Question

Give the first 5 terms of the series that is a solution to the given differential equation. $$y^{\prime}=5 y, \quad y(0)=5$$

Short answer

The first 5 terms are: 5, 25x, \( \frac{125x^2}{2} \), \( \frac{625x^3}{6} \), \( \frac{3125x^4}{24} \).

Step-by-step solution

Understand the differential equation

The differential equation given is a first-order linear ordinary differential equation (ODE) of the form \( y' = 5y \). This equation means that the rate of change of \( y \) is directly proportional to its current value, with a proportionality constant of 5.

Solve the differential equation

To solve \( y' = 5y \), we can use the method of separation of variables or recognize this as an exponential growth problem. Applying separation of variables, we have \( \frac{dy}{y} = 5 dx \). Integrating both sides, we get \( \ln |y| = 5x + C \). Exponentiating both sides gives \( y = e^{5x+C} = Ae^{5x} \), where \( A \) is an arbitrary constant.

Apply the initial condition

Given the initial condition \( y(0) = 5 \), substitute \( x = 0 \) and \( y = 5 \) into the solution \( y = Ae^{5x} \). This gives \( 5 = Ae^{0} = A \). Thus, \( A = 5 \), and the particular solution is \( y = 5e^{5x} \).

Expand the series for the solution

The solution \( y = 5e^{5x} \) can be expressed as a power series. The exponential function \( e^{5x} \) has the series expansion \( e^{5x} = \sum_{n=0}^{\infty} \frac{(5x)^n}{n!} \). Thus, \( 5e^{5x} = \sum_{n=0}^{\infty} 5 \frac{(5x)^n}{n!} \).

Write the first 5 terms of the series

Substitute the first few terms of the expansion: \( y = 5 (1 + 5x + \frac{(5x)^2}{2!} + \frac{(5x)^3}{3!} + \frac{(5x)^4}{4!} + \dots) \). Multiply this by 5 to get \( 5 + 25x + \frac{125x^2}{2} + \frac{625x^3}{6} + \frac{3125x^4}{24} \). Simplify these terms to get the series: \( 5 + 25x + \frac{125x^2}{2} + \frac{625x^3}{6} + \frac{3125x^4}{24} \).

Key concepts

Series Solution

When faced with solving differential equations, sometimes a series solution provides an insightful way to understand the behavior of the system. In our exercise, we approached the differential equation \( y' = 5y \) through its solution by expressing it in a series form. A series solution involves finding the solution to the differential equation as an infinite sum of terms. This method is particularly useful for complex equations where closed-form solutions are difficult to obtain.

In this specific problem, by transforming the exponential solution \( y = 5e^{5x} \) into a power series, we gain several advantages:

• We can approximate the function for small values of \( x \) by considering only a few terms.
• It describes the behavior of the solution in terms of known polynomials.
• It provides a format that's straightforward to evaluate computationally for various applications.

By considering only the first five terms of the power series, we simplify the computation while retaining a good approximation of the original function for small \( x \).

Exponential Growth

Exponential growth is a key feature of the ordinary differential equation given in our problem, \( y' = 5y \). In essence, exponential growth occurs when the growth rate of a mathematical function is proportional to its current value.

When you have a function like \( y = 5e^{5x} \), it grows exponentially because the exponent \( 5x \) causes the value of \( y \) to increase rapidly as \( x \) increases. Understanding exponential growth is essential because it appears in various real-world scenarios like population growth, investment growth, and radioactive decay.

• The constant factor in the exponent (in this case, 5) determines how quickly the growth occurs.
• In problems like this, having the solution as a rapidly growing exponential function can signify a fast-spreading phenomenon or trend in the modeled system.
• Recognizing exponential growth allows us to anticipate and effectively manage scenarios described by these functions.

It's integral for students to grasp this concept deeply, to recognize its implications in both theoretical and applied contexts.

Initial Value Problem

An initial value problem (IVP) in differential equations involves finding a unique solution to the equation by specifying the initial state of the system. In our example, we were given \( y' = 5y \) with the initial condition \( y(0) = 5 \). This information was crucial in determining the constant \( A \) in our general solution \( y = Ae^{5x} \).

IVP helps to pin down the single, unique solution from the infinite possibilities given by the general equation. Here's how the process works:

• The differential equation describes a family of solutions, each differing by a constant factor.
• The initial condition specifies where the solution lies in that family, particularly what the function equals at a certain point.
• By applying the initial condition, we fixed \( A = 5 \), thus simplifying our equation to describe the one function that starts at 5 when \( x = 0 \).

Understanding IVPs is fundamental for solving real-world situations where initial conditions are known, like determining future positions in physics or predicting financial outcomes based on starting capital.

Power Series Expansion

Power series expansion is a method in mathematical analysis where a function is represented as an infinite sum of terms calculated from its derivatives at a single point. This is useful for expressing complex functions in a simpler form, often to approximate values of the function near a specific point. In our task, we expanded the solution \( y = 5e^{5x} \) into a power series to better grasp and communicate its behavior.

Breaking down the power series expansion:

• It represents functions as sums of their polynomial components based on derivatives at a point, commonly zero.
• The function \( e^{5x} \) is expanded using its known series expansion: \( e^{5x} = \sum_{n=0}^{\infty} \frac{(5x)^n}{n!} \).
• By multiplying this infinite series by 5, we obtain the expansion of \( y \) into a comprehensible format usable for approximation.

This approach is powerful due to its applicability in numerical methods and analytical solutions, simplifying complex functions for various types of analysis or computation.