Problem 22 - Applied Calculus

Question

Match solutions and differential equations. (Note: Each equation may have more than one solution, or no solution.) (a) \(\frac{d y}{d x}=\frac{y}{x}\) (I) \(y=x^{3}\) (b) \(\frac{d y}{d x}=3 \frac{y}{x}\) (II) \(y=3 x\) (c) \(\frac{d y}{d x}=3 x\) (III) \(y=e^{3 x}\) (d) \(\frac{d y}{d x}=y\) (IV) \(y=3 e^{x}\) (e) \(\frac{d y}{d x}=3 y\) (V) \(y=x\)

Step-by-Step Solution

Answer

(a) matches V, (b) matches I, (c) has no match, (d) matches IV, (e) matches III.

Step 1: Match Equation (a)

The differential equation \( \frac{dy}{dx} = \frac{y}{x} \) is a separable equation. A typical solution for this form is \( y = Cx \), where \( C \) is a constant. Match this to one of the given functions. The function that fits this form is \( y = x \) (Option V).

Step 2: Match Equation (b)

The differential equation \( \frac{dy}{dx} = 3 \frac{y}{x} \) is also a separable equation and has a solution of the form \( y = Cx^3 \). Here, Option I, \( y = x^3 \), matches this equation.

Step 3: Match Equation (c)

The differential equation \( \frac{dy}{dx} = 3x \) suggests integrating the right-hand side to find the solution. Integrating, we get \( y = \frac{3}{2}x^2 + C \). There is no exact polynomial option that fits this, indicating a missing straightforward match with the given options. None of the provided solutions match perfectly.

Step 4: Match Equation (d)

The differential equation \( \frac{dy}{dx} = y \) is an exponential growth equation. A standard solution for this form is \( y = Ce^x \). Option IV, \( y = 3e^x \), matches this form.

Step 5: Match Equation (e)

The differential equation \( \frac{dy}{dx} = 3y \) is solved by \( y = Ce^{3x} \) for a constant \( C \). Therefore, Option III, \( y = e^{3x} \), fits this equation.

Key Concepts

Separable Equations

Separable equations are a special type of differential equation that can be expressed as the product of a function in terms of only one variable and another function in terms of another variable. Essentially, they allow us to "separate" the variables on either side of the equation. This makes finding solutions somewhat straightforward as we can integrate to solve for the variable of interest.

For example, consider the differential equation \( \frac{dy}{dx} = \frac{y}{x} \). This equation can be rewritten as \( \frac{1}{y} dy = \frac{1}{x} dx \), which shows separation of variables. When you integrate both sides, you get

1. \( \int \frac{1}{y} dy = \ln |y| + C \)
2. \( \int \frac{1}{x} dx = \ln |x| + C \)

Putting everything together, the general solution is \( y = Cx \), where \( C \) is a constant. Each specific problem might yield a unique \( C \), based on initial conditions or given solutions.

Separable equations are particularly useful in situations where relationships between two variables are directly proportional, which is typical in physics and other sciences.

Solutions of Differential Equations

To solve a differential equation means to find a function or a set of functions that satisfies the equation. Differential equations can often have multiple solutions due to the presence of arbitrary constants arising from integration. Let's go through the process using a simple example: solving \( \frac{dy}{dx} = 3 x \).

The basic approach involves integrating the differential equation. Here, we integrate the right-hand side of the equation:

• Integrate \( 3x \) to get \( \frac{3}{2}x^2 + C \)

Thus, the general solution is \( y = \frac{3}{2}x^2 + C \), where \( C \) is an integration constant.

The constants are determined by supplementary conditions called initial conditions. If an initial condition is given, like \( y(x_0) = y_0 \), you can substitute these values into the general solution to find the specific constant \( C \) for a particular solution.

Solving differential equations is not just about manipulating equations. It's about understanding the underlying processes they describe, whether it's growth patterns, decay, or other dynamic behaviors in science and engineering.

Exponential Growth Equations

Exponential growth equations are a specific type of differential equation that describe processes where the rate of change of a quantity is proportional to the quantity itself. This results in solutions that grow exponentially over time.

Consider the differential equation \( \frac{dy}{dx} = y \). The solution to this equation is \( y = Ce^x \), illustrating a classic case of exponential growth. Here, \( C \) is a constant that determines the initial amount or starting value of the function.

• For \( \frac{dy}{dx} = 3y \), the equation reflects accelerated growth, leading to a solution of the form \( y = Ce^{3x} \).

Such growth patterns are observed in various real-life situations, including population growth, radioactive decay, and compound interest. The key feature of exponential growth is its multiplicative nature; each unit of time, the quantity multiplies by a constant factor.

Understanding exponential growth is crucial in fields ranging from finance to biology. It demonstrates how systems can change dramatically over time, especially when they experience feedback processes that spur more growth or shrinkage.