution \( y(t) = -\frac{b}{a} + Ce^{at} \), the term \( Ce^{at} \) is what exhibits exponential decay or growth, depending on the sign of \( a \).
Since \( a \) is positive, if \( C > 0 \), this term decreases as \( t \) increases, indicating exponential decay.

• **Characteristic**: The rate at which it approaches zero is influenced by \( a \), where a larger \( a \) results in faster decay.
• **Behavior**: With \( t \) going to infinity, whether the initial condition started above or below, the decay ensures eventual approach to the equilibrium.

This natural tapering behavior is fundamental to many physical and natural processes, like cooling and depreciation.