Chapter 8: Problem 7
\(y^{\prime \prime \prime}+y^{\prime}=e^{t}, y(0)=y^{\prime}(0)=y^{\prime \prime}(0)=0\)
Short Answer
Expert verified
The solution is \( y(t) = -1 + \frac{1}{2} \text{cos}(t) - \frac{1}{2} \text{sin}(t) + \frac{1}{2} e^t \).
Step by step solution
01
Identify the Type of Differential Equation
Recognize that this is a third-order linear non-homogeneous differential equation with constant coefficients, of the form: \[ y''' + y' = e^t. \]
02
Find the Complementary Solution
To find the complementary solution, solve the associated homogeneous differential equation: \[ y''' + y' = 0. \]Assume a solution of the form \( y = e^{rt} \) and substitute into the homogeneous equation to get the characteristic equation: \[ r^3 + r = 0. \]Factoring out \( r \), we get: \[ r(r^2 + 1) = 0. \]This gives roots: \( r = 0, i, -i \). So the complementary solution is: \[ y_c(t) = C_1 + C_2 \text{cos}(t) + C_3 \text{sin}(t). \]
03
Find the Particular Solution
Now, find the particular solution, \( y_p(t) \), by using an appropriate form. Since the non-homogeneous term is \( e^t \), try: \[ y_p(t) = Ae^t. \]Take the derivatives: \[ y_p'(t) = Ae^t, \]\[ y_p'''(t) = Ae^t. \]Substitute into the non-homogeneous equation: \[ Ae^t + Ae^t = e^t, \]Simplify to get: \[ 2Ae^t = e^t. \]Thus, \( A = \frac{1}{2} \). Therefore, the particular solution is: \[ y_p(t) = \frac{1}{2} e^t. \]
04
Form the General Solution
Combine the complementary solution and the particular solution to get the general solution of the differential equation: \[ y(t) = y_c(t) + y_p(t) = C_1 + C_2 \text{cos}(t) + C_3 \text{sin}(t) + \frac{1}{2} e^t. \]
05
Apply Initial Conditions
Use the initial conditions \( y(0) = y'(0) = y''(0) = 0 \) to determine the constants. Start with \( y(0) = 0 \): \[ 0 = C_1 + C_2 \text{cos}(0) + C_3 \text{sin}(0) + \frac{1}{2} e^0, \]\[ 0 = C_1 + C_2 + \frac{1}{2}. \]So, \( C_1 + C_2 = -\frac{1}{2} \). Next, \( y'(t) = -C_2 \text{sin}(t) + C_3 \text{cos}(t) + \frac{1}{2} e^t \). Apply \( y'(0) = 0 \):\[ 0 = 0 + C_3 + \frac{1}{2}, \]\[ C_3 = -\frac{1}{2}. \]Lastly, \( y''(t) = -C_2 \text{cos}(t) - C_3 \text{sin}(t) + \frac{1}{2} e^t \). Apply \( y''(0) = 0 \):\[ 0 = -C_2 + 0 + \frac{1}{2}, \]\[ C_2 = \frac{1}{2}. \]Thus, \( C_1 = -1 \).
06
Write the Final Solution
Substitute the values of the constants back into the general solution: \[ y(t) = -1 + \frac{1}{2} \text{cos}(t) - \frac{1}{2} \text{sin}(t) + \frac{1}{2} e^t. \]
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Non-Homogeneous Differential Equations
A non-homogeneous differential equation includes a term that is not solely dependent on the function and its derivatives. In our problem, the differential equation is given by: \[ y''' + y' = e^t. \]Notice the presence of \(e^t\) which makes this equation non-homogeneous. Here's why understanding this is important:
The key to solving such equations is tackling two parts: the complementary solution (treating it as homogeneous) and the particular solution (factoring in the non-homogeneous term).
- Handles real-world scenarios: Many physical phenomena can't be explained by homogeneous equations because there's always some external influence or input, represented by the non-homogeneous term.
- Diverse solutions: Includes both a complementary solution and a particular solution to cover various influences.
The key to solving such equations is tackling two parts: the complementary solution (treating it as homogeneous) and the particular solution (factoring in the non-homogeneous term).
