Problem 43 - Precalculus: Functions and Graphs

Question

Solve the system for \(a\) and \(b\). (Hint: Treat terms such as \(e^{3 x}, \cos x,\) and \(\sin x\) as "constant coefficients.") $$\left\\{\begin{aligned} a e^{3 x}+b e^{-3 x} &=0 \\ a\left(3 e^{3 x}\right)+b\left(-3 e^{-3 x}\right) &=e^{3 x} \end{aligned}\right.$$

Step-by-Step Solution

Answer

\(a = \frac{1}{6}\) and \(b = -\frac{1}{6}\).

Step 1: Express the system as a matrix equation

Rearrange the given equations into a form that's easier to manipulate using matrices. You have two equations:\[ ae^{3x} + be^{-3x} = 0 \]\[ 3ae^{3x} - 3be^{-3x} = e^{3x} \]We can write this system as a matrix equation: \[\begin{bmatrix} e^{3x} & e^{-3x} \ 3e^{3x} & -3e^{-3x} \end{bmatrix}\begin{bmatrix} a \ b \end{bmatrix} = \begin{bmatrix} 0 \ e^{3x} \end{bmatrix}\]

Step 2: Simplify the matrix equation

Notice that both equations in the matrix form are already aligned and can be simplified without multiplying. Thus, we express the matrix equation by factoring out the coefficients:\[\begin{bmatrix} a + b \ 3a - 3b \end{bmatrix} = \begin{bmatrix} 0 \ 1 \end{bmatrix}\]

Step 3: Solve the first equation for one variable

From the first equation in the simplified system:\[ a + b = 0 \]Solving for \(a\) gives:\[ a = -b \]

Step 4: Substitute to find the values of both variables

Substitute \( a = -b \) into the second equation:\[ 3(-b) - 3b = 1 \]This simplifies to \(-3b - 3b = 1\) or \(-6b = 1\). Solve for \(b\) to get:\[ b = -\frac{1}{6} \]Use the expression \( a = -b \) to find \(a\):\[ a = \frac{1}{6} \]

Step 5: Verify the solution

Substitute \( a = \frac{1}{6} \) and \( b = -\frac{1}{6} \) back into the original equations to verify:For \( ae^{3x} + be^{-3x} = 0 \):\[ \frac{1}{6}e^{3x} - \frac{1}{6}e^{-3x} = 0 \] is satisfied.For \( 3a e^{3x} - 3b e^{-3x} = e^{3x} \):\[ \frac{1}{2}e^{3x} + \frac{1}{2}e^{-3x} = e^{3x} \] verifies the solution.Thus, the solutions for \(a\) and \(b\) are correct.

Key Concepts

Systems of Equations

A system of equations is a collection of two or more equations with the same set of variables. In our given exercise, the system consists of two equations involving the variables \(a\) and \(b\), both expressed with exponential terms \(e^{3x}\) and \(e^{-3x}\). These terms, while they might seem complex, actually act like coefficients since we treat them as constant factors for the purpose of solving the equation.

To find the solution to the system, we aim to determine a set of values for \(a\) and \(b\) that satisfy both equations simultaneously:
1. \(ae^{3x} + be^{-3x} = 0\)
2. \(3ae^{3x} - 3be^{-3x} = e^{3x}\)

Systems of equations can be solved using various methods, such as substitution, elimination, and matrix manipulation. In this case, we utilize a matrix approach for efficiency and clarity.

Linear Combination

Linear combination is a fundamental concept in algebra, often used in the context of solving systems of equations. It involves expressing a vector or function as a sum of multiples of other vectors or functions. For our specific problem, we can consider each equation as representing a linear combination of the terms involving \(e^{3x}\) and \(e^{-3x}\).

In matrix terms, the original system:
\[\begin{bmatrix} e^{3x} & e^{-3x} \ 3e^{3x} & -3e^{-3x} \end{bmatrix}\begin{bmatrix} a \ b \end{bmatrix} = \begin{bmatrix} 0 \ e^{3x} \end{bmatrix}\]
is an example of linear combinations of these exponential functions. The solution, expressed as \(a = \frac{1}{6}\) and \(b = -\frac{1}{6}\), represents the specific coefficients that balance these combinations to satisfy the equations simultaneously.
The process of finding these coefficients through simplification and substitution showcases the versatility of using linear combinations to find solutions in algebraic contexts.

Exponential Functions

Exponential functions are powerful mathematical expressions that involve exponents. An exponential function has the general form \(f(x) = a \, e^{bx}\), where \(e\) represents the base of natural logarithms, approximately equal to 2.718. These functions are characterized by their rapid growth or decay, depending on the sign of the exponent.

In the context of this problem, terms such as \(e^{3x}\) and \(e^{-3x}\) are used as constant-like coefficients for the purpose of solving the system. These terms allow us to illustrate how exponential growth and decay can interact within mathematical models. Despite their complex appearance, they are handled similarly to constant coefficients when conditioned to solve for other variables, like \(a\) and \(b\).

Understanding exponential functions is crucial as they are often applied in real-world phenomena, such as modeling population growth, radioactive decay, or any process exhibiting multiplicative change over time. In solutions, recognizing how to manipulate these can simplify the process, as demonstrated in writing them as part of matrix equations and systems.