Such equations are called homogeneous because of the zero on the right-hand side. In this chapter, we explore how to solve the more general case of nonhomogeneous equations, which include a non-zero function on the right-hand side, like so
For ease of discussion, will use the shorthand LNCC equations to refer to Linear Nonhomogeneous Constant Coefficient equations.
Recall, the solution of a homogeneous equation is made up of \(e^{r x}\) terms which have the unique property of being like terms with its derivatives. This is needed so the terms on the left-hand side can cancel out to zero.
Figure12.Comparison of a homogeneous equation (left) and a nonhomogeneous equation (right). In both, the solutions must simplify in a specific way when substituted into the equation.
In contrast, nonhomogeneous equations must have solutions that make the left-hand side simplify to \(f(x)\text{,}\) rather than zero. For this to happen, the solution should be “\(f(x)-\)like” in the sense that it shares terms with \(f(x)\text{.}\) For example, consider the equation
When the solution, \(y\text{,}\) is plugged into the equation, the left-side terms must simplify to \(9x\text{.}\) So, \(y\) must be “\(9x\) like”, but what is does “\(9x\) like” mean? As you will see in the sections that follow, it means \(y\) has the form \(y = Ax+B\text{.}\) For now, let’s just verify that \(y = 3x + 4\) is the “\(9x\) like” solution that corresponds to this equation.
1.Which of the following statements best describes the difference between a homogeneous and a nonhomogeneous LNCC equation?
Which of the following statements best describes the difference between a homogeneous and a nonhomogeneous LNCC equation?
The homogeneous equation has a zero constant term, while the nonhomogeneous equation has a non-zero constant term.
Correct! Homogeneous equations have a zero constant term, while nonhomogeneous equations include a non-zero function like \(f(x)\) on the right-hand side.
In a nonhomogeneous equation, all terms contain a dependent variable, but in a homogeneous equation, only one term can contain a dependent variable.
Incorrect. This is not an accurate description of either equation type.
A nonhomogeneous equation has a higher order than a homogeneous equation.
Incorrect. The order of the equation does not determine whether it is homogeneous or nonhomogeneous.
Homogeneous equations are always linear, while nonhomogeneous equations are nonlinear.
Incorrect. Both homogeneous and nonhomogeneous equations can be linear.
2.Which equation is nonhomogeneous?
Select the equation that is nonhomogeneous.
\(\ds y'' - 3y' + 2y = 0\)
Incorrect. This is a homogeneous equation because the right-hand side is zero.
\(\ds y'' - 4y' + 3y = 9x + 6\)
Correct! This equation is nonhomogeneous because it has a non-zero term, \(9x + 6\text{,}\) on the right-hand side.
\(\ds y' + 2y = 0\)
Incorrect. This is a homogeneous equation because the right-hand side is zero.
\(\ds y'' + y = 0\)
Incorrect. This is a homogeneous equation because the right-hand side is zero.
Incorrect, the particular solution is \(5x\text{.}\)
\(5x\)
Correct!
\(-5x^2\)
Incorrect, the particular solution is \(5x\text{.}\)
\(5\)
Incorrect, the particular solution is \(5x\text{.}\)
4.A solution of the equation...
A solution to the equation
\begin{equation*}
y''' + y'' + y' + y = x^3 + x
\end{equation*}
should contain a polynomial of what degree?
\(0\)
Incorrect, review row 1 of the table above.
\(1\)
Incorrect, plugging \(y=Ax+B\) (degree 1 polynomial) into the LHS would simplify to another degree 1 polynomial, but the RHS is a degree 3 polynomial.
\(2\)
Incorrect, plugging \(y=Ax^2+Bx+C\) (degree 2 polynomial) into the LHS would simplify to another degree 2 polynomial, but the RHS is a degree 3 polynomial.
\(3\)
Correct! Plugging \(y=Ax^3+Bx^2+Cx+D\) (degree 3 polynomial) into the LHS would simplify to a degree 3 polynomial, which is the degree of the polynomial on the RHS.
\(4\)
Incorrect, plugging \(y=Ax^4+Bx^3+Cx^2+Dx+E\) (degree 4 polynomial) into the LHS would leave you with a \(x^4\) term, not seen on the RHS.
