Question

verify that the given functions are solutions of the differential equation, and determine their Wronskian. $$ y^{\prime \prime \prime}+2 y^{\prime \prime}-y^{\prime}-2 y=0 ; \quad e^{t}, \quad e^{-t}, \quad e^{-2 t} $$

Short answer

Based on the step-by-step solution, verify that the given functions are solutions of the given differential equation, and find the Wronskian of these functions. The given functions are \(e^t\), \(e^{-t}\), and \(e^{-2t}\), and they are verified to be solutions to the given differential equation \(y^{\prime \prime \prime}+2 y^{\prime \prime}-y^{\prime}-2 y=0\). The Wronskian of the given functions is: $$ W(f,g,h) = -4e^t + 6e^{-t} - 2e^{-2t} $$

Step-by-step solution

Verify that given functions are solutions of the given differential equation

We will check if \(e^t\), \(e^{-t}\) and \(e^{-2t}\) are solutions by substituting them into the differential equation. For \(f(t) = e^t\): First, we find the derivatives of \(f(t)\). $$ f'(t) = e^t, \\ f''(t) = e^t, \\ f'''(t) = e^t $$ Now, substitute these into the differential equation: $$ (e^t) + 2(e^t) - (e^t) - 2(e^t) = 0 $$ We can see that the equation holds, so \(e^t\) is indeed a solution. For \(g(t) = e^{-t}\): We find derivatives of \(g(t)\). $$ g'(t) = -e^{-t}, \\ g''(t) = e^{-t}, \\ g'''(t) = -e^{-t} $$ Substitute these into the differential equation: $$ (-e^{-t}) + 2(e^{-t}) - (-e^{-t}) - 2(e^{-t}) = 0 $$ This equation holds true as well, so \(e^{-t}\) is also a solution. For \(h(t) = e^{-2t}\): We find derivatives of \(h(t)\). $$ h'(t) = -2e^{-2t}, \\ h''(t) = 4e^{-2t}, \\ h'''(t) = -8e^{-2t} $$ Substitute these into the differential equation: $$ (-8e^{-2t}) + 2(4e^{-2t}) - (-2e^{-2t}) - 2(e^{-2t}) = 0 \\ -8e^{-2t}+8e^{-2t}+2e^{-2t}-2e^{-2t} = 0 $$ This equation holds true, so \(e^{-2t}\) is a solution.

Calculate the Wronskian

Now that we have verified that all three functions are solutions to the differential equation, we will determine the Wronskian of these functions. The Wronskian is given by the determinant of the following matrix: $$ W(f,g,h)=\begin{vmatrix} f(t) & g(t) & h(t) \\ f'(t) & g'(t) & h'(t) \\ f''(t) & g''(t) & h''(t) \end{vmatrix} $$ Substitute our functions and their derivatives into the matrix: $$ W(f,g,h)=\begin{vmatrix} e^t & e^{-t} & e^{-2t} \\ e^t & -e^{-t} & -2e^{-2t} \\ e^t & e^{-t} & 4e^{-2t} \end{vmatrix} $$ Calculate the determinant: $$ W(f,g,h) = e^t(-e^{-t}(4e^{-2t}) - (-2e^{-2t})(-e^{-t})) + e^{-t}(e^t(4e^{-2t}) - (e^t)(-2e^{-2t})) \\ - e^{-2t}(e^t(-e^{-t}) - e^t(e^{-t})) \\ = e^t(-4) + e^{-t}(6) - e^{-2t}(2) \\ = -4e^t + 6e^{-t} - 2e^{-2t} $$ Hence, the Wronskian of the given functions is: $$ W(f,g,h) = -4e^t + 6e^{-t} - 2e^{-2t} $$

Key concepts

Differential Equations

Differential equations are equations that involve functions and their derivatives. These mathematical expressions relate a function with one or more of its derivatives and play a crucial role in modeling dynamic systems. The primary purpose of differential equations is to describe how a particular quantity changes over time, which is fundamental in disciplines like physics, engineering, biology, and economics.

There are various types of differential equations, but they all serve the purpose of solving for functions that satisfy the equation. In this exercise, we focus on a linear homogeneous differential equation of order three:

\[y''' + 2y'' - y' - 2y = 0\]

The goal is to verify whether given functions, such as exponential functions, are solutions to this type of differential equation. This means that we need to substitute these functions into the equation and check if they make the equation true. If they do, these functions are considered solutions.

Solution Verification

Solution verification in the context of differential equations involves substituting candidate functions into the equation to see if they satisfy it. The candidate functions we're considering in this exercise are exponential functions of the form \(e^{ct}\).

For each function, you first need to compute its derivatives. This is straightforward for exponential functions, as their derivatives follow a simple pattern. For instance, for \(f(t) = e^t\), we find:

• \(f'(t) = e^t\)
• \(f''(t) = e^t\)
• \(f'''(t) = e^t\)

Substituting these derivatives back into the differential equation verifies that \(f(t) = e^t\) satisfies it.

Repeating this process for \(g(t) = e^{-t}\) and \(h(t) = e^{-2t}\) involves calculating their derivatives
and ensuring each set of substitutions results in zero. As all three functions make the equation hold true,
they are affirmed as solutions.

Exponential Functions

Exponential functions are a crucial component in solving differential equations, particularly linear ones. These functions take the form \(e^{ct}\), where \(e\) is the base of natural logarithms, and \(c\) is a constant. They have properties that make them quite useful in solving both homogeneous and non-homogeneous differential equations.

For instance, exponential functions maintain their form when differentiated, which is a significant advantage. In our exercise, this trait simplifies the solution verification process for our given differential equation:

\[y''' + 2y'' - y' - 2y = 0\]

Exponential functions such as \(e^t\), \(e^{-t}\), and \(e^{-2t}\) were used. This is because their derivatives were straightforward to compute and substitute back into the original equation to verify the solutions. Their ability to capture growth or decay processes in a mathematical form further underlines their importance in various scientific fields.