Problem 2 - Calculus for Biology and Medicine

Question

\text { In Problems , solve each pure-time differential equation. } \frac{d y}{d t}=e^{-3 t}, \text { where } y(0)=10 .

Step-by-Step Solution

Answer

The solution is \( y(t) = -\frac{1}{3} e^{-3t} + \frac{31}{3} \).

Step 1: Understand the Given Differential Equation

The given equation is \( \frac{d y}{d t} = e^{-3t} \), which is a first-order differential equation with respect to \( t \). This equation gives how \( y(t) \) changes as a function of time \( t \).

Step 2: Integrate Both Sides of the Differential Equation

To solve for \( y(t) \), integrate both sides with respect to \( t \). The left side becomes \( \int \frac{dy}{dt} \, dt = y(t) \), and the right side \( \int e^{-3t} \, dt = \int e^{-3t} \, dt \).

Step 3: Solve the Integral

Calculate the integral \( \int e^{-3t} \, dt \). The antiderivative of \( e^{-3t} \) is \( -\frac{1}{3} e^{-3t} \), so:\[y(t) = -\frac{1}{3} e^{-3t} + C\] where \( C \) is the constant of integration.

Step 4: Apply the Initial Condition

Use the initial condition \( y(0) = 10 \) to find \( C \):\[10 = -\frac{1}{3} e^{0} + C \10 = -\frac{1}{3} + C \C = 10 + \frac{1}{3} = \frac{30}{3} + \frac{1}{3} = \frac{31}{3}\]

Step 5: Write the General Solution with the Constant

Substitute \( C = \frac{31}{3} \) back into the solution:\[y(t) = -\frac{1}{3} e^{-3t} + \frac{31}{3}\] This is the particular solution to the differential equation with the given initial condition.

Key Concepts

Integration in Differential Equations

Integration is a fundamental concept used extensively in the process of solving differential equations. In differential equations, you often need to integrate to find the solution to an equation that describes how a function changes. In our given problem, the differential equation \( \frac{d y}{d t}=e^{-3 t} \) is solved by integrating the right-hand side with respect to \( t \). This process allows us to determine the function \( y(t) \) that describes how \( y \) relates to \( t \).
When integrating an expression like \( e^{-3t} \), it's helpful to recognize it as an exponential function. The integral \( \int e^{-3t} \, dt \) is computed by finding the antiderivative, which in this case is \( -\frac{1}{3}e^{-3t} \). Here’s a quick breakdown of this integration:

• The formula for integrating \( e^{at} \) is \( \frac{1}{a}e^{at} \), where \( a \) is a constant.
• In our equation, \( a \) is -3, leading to \( -\frac{1}{3}e^{-3t} \) as the antiderivative.

Always remember to add the constant of integration \( C \), which accounts for any initial conditions or vertical shifts in the graph of the function. This ensures our solution is complete.

Understanding Initial Value Problems

An initial value problem involves a differential equation along with a specific given value at a point, called the initial condition. This initial condition is necessary to find a particular solution from the general solution of the differential equation. In simpler terms, while the integration gives us a family of curves or solutions,
the initial condition helps us find the exact curve out of that family that satisfies the conditions given in the problem.
In the equation \( \frac{d y}{d t} = e^{-3t} \) with the initial condition \( y(0) = 10 \), solving the differential equation leaves us with \( y(t) = -\frac{1}{3}e^{-3t} + C \). The role of the initial condition is crucial at this stage. We use it by substituting \( t=0 \) and \( y(0)=10 \) into the equation to solve for \( C \).

• This ensures that the solution curve passes through the point (0, 10) on the y-axis.
• By performing these calculations, we determine that \( C = \frac{31}{3} \), thus specifying the particular solution \( y(t) = -\frac{1}{3}e^{-3t} + \frac{31}{3} \).

This approach is what personalizes the general solution to fit the unique scenario presented by the problem.

First-Order Differential Equations

A first-order differential equation is an equation that involves the first derivative of the unknown function. The equation \( \frac{d y}{d t} = e^{-3t} \) is an example of a first-order differential equation because it contains the first derivative of \( y \) with respect to \( t \). This type of equation provides us with a rate of change, but not the actual function initially. Solving it introduces the actual function conveyed by the rate.
There are various methods to solve a first-order differential equation, with the most common being separation of variables, especially when one side of the equation can be expressed entirely in terms of one variable.

• To apply this, terms involving \( y \) (if present) and \( t \) are separated on either side of the equation.
• Then integration is applied on both sides to solve for the function \( y(t) \).

In our case, since the equation was already pure-time, involving only \( e^{-3t} \), the integration directly led to finding \( y(t) \). The simplicity of first-order equations makes them often easier to handle, serving as fundamental building blocks to understanding more complex differential equations.