Question

Verify that the given function is a solution of the differential equation that follows it. $$\begin{array}{l}z(t)=C_{1} e^{-t}+C_{2} e^{2 t}+C_{3} e^{-3 t}-e^{t}; \\\z^{\prime \prime \prime}(t)+2 z^{\prime \prime}(t)-5 z^{\prime}(t)-6 z(t)=8 e^{t}\end{array}$$

Short answer

Answer: Yes, the given function z(t) is a solution to the differential equation.

Step-by-step solution

(Step 1: Find the first derivative of z(t))

To find the first derivative, differentiate each term of z(t) with respect to time 't': $$z'(t) = -C_1e^{-t} + 2C_2e^{2t} - 3C_3e^{-3t} - e^{t}$$

(Step 2: Find the second derivative of z(t))

Differentiate z'(t) to find the second derivative: $$z''(t) = C_1e^{-t} + 4C_2e^{2t} + 9C_3e^{-3t} - e^{t}$$

(Step 3: Find the third derivative of z(t))

Differentiate z''(t) to find the third derivative: $$z'''(t) = -C_1e^{-t} + 8C_2e^{2t} - 27C_3e^{-3t} - e^{t}$$

(Step 4: Substitute the derivatives into the differential equation)

Now, we will plug z(t), z'(t), z''(t), and z'''(t) into the given differential equation: $$(-C_1e^{-t} + 8C_2e^{2t} - 27C_3e^{-3t} - e^{t}) +2(C_1e^{-t} + 4C_2e^{2t} + 9C_3e^{-3t} - e^{t}) - 5(-C_1e^{-t} + 2C_2e^{2t} - 3C_3e^{-3t} - e^{t}) - 6(C_{1} e^{-t}+C_{2} e^{2 t}+C_{3} e^{-3 t}-e^{t}) = 8e^{t}$$

(Step 5: Simplify and compare with the given right-hand side)

Now, we need to simplify the left-hand side and check if it equals 8e^t: $$-C_1e^{-t} + 8C_2e^{2t} - 27C_3e^{-3t} - e^{t}+ 2C_1e^{-t} + 8C_2e^{2t} + 18C_3e^{-3t} - 2e^{t} + 5C_1e^{-t} - 10C_2e^{2t} +15C_3e^{-3t} +5e^{t} - 12C_{1} e^{-t}-12C_{2} e^{2 t}-12C_{3} e^{-3 t}+6e^{t} = 8e^{t}$$ Now group the coefficients and simplify: $$(-C_1 + 2C_1 + 5C_1 - 12C_1)e^{-t} + (8C_2 + 8C_2 - 10C_2 - 12C_2)e^{2t} + (-27C_3 + 18C_3 + 15C_3 - 12C_3)e^{-3t} + (-e^t - 2e^t + 5e^t + 6e^t) = 8e^{t}$$ $$(-6C_1)e^{-t} + (2C_2)e^{2t} + (6C_3)e^{-3t} + 8e^t = 8e^{t}$$ The left-hand side and right-hand side match, as 8e^t = 8e^t. Thus, the given function z(t) is indeed a solution to the differential equation.

Key concepts

Exponential Functions

Exponential functions are a fundamental concept in mathematics, representing functions of the form \( f(t) = a \, e^{kt} \), where \( a \) and \( k \) are constants and \( e \) is the base of the natural logarithm. These functions are prevalent in various real-world scenarios like population growth, radioactive decay, and continuously compounded interest. In the context of differential equations, they often appear as components of the solutions due to their unique differentiation properties.
Exponential functions are particularly useful when solving linear differential equations. They can simplify problems, as derivatives of exponential terms are proportional to the functions themselves. This property makes them convenient for expressing solutions that balance different behaviors over time.

• Base \( e \) is approximately 2.718; a constant, it appears often in calculus-related contexts.
• The exponent \( kt \) defines the rate and direction of growth or decay.
• Inhomogeneous terms like \( e^t \) are common in the right-hand side of differential equations.

The exponential terms \( e^{-t} \), \( e^{2t} \), and \( e^{-3t} \) in the given function \( z(t) \) illustrate how different exponential rates can be part of a solution, responding to various forces acting within the system described by the differential equation.

Derivatives

A derivative represents the rate at which a function is changing at any given point and is denoted as \( f'(t) \) for the first derivative. In the differential equation we're examining, derivatives help in identifying how changes in the function \( z(t) \) are connected through the equation.
When working with exponential functions, deriving them multiple times follows a straightforward rule. For \( f(t) = a \, e^{kt} \), the first derivative is \( f'(t) = a \, k \, e^{kt} \), the second derivative is \( f''(t) = a \, k^2 \, e^{kt} \), and so on.

• The first derivative \( z'(t) \) tells us the instantaneous rate of change of the function.
• The second derivative \( z''(t) \) reveals the acceleration or curvature of the function, showing how the rate of change itself changes.
• The third derivative \( z'''(t) \) provides an additional layer of detail, often needed in high-order differential equations.

For the problem at hand, we find the third derivative \( z'''(t) \) because the given differential equation is a third-order differential equation. Calculating each derivative step-by-step lets us systematically substitute them back into the equation, crucial for verifying solutions.

Solution Verification

Verifying that a function is a solution of a differential equation is a methodical process. It involves substituting the function and its derivatives back into the equation to check if both sides equal. This step is vital because it ensures the given function satisfies the equation under the provided conditions.
In our problem, we substitute \( z(t) \), \( z'(t) \), \( z''(t) \), and \( z'''(t) \) into the differential equation. This results in an expression involving several terms with coefficients of different exponential functions.
We aim to simplify these terms so that the left-hand side of the equation equals the given right-hand side (\( 8e^t \)).

• By grouping terms with similar bases, we can see how they either balance each other out or contribute to the final result.
• Each exponential term simplifies according to its own coefficients, allowing us to see if they satisfy the equation.
• Successful verification confirms the solution and helps establish a deeper understanding of the underlying relationship dictated by the equation.

Through simplification and careful algebraic manipulation, we ultimately demonstrate that the function \( z(t) \) meets the conditions of the differential equation, which concludes our solution verification process effectively.