Problem 65 - A First Course in Differential Equations with Modeling Applications

Question

$$y^{\prime \prime \prime}-6 y^{\prime \prime}+2 y^{\prime}+y=0$$

Step-by-Step Solution

Answer

The general solution is \( y(t) = C_1 e^{t} + C_2 e^{\left(\frac{5 + \sqrt{29}}{2}\right)t} + C_3 e^{\left(\frac{5 - \sqrt{29}}{2}\right)t} \).

Step 1: Characteristic Equation

The given differential equation is a third-order linear homogeneous differential equation with constant coefficients: \(y''' - 6y'' + 2y' + y = 0\). To solve it, we first find the characteristic equation by assuming a solution of the form \(y = e^{rt}\). Substituting \(y = e^{rt}\), \(y' = re^{rt}\), \(y'' = r^2e^{rt}\), and \(y''' = r^3e^{rt}\) into the differential equation, we obtain:\[ r^3 e^{rt} - 6r^2 e^{rt} + 2r e^{rt} + e^{rt} = 0 \]Factoring out \(e^{rt}\) gives the characteristic equation: \[ r^3 - 6r^2 + 2r + 1 = 0 \].

Step 2: Solve the Characteristic Equation

The characteristic equation \( r^3 - 6r^2 + 2r + 1 = 0 \) needs to be solved for its roots.Let's try to factor the polynomial or find the roots using possible rational roots based on the Rational Root Theorem. After testing the possible roots, it turns out that \( r = 1 \) is a root. Now we perform synthetic division or polynomial long division with \( r = 1 \) to divide \( r^3 - 6r^2 + 2r + 1 \) by \( r - 1 \), resulting in the quotient \( r^2 - 5r - 1 \).Thus, the characteristic polynomial can be written as: \((r - 1)(r^2 - 5r - 1) = 0\).

Step 3: Solve the Quadratic Equation

Next, solve the quadratic equation \( r^2 - 5r - 1 = 0 \).Using the quadratic formula \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \], where \( a = 1, b = -5, c = -1 \), we have:\[ r = \frac{5 \pm \sqrt{25 + 4}}{2} = \frac{5 \pm \sqrt{29}}{2} \].This gives two distinct real roots: \[ r_1 = \frac{5 + \sqrt{29}}{2} \] and \[ r_2 = \frac{5 - \sqrt{29}}{2} \].

Step 4: Write the General Solution

From the roots, we obtain the solution for the differential equation:1. The root \( r = 1 \) provides a solution of the form \( C_1 e^{t} \).2. The roots \( r_1 = \frac{5 + \sqrt{29}}{2} \) and \( r_2 = \frac{5 - \sqrt{29}}{2} \) provide solutions of the forms \( C_2 e^{r_1 t} \) and \( C_3 e^{r_2 t} \), respectively. Therefore, the general solution to the differential equation is:\[ y(t) = C_1 e^{t} + C_2 e^{\left(\frac{5 + \sqrt{29}}{2}\right)t} + C_3 e^{\left(\frac{5 - \sqrt{29}}{2}\right)t} \].

Key Concepts

Characteristic equation

When dealing with third-order differential equations, the characteristic equation plays a crucial role. It's a polynomial equation derived by assuming a solution in an exponential form. Consider a differential equation like the one provided:

• The equation is initially given in the form of: \( y''' - 6y'' + 2y' + y = 0 \).
• To find the characteristic equation, we assume a solution of the form \( y = e^{rt} \).
• By substituting \( e^{rt} \) and its derivatives into the differential equation, we can isolate a polynomial in terms of \( r \).

For the given equation, the characteristic equation is \( r^3 - 6r^2 + 2r + 1 = 0 \). This polynomial is central to determining the nature of solutions of the differential equation. Identifying the characteristic equation correctly is the first step to solving homogeneous equations.

Homogeneous differential equation

A homogeneous differential equation, like the one in our example, is defined by the absence of any 'free' term not involving the dependent variable. What this means is that every term of the equation is multiplied by the function or its derivatives.

• For an equation to be homogeneous, it must fit the form \( L(y) = 0 \), where \( L \) is a differential operator.
• In this problem, each term involves either \( y \), \( y' \), \( y'' \), or \( y''' \), and there's no standalone constant or non-function term present.

This property simplifies the solution process since each term contributes to finding a characteristic equation that is set to zero. In practice, these equations reflect systems maintaining a balance or symmetry, such as electrical circuits or vibrating systems.

Quadratic formula

The quadratic formula is a handy tool for solving second-degree polynomial equations. It's especially useful when those equations arise as part of solving higher-order differential equations. From our example, after reducing the third-order characteristic equation, we ended up with the quadratic \( r^2 - 5r - 1 = 0 \).

Here’s how you apply the quadratic formula:

• The quadratic formula is expressed as: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
• In our example, plug in \( a = 1 \), \( b = -5 \), and \( c = -1 \).
• This yields:\[ r = \frac{5 \pm \sqrt{25 + 4}}{2} \]which simplifies to:\[ r = \frac{5 \pm \sqrt{29}}{2} \]

The quadratic formula helps solve for the roots efficiently, confirming whether they are real, complex, or repeated, which is vital for constructing the general solution.

General solution of differential equations

The general solution of a differential equation encapsulates all possible solutions of the equation. It is formed by combining solutions derived from the roots of the characteristic equation. In the context of the given third-order equation:

• Simple roots of the characteristic equation like \( r = 1 \) result in exponential solutions of the form \( C_1 e^{t} \).
• If the characteristic equation yields distinct real roots such as \( r_1 = \frac{5 + \sqrt{29}}{2} \) and \( r_2 = \frac{5 - \sqrt{29}}{2} \), we get exponential solutions \( C_2 e^{r_1 t} \) and \( C_3 e^{r_2 t} \).

The combination of these solutions gives us the general form:\[ y(t) = C_1 e^{t} + C_2 e^{\left(\frac{5 + \sqrt{29}}{2}\right)t} + C_3 e^{\left(\frac{5 - \sqrt{29}}{2}\right)t} \]The constants \( C_1, C_2, \) and \( C_3 \) are determined by initial conditions. Understanding this structure helps in identifying the form and behavior of solutions for any given system.