Problem 21 - Calculus

Question

Find all solutions of the form \(y=e^{t}\). $$ y^{\prime \prime \prime}-y^{\prime}=0 $$

Step-by-Step Solution

Answer

The solutions are \( y = C_1 + C_2 e^{t} + C_3 e^{-t} \).

Step 1: Substitute the assumed form of y

Since we are given that the solution is of the form \( y = e^{t} \), we assume \( y = e^{kt} \). We need to find values of \( k \) that satisfy the differential equation \( y^{\prime \prime \prime}-y^{\prime}=0 \).

Step 2: Calculate derivatives of y

First, calculate the first derivative: \( y' = ke^{kt} \).Next, calculate the second derivative: \( y'' = k^2e^{kt} \).Finally, calculate the third derivative: \( y''' = k^3e^{kt} \).

Step 3: Substitute derivatives into the differential equation

Substitute \( y' \) and \( y''' \) into the equation \( y^{\prime \prime \prime}-y^{\prime}=0 \):\[ k^3e^{kt} - ke^{kt} = 0 \].

Step 4: Factor the expression

Factor out \( e^{kt} \) from the equation:\[ e^{kt}(k^3 - k) = 0 \].Since \( e^{kt} eq 0 \) for any real \( k \), we set the remaining expression equal to zero: \( k^3 - k = 0 \).

Step 5: Solve for k

Factor the equation \( k^3 - k = 0 \) to \( k(k^2 - 1) = 0 \).This gives factors: \( k = 0 \) and \( k^2 - 1 = 0 \), which further factors to \( k = \pm 1 \).

Step 6: Write the general solution

Using the values of \( k \), the solutions for \( y \) are: \( y = e^{0 \cdot t}, y = e^{t}, y = e^{-t} \). Thus, the general solution is:\[ y = C_1e^{0} + C_2e^{t} + C_3e^{-t} \].This simplifies to:\[ y = C_1 + C_2e^{t} + C_3e^{-t} \], where \( C_1, C_2, \) and \( C_3 \) are constants.

Key Concepts

Exponential Solutions

Exponential solutions play an important role in solving differential equations, especially when the spectrum of possible solutions is sought. In this context, an exponential solution means that the function being solved for, often denoted as \(y(t)\), can be expressed as \(y = e^{kt}\). Here, \(k\) is a constant that we need to determine in order to satisfy the original differential equation.
When solving differential equations, assuming exponential solutions can greatly simplify the process. This is because the differentiation of exponential functions follows a predictable pattern:

• First derivative of \(e^{kt}\) becomes \(ke^{kt}\).
• Second derivative becomes \(k^2e^{kt}\).
• Third derivative becomes \(k^3e^{kt}\).

These predictable transformations allow one to substitute these derivatives into the differential equation. This substitution transforms the problem into a simpler algebraic one, which is easier to solve for \(k\). Once \(k\) is found, the corresponding exponential functions form the basis of the solution.

Third Order Differential Equation

A third order differential equation involves derivatives up to the third order. In this case, the given differential equation is \( y''' - y' = 0 \).
Such equations describe systems with more complex dynamics compared to first or second order equations. They can appear in contexts like fluid dynamics, mechanical systems, and electrical circuits.
To solve third order differential equations, one approach is to assume a particular form for the solution, like exponential solutions. By undergoing the process of:

• Finding derivatives up to the third order.
• Substituting these into the differential equation.
• Solving the resulting algebraic equation for constants.

Students can determine specific exponential functions that satisfy the equation.Solving these equations piece-by-piece reveals each possible behavior of the system described by the differential equation, which is represented by different constants \(k\) found during the process.

General Solution of Differential Equations

The general solution of a differential equation is a formula that includes all possible solutions. For the third order differential equation \( y''' - y' = 0 \), the general solution is expressed as a linear combination of solutions derived from different values of \(k\).
The previous steps led us to find particular solutions associated with specific values of \(k\), which are \( k = 0, 1, -1 \). Accordingly, the exponential solutions are \(e^{0}, e^{t},\) and \(e^{-t}\). The general solution, then, is:

• \( y = C_1 + C_2e^{t} + C_3e^{-t} \)

where \(C_1, C_2, \, \text{and} \, C_3\) are arbitrary constants that can be determined if additional initial conditions are provided.
The expression of the general solution reflects the linearity of differential equations, allowing these arbitrary constants to scale and combine the exponential solutions into infinite forms of \(y(t)\). This robust method ensures plotting any behavior the original equation can describe by adjusting \(C_1, C_2, \, \text{and} \, C_3\).