Problem 51 - Calculus for Biology and Medicine

Question

Solve each differential equation with the given initial condition. \(\frac{d r}{d t}=r e^{-t}\), with \(r_{0}=1\) if \(t_{0}=0\)

Step-by-Step Solution

Answer

The solution is \( r(t) = e^{1-e^{-t}} \).

Step 1: Set Up the Equation

This is a separable differential equation, so we start by rearranging it so that all terms involving \(r\) are on one side and all terms involving \(t\) are on the other. This gives us \( \frac{1}{r} \, dr = e^{-t} \, dt \).

Step 2: Integrate Both Sides

Integrate both sides of the equation separately. The left side with respect to \(r\), and the right side with respect to \(t\). This yields \( \int \frac{1}{r} \, dr = \int e^{-t} \, dt \).

Step 3: Calculate the Integrals

The integral on the left is \( \ln |r| \) and the integral on the right is \(-e^{-t}\). So we have \( \ln |r| = -e^{-t} + C \), where \(C\) is the constant of integration.

Step 4: Solve for r

Solve the equation \( \ln |r| = -e^{-t} + C \) for \(r\) by exponentiating both sides to eliminate the natural logarithm. This gives us \( |r| = e^{-e^{-t} + C} = e^C e^{-e^{-t}} \).

Step 5: Apply Initial Conditions

Use the initial condition \( r(0) = 1 \) to solve for \(C\). Substituting \(t = 0\) and \(r = 1\) gives \( 1 = e^C e^{-1} \). This implies \( e^C = e \), so \(C = 1\).

Step 6: Write the Particular Solution

Substitute \(C = 1\) into the general solution. This gives us \( r = e^{1} e^{-e^{-t}} = e \cdot e^{-e^{-t}} = e^{1-e^{-t}} \). Thus, the particular solution is \( r = e^{1-e^{-t}} \).

Key Concepts

Understanding Separable Equations

Separable equations are a special type of differential equation that are relatively straightforward to solve given their unique structure. In these equations, we can separate the variables to each side, isolating the terms involving one variable on one side of the equation and the terms involving the other variable on the opposite side. This makes them easier to integrate.

For example, consider the differential equation given in our exercise: \(\frac{d r}{d t}=r e^{-t}\). To separate the variables, we rearrange it so that all terms involving \(r\) are on one side, and all terms involving \(t\) are on the other. This manipulation gives us: \( \frac{1}{r} \, dr = e^{-t} \, dt \). Now each side of the equation only contains terms involving a single variable, making it ready for the next step: integration.

This technique, applicable only to separable equations, simplifies the integration process, as we no longer need to deal with mixed variables. Remember, not all differential equations can be separated, which makes recognizing when an equation is separable a useful skill.

Applying Initial Conditions

Initial conditions are pivotal in finding a particular solution from a general solution, especially when solving differential equations. They are provided information about the state of the system at a specific point in time, allowing us to determine the constant of integration resulting from the indefinite integrals.In our exercise solution, after obtaining the general solution of the form \( \ln |r| = -e^{-t} + C \), we used the initial condition \( r(0) = 1 \) to solve for \(C\). Here, substituting \(t = 0\) and \(r = 1\) into the equation allows us to plug and solve for the unknown constant:

• Substitute into general equation: \( 1 = e^C e^{-1} \)
• Solve this expression to get \( e^C = e \).

This gives us \(C = 1\), which is then used to tailor the general solution to become a specific solution that fits the initial condition. Initial conditions are crucial because they ensure that the solution accurately reflects specific, real-world scenarios at a given initial state.

Integration in Solving Differential Equations

Integration plays a core role in solving differential equations, primarily because it reverses differentiation. When we integrate, we essentially 'undo' the process represented by the original differential equation, recovering the original function or a family of functions.In the problem we tackled, once the equation was separated into \( \frac{1}{r} \, dr = e^{-t} \, dt \), the next step was to integrate both sides:

• Left side: The integral of \(\frac{1}{r}\) with respect to \(r\) is \( \ln |r| \).
• Right side: The integral of \(e^{-t}\) with respect to \(t\) is \(-e^{-t}\).

These integrations lead to \( \ln |r| = -e^{-t} + C \), a general solution which includes an arbitrary constant \(C\).This constant appears because the integral of a function has an infinite number of antiderivatives, differing only by a constant. It's why we obtain a family of solutions unless we apply an initial condition to determine the specific solution. Integration transforms the separated equation into a solvable form, bringing us closer to the solution of the original differential problem.