Problem 36 - Calculus

Question

Solve the differential equations in Problems \(34-43 .\) Assume \(a, b,\) and \(k\) are nonzero constants. $$\frac{d P}{d t}=P-a$$

Step-by-Step Solution

Answer

The solution is \( P(t) = a + Ce^{t} \).

Step 1: Recognize the Type of Differential Equation

The given differential equation \( \frac{dP}{dt} = P - a \) is a first-order linear differential equation where the derivative \( \frac{dP}{dt} \) is expressed in terms of \( P \) and \( a \). The equation can be rewritten as \( \frac{dP}{dt} - P = -a \).

Step 2: Find the Integrating Factor

To solve \( \frac{dP}{dt} - P = -a \), we first identify the coefficient of \( P \), which is \(-1\). The integrating factor \( \mu(t) \) is found using \( e^{\int -1 \, dt} = e^{-t} \).

Step 3: Multiply by the Integrating Factor

Multiply each term of the differential equation by the integrating factor \( e^{-t} \). The equation becomes \( e^{-t} \frac{dP}{dt} - e^{-t} P = -a e^{-t} \).

Step 4: Simplify to a Perfect Derivative

Notice that the left-hand side becomes \( \frac{d}{dt}(e^{-t}P) \) upon simplifying. Therefore, the equation is now \( \frac{d}{dt}(e^{-t}P) = -a e^{-t} \).

Step 5: Integrate Both Sides

Integrate both sides with respect to \( t \). The left side becomes \( e^{-t}P \) and the right side integrates to \( -a \int e^{-t} \, dt = a e^{-t} + C \), where \( C \) is the constant of integration.

Step 6: Solve for \( P \)

To isolate \( P \), multiply through by \( e^{t} \) yielding \( P = a + Ce^{t} \).

Step 7: Conclusion

The general solution to the differential equation \( \frac{dP}{dt} = P - a \) is \( P(t) = a + Ce^{t} \), where \( C \) is an arbitrary constant based on initial conditions.

Key Concepts

First-Order Linear Differential Equation

In mathematics, a first-order linear differential equation is an equation that involves the derivatives of a function and the function itself, with both appearing in a linear fashion. The general form for such an equation is \( \frac{dy}{dx} + P(x)y = Q(x) \), where \( P(x) \) and \( Q(x) \) are functions of \( x \). For our specific problem, the equation \( \frac{dP}{dt} = P - a \) can be rewritten to conform to this general form as \( \frac{dP}{dt} - P = -a \). Here:

• \( y \) corresponds to \( P \)
• \( P(t) = -1 \) is a constant coefficient of \( y \) (here \( P \) is a function, not to confuse with the variable \( P \) in the differential equation)
• \( Q(t) = -a \) is a constant function

Understanding this form allows us to identify the structure and method required for solving the equation. The key is that the equation is linear, making it appropriate to use an integrating factor, a technique specific to this type of equation.

Integrating Factor

The integrating factor is a crucial step in solving first-order linear differential equations. It transforms the original differential equation into a form that can be easily integrated. To determine the integrating factor \( \mu(t) \), we use the coefficient of \( P \), which is \(-1\), and compute:\[ \mu(t) = e^{\int -1 \,dt} = e^{-t} \]This exponential function simplifies the equation when each term is multiplied by it. Essentially, it turns the left-hand side into a derivative of the product \( e^{-t}P \), allowing you to integrate the entire equation straightforwardly. Applying the integrating factor results in:\[ e^{-t}\frac{dP}{dt} - e^{-t}P = -ae^{-t} \]This makes the equation more manageable, as you can now express it as a derivative of a single expression \( \frac{d}{dt}(e^{-t}P) \), enabling easier integration.

General Solution

Upon transforming the original differential equation using the integrating factor, the next step is to integrate both sides to find the general solution. When we integrate \( \frac{d}{dt}(e^{-t}P) = -ae^{-t} \), the left side simplifies to \( e^{-t}P \), and the right side becomes:\[ -a\int e^{-t} \, dt = ae^{-t} + C \]Here, \( C \) is the constant of integration. By multiplying through by \( e^{t} \) to eliminate the exponential on the left side, you isolate \( P \), arriving at:\[ P(t) = a + Ce^{t} \]This equation represents the general solution to the differential equation. It reveals how the function evolves over time based on initial conditions. The constant \( C \) can be determined if an initial value is provided, tailoring the solution to specific scenarios.