Question

In each of Problems 1 through 10 find the general solution of the given differential equation. \(9 y^{\prime \prime}+6 y^{\prime}+y=0\)

Short answer

Question: Find the general solution of the given second-order linear homogeneous differential equation: \(9y^{\prime\prime} + 6y^{\prime} + y = 0\). Solution: The general solution is \(y(x) = C_1 e^{-\frac{1}{3}x} + C_2 xe^{-\frac{1}{3}x}\), where \(C_1\) and \(C_2\) are arbitrary constants.

Step-by-step solution

Identify the given equation

The given equation is a second-order linear homogeneous differential equation: \(9y^{\prime\prime} + 6y^{\prime} + y = 0\).

Formulate the characteristic Equation

We derive a characteristic equation from the given differential equation. To do this, we replace \(y^{\prime\prime}\) with \(m^2\), \(y^{\prime}\) with \(m\), and \(y\) with \(1\). This results in the following quadratic equation: \(9m^2 + 6m + 1 = 0\).

Solve the characteristic equation

Now, we need to solve the quadratic equation \(9m^2 + 6m + 1 = 0\). This equation can be factored into \((3m + 1)^2 = 0\), so we have \((3m + 1)(3m + 1) = 0\). Thus, the double root is \(m = -\dfrac{1}{3}\).

Formulate the general solution

Since we have a double root \(m = -\dfrac{1}{3}\), the general solution of the given differential equation is of the form \(y(x) = C_1 e^{-\frac{1}{3}x} + C_2 xe^{-\frac{1}{3}x}\), where \(C_1\) and \(C_2\) are arbitrary constants.

Key concepts

General Solution

In mathematics, when we talk about the general solution of a differential equation, we refer to a formula that expresses all possible solutions of that equation. A general solution generally consists of arbitrary constants that can be adjusted to fit initial conditions or specific requirements of a problem. For the equation given in the ORIGINAL EXERCISE, the general solution is represented as:

• \( y(x) = C_1 e^{-rac{1}{3}x} + C_2 xe^{-rac{1}{3}x} \)

Here, \(e^{-rac{1}{3}x}\) is an exponential function reflecting the nature of the solution, and \(C_1\) and \(C_2\) are constants which are determined by the initial or boundary conditions.
The presence of both \(e^{-rac{1}{3}x}\) and \(xe^{-rac{1}{3}x}\) arises due to a special feature called a double root. These components mix in both simple exponential and modified exponential forms to account for the behavior of solutions to this type of differential equation.

Second-Order Linear Homogeneous Equation

A second-order linear homogeneous differential equation is a specific type of differential equation. It has the generical form:

• \( a y'' + b y' + c y = 0 \)

Where \(a\), \(b\), and \(c\) are constants, and the coefficients of all derivative terms are linear (no variables other than the derivative itself). Importantly, it is homogeneous because the right side of the equation is zero.
The example from the ORIGINAL EXERCISE fits into this category, defined as \(9y'' + 6y' + y = 0\). This characteristic zero on the right side indicates no external force or function applied to the system, representing a natural behavior or response of the system.

Characteristic Equation

The characteristic equation is a crucial tool when solving linear differential equations. To form a characteristic equation from a differential equation, we employ a process called 'characteristic substitution.' This involves replacing:

• \(y''\) with \(m^2\)
• \(y'\) with \(m\)
• \(y\) with 1

These substitutions turn the differential equation into a quadratic algebraic equation. For the given differential equation \(9y'' + 6y' + y = 0\), the characteristic equation becomes:

• \(9m^2 + 6m + 1 = 0\)

The solutions to the characteristic equation determine the roots \(m\), which then guide us to the general solution for the differential equation. Solving this quadratic will indicate whether there are distinct roots, a double root, or complex roots.

Double Root Solution

A double root occurs when the characteristic equation yields the same root twice. For the equation \(9m^2 + 6m + 1 = 0\), factoring reveals that it simplifies to

• \((3m + 1)^2 = 0\)

This results in a double root, \(m = -\frac{1}{3}\). In such cases, the general solution of the differential equation must take into account both the double occurrence of the root and the resulting form of the solution.
The special feature of a double root introduces an additional component, leading to solutions of the form:

• \(y(x) = C_1 e^{mx} + C_2 x e^{mx}\)

This modified form ensures that it captures all potential solutions of our second-order linear homogeneous differential equation. In conclusion, when confronted with double roots, adding an extra \(x\) factor to one component provides the necessary adjustment to encompass the full solution space. This results in the more enriched and broadened form observed in the general solution explained.