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Magazine Chapter 4: Problem 1
\(y^{\prime \prime \prime}+y^{\prime \prime}=e^{t}\)
Short Answer
Expert verified
y(t) = C_1 + C_2 e^{-t} + 1/2 e^t
Step by step solution
01
- Solve the homogeneous equation
Find the characteristic equation of the homogeneous part: \[ y''' + y'' = 0. \] Assume a solution of the form \(y = e^{rt}\), which gives the characteristic equation \[ r^3 + r^2 = 0. \] Solve this by factoring: \[ r^2(r + 1) = 0. \] This gives roots: \[ r = 0, -1 \] So the general solution to the homogeneous equation is \[ y_h(t) = C_1 + C_2 e^{-t} \].
02
- Find a particular solution
For the inhomogeneous equation, use the method of undetermined coefficients. Since the non-homogeneous term is \(e^t\), try a particular solution of the form \[ y_p(t) = Ae^t \]. Calculate its derivatives: \[ y_p' = Ae^t \] and \[ y_p'' = Ae^t \] and \[ y_p''' = Ae^t \]. Insert these into the original differential equation: \[ Ae^t + Ae^t = e^t \]. This simplifies to \[ 2Ae^t = e^t \]. Thus, \[ A = \frac{1}{2} \], giving a particular solution \[ y_p(t) = \frac{1}{2}e^t \].
03
- Form the general solution
Combine the homogeneous and particular solutions: \[ y(t) = y_h(t) + y_p(t) = C_1 + C_2 e^{-t} + \frac{1}{2}e^t \].
04
- Simplify the final solution
The final solution simplifies to: \[ y(t) = C_1 + C_2 e^{-t} + \frac{1}{2}e^t \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Characteristic Equation
In solving a differential equation, the characteristic equation is a crucial step. It helps determine the nature of the solution. To find it, start by assuming a solution of the form \( y = e^{rt} \). This simplifies the differential equation into a polynomial equation in terms of \( r \). For the given problem, we begin with:
\[ y''' + y'' = 0 \]
Assuming \( y = e^{rt} \), we plug this into the equation and get:
\[ r^3 + r^2 = 0 \]
By factoring, we obtain:
\[ r^2(r + 1) = 0 \]
This results in the roots \( r = 0 \) and \( r = -1 \), which are critical for constructing the general solution of the homogeneous equation. Understanding how to derive and factor the characteristic equation is key to solving higher-order differential equations.
\[ y''' + y'' = 0 \]
Assuming \( y = e^{rt} \), we plug this into the equation and get:
\[ r^3 + r^2 = 0 \]
By factoring, we obtain:
\[ r^2(r + 1) = 0 \]
This results in the roots \( r = 0 \) and \( r = -1 \), which are critical for constructing the general solution of the homogeneous equation. Understanding how to derive and factor the characteristic equation is key to solving higher-order differential equations.
Homogeneous Equation
The homogeneous equation is part of a differential equation that excludes non-homogeneous terms (i.e., terms not involving the derivative of the function). In our example, this means solving:
\[ y''' + y'' = 0 \]
The general solution to the homogeneous equation uses the results from the characteristic equation. Here, the roots are \( r = 0 \) and \( r = -1 \). Therefore, the general solution to the homogeneous equation is:
\[ y_h(t) = C_1 + C_2 e^{-t} \]
Where \( C_1 \) and \( C_2 \) are constants determined by initial or boundary conditions. This forms the base of our overall solution when combined with a particular solution.
\[ y''' + y'' = 0 \]
The general solution to the homogeneous equation uses the results from the characteristic equation. Here, the roots are \( r = 0 \) and \( r = -1 \). Therefore, the general solution to the homogeneous equation is:
\[ y_h(t) = C_1 + C_2 e^{-t} \]
Where \( C_1 \) and \( C_2 \) are constants determined by initial or boundary conditions. This forms the base of our overall solution when combined with a particular solution.
Particular Solution
For non-homogeneous differential equations, we need to find a particular solution that satisfies the entire equation. With the non-homogeneous term \( e^t \), we try a particular solution of the form:
\[ y_p(t) = Ae^t \]
By computing its derivatives: \( y_p' = Ae^t \) and \( y_p'' = Ae^t \) and \( y_p''' = Ae^t \). Substituting these into the original equation yields:
\[ Ae^t + Ae^t = e^t \]
Simplifying, we get:
\[ 2Ae^t = e^t \]
Therefore, \( A = \frac{1}{2} \). Our particular solution becomes:
\[ y_p(t) = \frac{1}{2}e^t \]
This particular solution allows us to address the specific non-homogeneous part of the differential equation.
\[ y_p(t) = Ae^t \]
By computing its derivatives: \( y_p' = Ae^t \) and \( y_p'' = Ae^t \) and \( y_p''' = Ae^t \). Substituting these into the original equation yields:
\[ Ae^t + Ae^t = e^t \]
Simplifying, we get:
\[ 2Ae^t = e^t \]
Therefore, \( A = \frac{1}{2} \). Our particular solution becomes:
\[ y_p(t) = \frac{1}{2}e^t \]
This particular solution allows us to address the specific non-homogeneous part of the differential equation.
Method of Undetermined Coefficients
The method of undetermined coefficients is a powerful tool for finding particular solutions to non-homogeneous linear differential equations. It involves guessing the form of the particular solution based on the non-homogeneous term and then finding appropriate coefficients. Here's how it works:
For our exercise, the non-homogeneous term is \( e^t \). We guessed a solution of the form \( y_p(t) = Ae^t \) and found that \( A = \frac{1}{2} \). This method is systematic and works well for a wide range of non-homogeneous terms.
- Identify the type of the non-homogeneous term (e.g., exponential, polynomial, sine/cosine).
- Guess a solution form that resembles the non-homogeneous term.
- Determine the coefficients by substituting the guessed solution into the differential equation.
For our exercise, the non-homogeneous term is \( e^t \). We guessed a solution of the form \( y_p(t) = Ae^t \) and found that \( A = \frac{1}{2} \). This method is systematic and works well for a wide range of non-homogeneous terms.