Particular Solution
The particular solution, \(y_p(t)\), is a specific solution that satisfies the entire non-homogeneous differential equation. For our equation, \( y''' + y' = e^t \), the non-homogeneous term is \( e^t \). To find the particular solution, guess a form that mimics the non-homogeneous term.
Since the non-homogeneous term is \( e^t \), we try a solution form: \[ y_p(t) = Ae^t. \]Taking derivatives, we get:\[ y_p'(t) = Ae^t, \]\[ y_p'''(t) = Ae^t. \]Substitute these into the differential equation:\[ Ae^t + Ae^t = e^t, \]Simplifying, we find:\[ 2Ae^t = e^t, \]which yields:\[ A = \frac{1}{2}. \]Thus, the particular solution is:\[ y_p(t) = \frac{1}{2} e^t. \]This particular solution adds to our understanding of how the non-homogeneous term \(e^t\) influences the system.
Since the non-homogeneous term is \( e^t \), we try a solution form: \[ y_p(t) = Ae^t. \]Taking derivatives, we get:\[ y_p'(t) = Ae^t, \]\[ y_p'''(t) = Ae^t. \]Substitute these into the differential equation:\[ Ae^t + Ae^t = e^t, \]Simplifying, we find:\[ 2Ae^t = e^t, \]which yields:\[ A = \frac{1}{2}. \]Thus, the particular solution is:\[ y_p(t) = \frac{1}{2} e^t. \]This particular solution adds to our understanding of how the non-homogeneous term \(e^t\) influences the system.
Complementary Solution
The complementary solution, \(y_c(t)\), solves the associated homogeneous equation: \[ y''' + y' = 0. \]Assuming a solution of the form \( y = e^{rt} \) transforms the equation into a characteristic equation:\[ r^3 + r = 0. \]By factoring out \( r \), we get:\[ r(r^2 + 1) = 0. \]This gives the roots: \( r = 0, \, i, \, -i \).
Thus, the complementary solution is:\[ y_c(t) = C_1 + C_2 \, \text{cos}(t) + C_3 \, \text{sin}(t). \]Here's why the complementary solution is vital:
Thus, the complementary solution is:\[ y_c(t) = C_1 + C_2 \, \text{cos}(t) + C_3 \, \text{sin}(t). \]Here's why the complementary solution is vital:
- Forms base: It provides the fundamental solutions under no external influence.
- Completeness: Combined with the particular solution, it covers all scenarios.
Initial Conditions
Initial conditions define the specific behavior of the differential equation's solution at the beginning. In our equation, we have:\[ y(0) = y'(0) = y''(0) = 0. \]To incorporate these, substitute the initial conditions into the general solution. Here's the general solution formed by adding both the complementary and particular solutions:\[ y(t) = C_1 + C_2 \, \text{cos}(t) + C_3 \, \text{sin}(t) + \frac{1}{2} e^t. \]Using the initial conditions:\[ y(0) = 0 \Rightarrow C_1 + C_2 + \frac{1}{2} = 0 \Rightarrow C_1 + C_2 = -\frac{1}{2}, \]\[ y'(0) = 0 \Rightarrow C_3 + \frac{1}{2} = 0 \Rightarrow C_3 = -\frac{1}{2}, \]\[ y''(0) = 0 \Rightarrow -C_2 + \frac{1}{2} = 0 \Rightarrow C_2 = \frac{1}{2}. \]Solving these, we get:\[ C_1 = -1. \]So, our final solution respecting initial conditions is:\[ y(t) = -1 + \frac{1}{2} \, \text{cos}(t) - \frac{1}{2} \, \text{sin}(t) + \frac{1}{2} e^t. \]Initial conditions are critical to tailoring the general solution to the specifics you'd encounter in real-world situations.
Constant Coefficients
In our differential equation, the coefficients of y and its derivatives are constants, as seen in \[ y''' + y' = e^t. \]Having constant coefficients simplifies the process:
- Predictable solutions: Making it easier to apply standard methods like characteristic equations.
- Ease of factoring: Facilitates the derivation of characteristic equations.