Problem 18 - Calculus and its applications

Question

(a) find the general solution of each differential equation, and (b) check the solution by substituting into the differential equation. \(\frac{d V}{d t}=1.33 \mathrm{~V}\)

Step-by-Step Solution

Answer

The general solution is \(V = Ce^{1.33t}\) and it satisfies the differential equation.

Step 1: Recognize the Type of Differential Equation

The given differential equation is \(\frac{dV}{dt} = 1.33V\). This is a first-order linear differential equation, which can be solved using the method of separation of variables or recognizing it as an exponential growth equation.

Step 2: Rewrite the Equation Using Separation of Variables

To use separation of variables, rewrite the differential equation as \(\frac{dV}{V} = 1.33 \, dt\). This separates the variables \(V\) and \(t\) on opposite sides.

Step 3: Integrate Both Sides

Integrate both sides of the equation: \(\int \frac{1}{V} \, dV = \int 1.33 \, dt\). The integral of \(\frac{1}{V} \, dV\) is \(\ln |V|\), and the integral of \(1.33 \, dt\) is \(1.33t\). So, \(\ln |V| = 1.33t + C\), where \(C\) is the constant of integration.

Step 4: Solve for V

Solve for \(V\) by exponentiating both sides: \(|V| = e^{1.33t + C}\). Simplifying, \(V = Ce^{1.33t}\), where \(C = e^C\) is a new constant, representing any real number (positive, negative, or zero). This is the general solution.

Step 5: Verify the Solution

Substitute \(V = Ce^{1.33t}\) into the original differential equation: \(\frac{d}{dt}(Ce^{1.33t}) = Ce^{1.33t} \cdot 1.33\). The derivative is \(1.33Ce^{1.33t}\), which matches the original equation \(1.33V\). This confirms the solution is correct.

Key Concepts

First-order Linear Differential Equation

A first-order linear differential equation is a type of equation that involves the first derivative of a function and can be written in the standard form:
\( \frac{dy}{dx} + P(x)y = Q(x) \).
This form shows us that the equation depends on the function \(y\) itself and its first derivative. In many scenarios, these equations arise naturally in processes that change with time, such as cooling rates, population growth, or changes in investment accounts.

• P(x) and Q(x) are functions of the independent variable \(x\).
• dy/dx represents the rate of change of \(y\) with respect to \(x\).

Understanding this foundational equation type sets the stage for solving many real-world problems by finding a function \(y(x)\) that satisfies the equation. In our exercise, \( \frac{dV}{dt} = 1.33V \) is already in a similar format where \(P(x) = -1.33\) simplifying our problem to focus on solving via suitable techniques like separation of variables or identifying it as an exponential form.

Separation of Variables

Separation of variables is a fundamental technique used for solving differential equations. The goal is to rearrange the terms of an equation allowing each variable and its differential to be on separate sides of the equation. This method turns the original differential equation into two integrals, which can often be solved more easily.
To employ this method, start by ensuring all occurrences of one variable are on one side and all occurrences of the other variable are on the opposite side. For the exercise \( \frac{dV}{dt} = 1.33V \):

• You rewrite it to completion: \( \frac{dV}{V} = 1.33 \, dt \).
• By doing this, you have 'separated' the variables \(V\) and \(t\).

Next, integrate both sides: the integral of \( \frac{1}{V} \, dV \) is \( \ln |V| \), and the integral of \( 1.33 \, dt \) is \( 1.33t+C \). This gives the intermediary step crucial for finding the solution:
\( \ln |V| = 1.33t + C \).
Ultimately, the method of separation of variables is valuable because it simplifies complex relationships into manageable forms.

Exponential Growth Equation

The concept of exponential growth arises frequently in natural and social sciences, particularly describing situations where the rate of change of a quantity is proportional to the current amount. Such growth is represented mathematically by the equation of the form:
\( \frac{dN}{dt} = kN \).
This showcases an exponential growth model, with \(N\) as the quantity, \(t\) as time, and \(k\) as the constant growth rate.
In our example, \( \frac{dV}{dt} = 1.33V \) represents an exponential growth situation where the quantity \(V\) grows at a rate proportional to its current value. Solving for \(V\) involves finding a function that satisfies this equation, often resulting in the formula:
\( V(t) = Ce^{kt} \), where \(C\) is the constant derived from initial conditions.

• Exponential growth is characterized by rapid increase, common within populations, capital investments, and more.
• The solution form \( V(t) = Ce^{1.33t} \) illustrates how solutions align with the exponential growth paradigm.

Recognizing when a differential equation reflects exponential growth helps in predicting future behavior of the modeled scenario effectively.