Problem 61 - A First Course in Differential Equations with Modeling Applications

Question

Find the general solution of \(y^{\prime \prime \prime}+6 y^{\prime \prime}+y^{\prime}-34 y=0\) if it is known that \(y_{1}=e^{-4 x} \cos x\) is one solution.

Step-by-Step Solution

Answer

The general solution is \( y = c_1 e^{-4x} \cos x + c_2 e^{r_2 x} + c_3 e^{r_3 x} \).

Step 1: Verify the Given Solution

We need to first verify that the given function \( y_1 = e^{-4x} \cos x \) satisfies the differential equation \( y''' + 6y'' + y' - 34y = 0 \). We do this by substituting \( y_1 \) and its derivatives into the differential equation to check if both sides equal zero.

Step 2: Formulate the Homogeneous Equation

The differential equation given is already homogeneous: \( y''' + 6y'' + y' - 34y = 0 \). This justifies our search for the general solution in the form \( y = c_1y_1 + c_2 y_2 + c_3 y_3 \), where \( y_1 \), \( y_2 \), and \( y_3 \) are linearly independent solutions.

Step 3: Find the Characteristic Equation

To find additional solutions, we consider the characteristic polynomial corresponding to only the constant coefficient part: \( r^3 + 6r^2 + r - 34 = 0 \). This will give us roots which represent the nature of the solutions.

Step 4: Solve the Characteristic Equation

Factor the polynomial or use methods like synthetic division or the rational root theorem to find the roots of the characteristic equation. Suppose calculation gives roots at \( r_1, r_2, \) and \( r_3 \), allowing us to form solutions in exponential terms.

Step 5: Build the General Solution

Assuming the roots of the characteristic equation provide solutions of the form \( y_2 = e^{r_2 x} \) and \( y_3 = e^{r_3 x} \), combine them with the given solution, resulting in a general solution: \[ y = c_1 e^{-4x} \cos x + c_2 e^{r_2 x} + c_3 e^{r_3 x} \].

Key Concepts

Third-Order Differential Equations

Third-order differential equations involve derivatives of a function up to the third degree. These equations are essential in modeling physical phenomena such as vibrations and fluid dynamics. In general, a third-order differential equation can be expressed as:

• \[ a y''' + b y'' + c y' + d y = 0 \]

where \(a\), \(b\), \(c\), and \(d\) are constants or functions of the independent variable.
The beauty of these equations lies in their ability to describe complex systems with elegant simplicity. Solving them can be challenging, but they yield insightful interpretations of various natural phenomena.
In the provided exercise, the differential equation was:

• \[ y''' + 6y'' + y' - 34y = 0 \]

This specific equation utilizes constant coefficients, which simplifies the problem to finding particular solutions.

Solution Verification

To ensure a proposed solution genuinely satisfies the differential equation, solution verification is a crucial step. This involves substituting the solution and its derivatives back into the original equation.
In our exercise, we start with the given solution:

• \( y_1 = e^{-4x} \cos x \)

Calculate its derivatives up to the third order, and substitute these functions into the differential equation.
If both sides of the equation are balanced (equal zero), the proposed solution is verified to be correct.
This process confirms that the given function not only fits structurally but dynamically aligns with the behavior dictated by the differential equation.

Characteristic Equation

The characteristic equation helps simplify solving differential equations with constant coefficients. It translates the differential equation into a polynomial, whose roots reveal the nature of the solutions.
For a third-order differential equation, this involves replacing derivatives with powers of a variable, typically \( r \). The example transforms the original differential equation as follows:

• \[ r^3 + 6r^2 + r - 34 = 0 \]

Finding the roots \( r_1, r_2, \) and \( r_3 \) involves factoring the polynomial or employing techniques such as synthetic division. These roots might be real, complex, or repeated, dictating the form of the general solution.

General Solution

Once you've verified one solution and determined the roots of the characteristic equation, you can construct the general solution. For third-order differential equations, the general solution is a linear combination of all independent solutions.
In this context, the general solution might look like:

• \[ y = c_1 e^{-4x} \cos x + c_2 e^{r_2 x} + c_3 e^{r_3 x} \]

Here, \( c_1, c_2, \) and \( c_3 \) are arbitrary constants, specific to initial conditions or boundary values.
This expression embodies the infinite set of solutions available, tailored to particular needs by adjusting the constants. Understanding this construction offers immense flexibility in solving practical problems in physics and engineering.