Question

Give the first 5 terms of the series that is a solution to the given differential equation. $$y^{\prime}=y^{2}, \quad y(0)=1$$

Short answer

First 5 terms: -1, -x, -x^2, -x^3, -x^4.

Step-by-step solution

Identify the type of differential equation

We are given a first-order differential equation: \( y' = y^2 \). This is a separable differential equation, where the right-hand side can be expressed solely in terms of \( y \).

Write the equation in a separable form

To solve a separable differential equation, write it as \( \frac{dy}{y^2} = dx \). This allows us to separate the variables, with all \( y \)-terms on one side and all \( x \)-terms on the other side.

Integrate both sides

Integrate both sides of the equation: \( \int \frac{dy}{y^2} = \int dx \). The left-hand side becomes \( -\frac{1}{y} \), while the right-hand side becomes \( x + C \), where \( C \) is the constant of integration.

Solve for y

Re-arrange \( -\frac{1}{y} = x + C \) to solve for \( y \): \( y = -\frac{1}{x + C} \).

Apply the initial condition

Using the initial condition \( y(0) = 1 \), substitute into the equation: \( 1 = -\frac{1}{0 + C} \) gives \( C = -1 \). Thus, \( y = -\frac{1}{x - 1} \).

Find the series expansion

Now, express \( y(x) = -\frac{1}{x - 1} \) as a power series centered at \( x = 0 \). Recognize that this is a geometric series \( -\sum_{n=0}^{\infty} x^n \).

Write the first 5 terms of the series

The series expansion is \( y(x) = -1 - x - x^2 - x^3 - x^4 - \ldots \). The first 5 terms are: \(-1 - x - x^2 - x^3 - x^4\).

Key concepts

Separable Differential Equation

A separable differential equation is a type of ordinary differential equation that can be expressed in such a way that all the terms involving one variable are on one side of the equation and all the terms involving the other variable are on the other side. In other words, it can be written in the form \( \frac{dy}{g(y)} = f(x) \, dx \). This makes these equations easier to solve, as we can integrate both sides separately.
To solve a separable differential equation, begin by rearranging the equation to separate the variables. In our example, we started with \( y' = y^2 \) and rewrote it as \( \frac{dy}{y^2} = dx \). Once separated, both sides of the equation can be integrated independently. This approach streamlines finding the solution by leveraging the properties of integration.
Separable differential equations are commonly found in mathematical models across biology, chemistry, and physics, where factors can be distinctly classified and isolated.

Initial Conditions

Initial conditions are specific values that allow us to find the particular solution to a differential equation rather than just a general one. Typically, this involves fixing the constants of integration that arise when solving differential equations.
In our problem, the initial condition provided is \( y(0) = 1 \). This tells us the value of the solution at \( x = 0 \), which is critical for determining the exact form of the solution. After solving the differential equation, we found \( y = -\frac{1}{x + C} \). By plugging in the initial condition, \( y(0) = 1 \), we determined that \( C = -1 \). This gave us the specific solution \( y = -\frac{1}{x - 1} \).
Initial conditions are crucial in initial value problems and help in modeling real-world scenarios where specific starting conditions are known or can be imposed.

Series Expansion

Series expansion is a method of expressing a function as an infinite sum of terms, often powers of a variable or its derivatives. It is particularly useful for approximating functions and for solving differential equations where a closed form is difficult or impossible to find.
In the context of our exercise, the function \( y = -\frac{1}{x - 1} \) is expressed as a power series centered at \( x = 0 \). Recognizing it as a geometric series allowed us to expand it:

• Start with the expression: \( y(x) = -\frac{1}{x - 1} \)
• Recognize it matches the form of the geometric series \( \frac{1}{1-z} = 1 + z + z^2 + z^3 + \ldots \) for \( |z| < 1 \).
• Thus, we expressed it as a series \(-\sum_{n=0}^{\infty} x^n \).
• Consequently, the first 5 terms are: \(-1 - x - x^2 - x^3 - x^4\).

Series expansion is not only useful for analysis within calculus but also plays a critical role in physics and engineering, where approximations simplify complex problems.

Power Series

A power series is a type of series where each term involves a power of a variable, usually represented as \( \sum_{n=0}^{\infty} a_n (x - c)^n \). It converges within a certain radius around a point \( c \), known as the center, and behaves like a polynomial of infinite degree within this interval.
The power series is an important tool in calculus and is used to express functions near a point. It helps in approximations and in solving equations where finding an exact solution is cumbersome.
In our exercise, the function \( y(x) = -\frac{1}{x - 1} \) was expanded into a power series around \( x = 0 \). Notably, this expansion was possible because it mimics a geometric series. We wrote the expansion as:

• \( y(x) = -1 - x - x^2 - x^3 - x^4 - \ldots \)

This format is particularly user-friendly when calculating function values at points near zero, enhancing computational simplicity and effectiveness. Power series are foundational in fields such as analysis, where they provide insight into the behavior of functions.