Problem 13 - A First Course in Differential Equations with Modeling Applications

Question

Find the general solution of the given second-order differential equation. $$3 y^{\prime \prime}+2 y^{\prime}+y=0$$

Step-by-Step Solution

Answer

The general solution is \(y(t) = e^{-t/3}(C_1 \cos(\sqrt{2}t/3) + C_2 \sin(\sqrt{2}t/3))\).

Step 1: Identify the Differential Equation Form

The given differential equation is \(3y'' + 2y' + y = 0\). It is a second-order linear homogeneous differential equation with constant coefficients.

Step 2: Write the Characteristic Equation

To solve the differential equation, we first find the characteristic equation corresponding to the homogeneous equation. For the given equation, the characteristic equation is derived by replacing \(y''\) with \(r^2\), \(y'\) with \(r\), and \(y\) with 1. Thus, the characteristic equation is: \(3r^2 + 2r + 1 = 0\).

Step 3: Solve the Characteristic Equation

We solve the quadratic equation \(3r^2 + 2r + 1 = 0\) using the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=3\), \(b=2\), \(c=1\). Substituting the values gives: \[r = \frac{-2 \pm \sqrt{2^2 - 4\cdot3\cdot1}}{2\cdot3} = \frac{-2 \pm \sqrt{4 - 12}}{6} = \frac{-2 \pm \sqrt{-8}}{6}\].

Step 4: Simplify the Roots

Since the discriminant \(b^2 - 4ac = -8\) is negative, the roots are complex. Simplifying further: \[r = \frac{-2 \pm 2i\sqrt{2}}{6} = \frac{-1 \pm i\sqrt{2}}{3}\], which gives the roots as \(r_1 = \frac{-1}{3} + \frac{i\sqrt{2}}{3}\) and \(r_2 = \frac{-1}{3} - \frac{i\sqrt{2}}{3}\).

Step 5: Write the General Solution

The general solution for a differential equation with complex roots \(r = \alpha \pm \beta i\) is given by \(y(t) = e^{\alpha t}(C_1 \cos(\beta t) + C_2 \sin(\beta t))\). Here, \(\alpha = -\frac{1}{3}\) and \(\beta = \frac{\sqrt{2}}{3}\). Therefore, the general solution is: \[y(t) = e^{\frac{-t}{3}}(C_1 \cos(\frac{\sqrt{2}}{3} t) + C_2 \sin(\frac{\sqrt{2}}{3} t))\].

Key Concepts

Linear Homogeneous Differential Equations

A linear homogeneous differential equation is a type of differential equation where the dependent variable and its derivatives appear linearly. This means each term is a multiple of either the function or its derivatives, but there's no constant term or nonlinear functions like products of derivatives. The general form for a second-order linear homogeneous differential equation with constant coefficients is:

• \( ay'' + by' + cy = 0 \)

where \( a \), \( b \), and \( c \) are constants.
This kind of equation describes systems that return to balance after being disturbed, like electrical circuits or mechanical oscillators without forcing functions.
In the example given, the equation \( 3y'' + 2y' + y = 0 \) is linear and homogeneous, with constant coefficients 3, 2, and 1.

Characteristic Equation

The characteristic equation is a crucial part of solving linear homogeneous differential equations. It is derived by substituting the derivatives in the differential equation with powers of a variable, typically \( r \). This transformation turns the differential equation into an algebraic equation, which is often easier to solve.
In our specific example, the differential equation \( 3y'' + 2y' + y = 0 \) turns into the characteristic equation:

• \( 3r^2 + 2r + 1 = 0 \)

This quadratic equation is solved to find the characteristic roots \( r \), which are used to determine the behavior of the differential equation's solutions.
The process essentially transforms the solution of a differential equation into the solution of a polynomial, which is particularly useful for equations with constant coefficients.

Complex Roots

When solving the characteristic equation, sometimes the roots are not real numbers. This happens when the discriminant \( b^2 - 4ac \) is negative, meaning no real solutions exist, indicating complex roots.
For the quadratic equation \( 3r^2 + 2r + 1 = 0 \), the discriminant can be calculated and was found to be \( -8 \), a negative value. Therefore, the roots are complex and are:

• \( r_1 = \frac{-1}{3} + \frac{i\sqrt{2}}{3} \)
• \( r_2 = \frac{-1}{3} - \frac{i\sqrt{2}}{3} \)

Complex roots, generally expressed as \( \alpha \pm \beta i \), lead to solutions involving sine and cosine functions due to Euler's formula.
These solutions signify oscillations or periodic behaviors, common in physical systems like vibrations.

General Solution

Once the roots of the characteristic equation are known, we can write the general solution of the differential equation. With complex roots \( \alpha \pm \beta i \), the general solution is expressed using exponential, sine, and cosine functions:

• \( y(t) = e^{\alpha t} (C_1 \cos(\beta t) + C_2 \sin(\beta t)) \)

In the worked example, \( \alpha = \frac{-1}{3} \) and \( \beta = \frac{\sqrt{2}}{3} \), which leads to:

• \( y(t) = e^{\frac{-t}{3}} (C_1 \cos(\frac{\sqrt{2}}{3} t) + C_2 \sin(\frac{\sqrt{2}}{3} t)) \)

This general solution shows how the function behaves over time, describing a damped oscillation due to the negative exponential term, \( \alpha \), reducing the amplitude as \( t \) increases.
Constants \( C_1 \) and \( C_2 \) represent the contribution of the cosine and sine components, determined by initial conditions or system specifics.