Problem 1 - A First Course in Differential Equations with Modeling Applications

Question

Find the general solution of the given second-order differential equation. $$4 y^{\prime \prime}+y^{\prime}=0$$

Step-by-Step Solution

Answer

The general solution is \( y(t) = C_1 + C_2 e^{-\frac{1}{4}t} \).

Step 1: Identify the Type of Differential Equation

The equation given is a second-order linear homogeneous differential equation of the form \( a y'' + b y' + c y = 0 \) with constant coefficients. Here, we are given \( a = 4 \), \( b = 1 \), and \( c = 0 \).

Step 2: Write the Characteristic Equation

For a second-order linear homogeneous differential equation with constant coefficients, the characteristic equation is obtained by replacing \( y'' \) with \( r^2 \), \( y' \) with \( r \), and \( y \) with \( 1 \). The characteristic equation for \( 4y'' + y' = 0 \) is: \[ 4r^2 + r = 0 \]

Step 3: Solve the Characteristic Equation

Factor the characteristic equation: \[ r(4r + 1) = 0 \]This gives two solutions: \[ r_1 = 0 \] and \[ r_2 = -\frac{1}{4} \]

Step 4: Form the General Solution

Since the characteristic equation gives distinct real roots, the general solution to the differential equation is formed using the roots \( r_1 \) and \( r_2 \). The general solution is: \[ y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \] Substituting the values of \( r_1 \) and \( r_2 \): \[ y(t) = C_1 + C_2 e^{-\frac{1}{4}t} \]

Step 5: Finalize the General Solution

The general solution for the given differential equation \( 4y'' + y' = 0 \) is: \[ y(t) = C_1 + C_2 e^{-\frac{1}{4}t} \]Here, \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial conditions if they are provided.

Key Concepts

Linear Homogeneous Differential Equations

When we talk about linear homogeneous differential equations, we are looking at equations where each term is either a constant or a product of a constant with a derivative of the unknown function. There is no standalone term without the function or its derivatives, hence the term 'homogeneous'. In simple terms, when you set up the equation, everything on the left should sum up to zero.
A linear homogeneous differential equation of the second order typically looks like this:
\[ a y'' + b y' + c y = 0 \]

• Here, \(a\), \(b\), and \(c\) are constants.
• \(y''\) is the second derivative of the function \(y\).
• \(y'\) is the first derivative, and \(y\) is the function itself.

In these types of equations, the focus is on finding solutions for \(y\) that satisfy the equation under given conditions.

Constant Coefficients

In the realm of differential equations, when we speak of constant coefficients, it implies that the multipliers of the derivatives in the equation do not change. They remain constant, making such equations easier to handle mathematically.
This concept is vital because it allows us to convert the differential equation into an algebraic equation, known as the characteristic equation. With the equation:
\[ 4 y'' + y' = 0 \]
The coefficients for \(y''\), \(y'\), and \(y\) are 4, 1, and 0 respectively. The presence of constant coefficients means that traditional methods to solve these equations, like factoring, can be applied straightforwardly.

Characteristic Equation

The characteristic equation bridges the gap between differential and algebraic equations. It's formed by substituting polynomial-like terms for each derivative in the differential equation. This transforms our complex differential problem into a simpler algebraic one.
For our example, with the differential equation \[ 4y'' + y' = 0 \], the characteristic equation is created by replacing \(y''\) with \(r^2\), \(y'\) with \(r\), and \(y\) with 1. This leads to:
\[ 4r^2 + r = 0 \]

• We factor this to solve for \(r\): \(r(4r + 1) = 0\).
• This gives us the roots, \(r_1 = 0\) and \(r_2 = -\frac{1}{4}\).

These roots provide crucial information needed to form the general solution.

General Solution

The general solution to a differential equation provides the broadest expression for all possible solutions, utilizing the roots obtained from the characteristic equation. In essence, the general solution is a formula that incorporates constants that can be adjusted to fit specific initial or boundary conditions.
For our characteristic equation with roots \( r_1 = 0 \) and \( r_2 = -\frac{1}{4} \), the general solution expresses the behavior of \(y(t)\) over time. It can be written as:
\[ y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \]

• By substituting the roots, it becomes: \( y(t) = C_1 + C_2 e^{-\frac{1}{4}t} \).
• Here, \(C_1\) and \(C_2\) are constants determined using additional information, such as initial conditions.

This solution reflects both a constant solution and an exponential decay, covering the complete set of possible solutions for the differential equation.