Problem 5 - Applied Calculus

Question

Find solutions to the differential equations, subject to the given initial condition. $$ \frac{d Q}{d t}=\frac{Q}{5}, \quad Q=50 \text { when } t=0 $$

Step-by-Step Solution

Answer

The solution is \( Q(t) = 50e^{t/5}.\)

Step 1: Recognize the Type of Differential Equation

Identify the type of given differential equation. Here, \( \frac{dQ}{dt} = \frac{Q}{5} \) is a first-order linear differential equation.

Step 2: Formulate the General Solution

A first-order linear differential equation \( \frac{dQ}{dt} = kQ \) has the general solution \( Q(t) = Ce^{kt} \). In this problem, \( k = \frac{1}{5} \), so the general solution becomes \( Q(t) = Ce^{t/5} \).

Step 3: Apply the Initial Condition

The initial condition provided is \( Q(0) = 50 \). Substitute \( t = 0 \) and \( Q = 50 \) into the general solution to find \( C \): \[ 50 = Ce^{0} \Rightarrow C = 50. \]

Step 4: Write the Particular Solution

Substitute the value of \( C \) back into the general solution. The particular solution is \( Q(t) = 50e^{t/5} \).

Key Concepts

First-order Linear Differential Equation

Differential equations are mathematical equations that involve functions and their derivatives. A **first-order linear differential equation** is the simplest type of ordinary differential equation one can encounter. In our example \[ \frac{dQ}{dt} = \frac{Q}{5} \] this is clearly a first-order equation because it involves the first derivative of the function \( Q \) with respect to the variable \( t \).
This type of equation can always be written in the form \( \frac{dy}{dx} + P(x)y = Q(x) \), where \( P(x) \) and \( Q(x) \) are continuous functions. In the special case where P(x) = 0, like our case, the equation reduces to \( \frac{dy}{dx} = ky \).

First-order linear differential equations are widely used in various fields such as physics, engineering, and biology because they model many natural phenomena. Solving these equations involves techniques such as integrating factors or directly identifying the structure that allows them to be rewritten in an easily integrable form.

Initial Condition

Initial conditions are a key aspect when working with differential equations. They provide specific values that allow us to discover unique solutions among the many possible ones from a differential equation. In our example, we know that when \( t = 0 \), \( Q = 50 \).
This piece of information is crucial because it allows us to find the particular value of any constant in our general solution.

Think of initial conditions as the starting point of a journey. Without knowing where you start, it would be challenging to chart the course of your journey accurately. Similarly, in mathematics, initial conditions help us personalize the general solution to fit a specific scenario or real-world problem.

• Apply the initial condition by substituting it into the general solution.
• Solve for the constant that accompanies the general solution.

This process gives us a particular solution that perfectly suits the given context.

Exponential Growth

Exponential growth describes processes that increase rapidly over time. It occurs when the rate of growth of a mathematical function is proportional to the function's current value. Our differential equation \[ \frac{dQ}{dt} = \frac{Q}{5} \] is a classic representation of exponential growth.

The solution, \( Q(t) = 50e^{t/5} \), reflects this growth pattern. The variable \( t \) (time) affects the function's value by being in the exponent, causing the solution to grow (or shrink) exponentially.

Important characteristics of exponential growth include:

• Constant relative growth rate, meaning the rate of increase is continually proportional to the current value.
• Doubling time, the period it takes for a quantity to double in size, which can be computed given the growth rate.

Understanding exponential growth helps in predicting population growth, radioactive decay, and other phenomena where there's a constant proportionality in the growth rate.