Problem 14 - Calculus for Biology and Medicine

Question

In Problems , solve each autonomous differential equc tion. $$ \frac{d y}{d t}=2(1-y), \text { where } y(0)=0 $$

Step-by-Step Solution

Answer

The solution is \( y = 1 - e^{-2t} \).

Step 1: Separate Variables

First, we need to separate the variables in the given differential equation. The given equation is \( \frac{dy}{dt} = 2(1-y) \). We can rewrite this as \( \frac{dy}{1-y} = 2dt \). This allows us to integrate both sides separately.

Step 2: Integrate Both Sides

Integrate both sides of the equation to find the general solution. The integral of the left side is \( \int \frac{dy}{1-y} = -\ln|1-y| \) and the integral of the right side is \( \int 2dt = 2t + C \), where \( C \) is the constant of integration.

Step 3: Solve for y

To solve for \( y \), we rearrange the equation \( -\ln|1-y| = 2t + C \). First, remove the negative sign: \( \ln|1-y| = -2t - C \). Apply the exponential function to both sides: \( |1-y| = e^{-2t-C} = Ce^{-2t} \). As \( 1-y \) must be positive, we can drop the absolute value to get: \( 1-y = Ce^{-2t} \). Finally, solve for \( y \): \( y = 1 - Ce^{-2t} \).

Step 4: Apply Initial Condition

To determine \( C \), use the initial condition \( y(0) = 0 \). Substitute \( t = 0 \) and \( y = 0 \) into the equation: \( 0 = 1 - Ce^{0} \). Solving gives \( C = 1 \).

Step 5: Write the Particular Solution

Substitute \( C = 1 \) back into the equation from Step 3 to find the particular solution: \( y = 1 - e^{-2t} \).

Key Concepts

Separation of Variables

The technique of separating variables is crucial when solving differential equations, especially in autonomous differential equations like the one given: \( \frac{dy}{dt} = 2(1-y) \). In separating variables, we aim to shuffle the equation around so each variable is isolated with its differential on different sides of the equation.

Here's how it works: with the original equation, we can move terms to achieve \( \frac{dy}{1-y} = 2dt \). This step effectively means we've separated the terms so that they only involve one variable per side. This isolation is useful because it allows us to integrate each side independently of the other.

Why is this method important? Well, separating variables transforms the process of solving complex differential equations into simpler integration tasks, making it easier to find a solution. The idea is to manipulate the equation so that you have all the \( y \) terms with \( dy \) and all the \( t \) terms with \( dt \).
In practice, this means that each side of the equation can be integrated to give us functions of \( y \) and \( t \), which ultimately helps in solving the differential equation for \( y \).

Initial Condition

After deriving a general solution from an autonomous differential equation, we use an initial condition to pin down a specific solution. In the given problem, the initial condition is \( y(0) = 0 \).

The initial condition is vital because while the general solution \( y = 1 - Ce^{-2t} \) is correct, it contains a constant "C" that must be determined. The initial condition tells us exactly how the function behaves at a specific point. By substituting \( t = 0 \) and \( y = 0 \) into the general equation, we can solve for \( C \).

• This step ensures that our solution meets real-world conditions or constraints.
• It provides the exact form of the function \( y(t) \) that fits the initial scenario described in the problem.

Without applying an initial condition, a differential equation might have infinitely many solutions. Therefore, initial conditions are essential for finding a unique solution that applies to the specific problem scenario.

Particular Solution

The particular solution of a differential equation is what we obtain after applying the initial condition to the general solution. For our example, after determining the constant \( C \) using the initial condition \( y(0) = 0 \), we substitute back to obtain the particular solution.

In this case, the general solution \( y = 1 - Ce^{-2t} \) becomes \( y = 1 - e^{-2t} \) once \( C = 1 \) is applied.

• The particular solution is specifically tailored to the conditions set by the problem.
• It describes the unique behavior of the function \( y \) across all values of \( t \).
• Particular solutions provide insight into the specific dynamics and response of a system described by the differential equation, reflecting how it starts at a given moment.

In simple terms, finding the particular solution is the final step that fine-tunes the answer to exactly match the problem's requirements. It's not just mathematically correct, but contextually relevant as well.