Problem 49 - A First Course in Differential Equations with Modeling Applications

Question

Find a homogeneous linear differential equation with constant coefficients whose general solution is given. $$y=c_{1} e^{x}+c_{2} e^{5 x}$$

Step-by-Step Solution

Answer

The differential equation is \( y'' - 6y' + 5y = 0 \).

Step 1: Understand the Form of the Solution

We are given the general solution of the form \( y = c_1 e^{x} + c_2 e^{5x} \). This suggests that the roots of the characteristic equation of the differential equation are \( r_1 = 1 \) and \( r_2 = 5 \).

Step 2: Construct the Characteristic Equation

The roots \( r_1 = 1 \) and \( r_2 = 5 \) imply that the characteristic equation is \((r - 1)(r - 5) = 0\). Expanding this, we get the characteristic polynomial: \( r^2 - 6r + 5 = 0 \).

Step 3: Form the Homogeneous Differential Equation

The characteristic polynomial corresponds to the differential operator \(D^2 - 6D + 5\) where \(D\) represents the derivative operator. Therefore, the homogeneous differential equation is: \( (D^2 - 6D + 5)y = 0 \), or explicitly, \( y'' - 6y' + 5y = 0 \).

Key Concepts

Homogeneous Equations

A homogeneous linear differential equation is a differential equation where every term is a derivative of the unknown function. In simpler terms, this means all terms either involve the function or its derivatives.
An equation can be called homogeneous when it can be written in the form \( a_n y^{(n)} + a_{n-1} y^{(n-1)} + \ldots + a_1 y' + a_0 y = 0 \) where \(a_0, a_1, \ldots, a_n\) are constants or functions of the independent variable.

What makes homogeneous equations notable is the center of their structure: there is no standalone term without the unknown function or its derivatives, which is why we see = 0 at the end. This aspect lets us use unique methods to solve them, often involving characteristic equations and exponential functions as part of the solution.

Constant Coefficients

Constant coefficients in differential equations refer to when the coefficients of the terms in the equation remain the same throughout. This means that the numbers multiplying the function and its derivatives do not change, always holding a constant value.
For example, in the differential equation \( y'' - 6y' + 5y = 0 \), the coefficients are -6 and 5, which remain the same no matter what value \(y\) or its derivatives take.

Having constant coefficients simplifies solving the equation significantly because it reduces the procedure to finding the roots of a polynomial, rather than dealing with complex functions of the independent variable. Constant coefficients often lead to characteristic equations, which greatly aid in finding the solution.

Characteristic Equation

A characteristic equation is a key concept when solving linear differential equations with constant coefficients.
It's a polynomial equation obtained from a differential equation by replacing the derivatives with powers of an unknown variable, typically \(r\).

For example, starting with terms like \( y'' - 6y' + 5y = 0 \), translating each derivative to \( r \), you get \( r^2 - 6r + 5 = 0 \). Solving this characteristic equation provides the roots, which dictate the form of the solution.

• Each distinct real root corresponds to an exponential term in the general solution.
• Repeated roots would require a multiple of \(x\) in the solution.

Thus, for the equation \( r^2 - 6r + 5 = 0 \), the roots \( r_1 = 1 \) and \( r_2 = 5 \) give us the solution \( y = c_1 e^{x} + c_2 e^{5x} \), highlighting the direct link from characteristic equation to the solution's structure.