Question

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

Short answer

Answer: The general solution is \(y(x) = C_1 e^{3x} + C_2 xe^{3x}\), where \(C_1\) and \(C_2\) are constants.

Step-by-step solution

Write the characteristic equation

In order to find the general solution of the given homogeneous linear equation, we need to write the characteristic equation first. The characteristic equation is given by: \(r^2 - 6r + 9 = 0\)

Solve the characteristic equation

Now we need to solve the quadratic equation for r: \(r^2-6r+9=(r-3)(r-3)=0\) This quadratic has a double root at \(r=3\) (since the two distinct roots are equal).

Write the general solution

Since we have a double root at \(r=3\), the general solution for the differential equation is given by the following form: \(y(x)=C_1 e^{3x} + C_2 xe^{3x}\) where \(C_1\) and \(C_2\) are constants to be determined by the initial conditions if provided. Final answer: \(\boxed{y(x) = C_1 e^{3x} + C_2 xe^{3x}}\)

Key concepts

Characteristic Equation

The characteristic equation is a mathematical tool used to solve linear differential equations, especially when dealing with second-order equations. It allows us to transform a differential equation into an algebraic form that is often easier to solve. In this exercise, the equation was given as

• \( y'' - 6y' + 9y = 0 \)

The corresponding characteristic equation is obtained by replacing the derivatives with powers of a variable, usually "r," such as:

• \( r^2 - 6r + 9 = 0 \)

This characteristic equation is a quadratic equation, whose roots will help us to find the general solution of the differential equation.

General Solution

The general solution of a differential equation includes all possible solutions. It depends on the nature of the characteristic roots. For our quadratic characteristic equation, the roots allow us to determine the form of the general solution. When solving

• \( y'' - 6y' + 9y = 0 \)

we found that the roots are not distinct, meaning we have a double root \(r = 3\). The general solution for an equation with a double root takes the special form:

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

where \(C_1\) and \(C_2\) are arbitrary constants that can be adjusted based on initial conditions provided in a particular problem.

Homogeneous Linear Equation

A homogeneous linear equation is a type of differential equation where the sum of the terms equals zero. In its simplest form, a second-order homogeneous linear differential equation looks like:

• \( a y'' + b y' + c y = 0 \)

In this case, our equation is

• \( y'' - 6y' + 9y = 0 \)

Homogeneous equations imply that the solutions are derived purely from the relationships between the terms themselves with no external forces acting on the system. Solving such equations involves finding a general solution that satisfies all possible conditions of the equation.

Double Root

The double root scenario arises when a quadratic equation, like our characteristic equation \(r^2 - 6r + 9 = 0\), results in one repeated solution, or root, rather than two distinct ones. Through factoring, the characteristic equation simplifies to:

• \((r-3)(r-3) = 0\)

This indicates a double root at \( r = 3 \). The presence of a double root affects the form of the general solution and is crucial for the structure:

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

The term \( xe^{3x} \) accounts for the repeat of the single root and ensures that the equations describe a full set of solutions even when distinct roots are not available.