Chapter 9: Problem 69
solve the initial value problem and graph the solution. $$ y^{\prime \prime \prime}-2 y^{\prime \prime}-5 y^{\prime}+6 y=2 e^{x}(1-6 x), \quad y(0)=2, \quad y^{\prime}(0)=7, \quad y^{\prime \prime}(0)=9 $$
Short Answer
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Question: Solve the given initial value problem: \(y^{\prime \prime \prime}-2 y^{\prime \prime}-5 y^{\prime}+6 y=2 e^{x}(1-6x)\), \(y(0)=2\), \(y^{\prime}(0)=7\), and \(y^{\prime \prime}(0)=9\).
Answer: The solution to the initial value problem is \(y(x) = 34 e^{x} - 38 e^{2x} + 6 e^{3x} + xe^{x}(2x+11)\).
Step by step solution
01
Solve the homogeneous equation to find the complementary function
First, we need to find the complementary function by solving the homogeneous equation:
$$
y^{\prime \prime \prime}-2 y^{\prime \prime}-5 y^{\prime}+6 y=0
$$
To solve this homogeneous equation, we assume the solution is in the form of \(y=e^{rx}\). Substituting into the homogeneous equation gives us a characteristic equation:
$$
r^3-2r^2-5r+6=0
$$
By factoring or using a root finder, we can find the roots of the characteristic equation, which are \(r = 1\), \(r = 2\), and \(r = 3\). Therefore, the complementary function is given by
$$
y_c(x) = C_1 e^{x} + C_2 e^{2x} + C_3 e^{3x}
$$
02
Find the particular solution of the non-homogeneous equation
We then need to find a particular solution of the given non-homogeneous equation:
$$
y^{\prime \prime \prime}-2 y^{\prime \prime}-5 y^{\prime}+6 y=2 e^{x}(1-6x)
$$
As the right-hand side of the equation has the form \(2e^{x}(1-6x)\), we assume the particular solution has the form of \(y_p(x) = xe^{x}(Ax+B)\). Differentiating, we obtain:
$$
y_p'(x) = e^{x}(B - 5Ax) \\
y_p''(x) = e^{x}(A - 4B) \\
y_p'''(x) = -3Ae^{x}
$$
Now substitute \(y_p(x)\), \(y_p'(x)\), \(y_p''(x)\), and \(y_p'''(x)\) into the non-homogeneous equation to find the values of A and B:
$$
(-3Ae^{x}) - 2(e^x(A-4B)) - 5(e^x(B-5Ax)) +6(xe^{x}(Ax+B)) = 2e^x(1-6x)
$$
By comparing coefficients and solving the resulting system of equations, we find that \(A = 2\) and \(B = 11\). So the particular solution is given by
$$
y_p(x) = xe^{x}(2x+11)
$$
03
Combine the complementary function and the particular solution
The general solution of the given non-homogeneous equation is the sum of the complementary function and the particular solution:
$$
y(x) = y_c(x) + y_p(x) = C_1 e^{x} + C_2 e^{2x} + C_3 e^{3x} + xe^{x}(2x+11)
$$
04
Apply the given initial conditions
Given the initial conditions \(y(0)=2\), \(y^{\prime}(0)=7\), and \(y^{\prime \prime}(0)=9\), we can determine the constants \(C_1\), \(C_2\), and \(C_3\):
1. Using \(y(0)=2\):
$$
2= C_1 + C_2 + C_3
$$
2. Using \(y^{\prime}(0)=7\):
$$
7 = e^{0}(B - 5A) = 11 - 10 \\
C_3 = 6 \\
C_1 + C_2 + 6 = 2 \\
C_1 + C_2 = -4
$$
3. Using \(y^{\prime \prime}(0)=9\):
$$
9 = e^{0}(A - 4B) = 2 - 44 \\
C_2 = -38 \\
C_1 = 34
$$
So, the final solution is:
$$
y(x) = 34 e^{x} - 38 e^{2x} + 6 e^{3x} + xe^{x}(2x+11)
$$
Now we have the solution to the initial value problem, and we can graph it if desired.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Initial Value Problem
An initial value problem in differential equations involves finding a function that satisfies a given differential equation and meets specified initial conditions. In this scenario, we are given a third-order differential equation alongside initial conditions: \( y(0) = 2 \), \( y'(0) = 7 \), and \( y''(0) = 9 \).
Initial conditions are values assigned to the function and its derivatives at a specific point, usually \( x = 0 \). They are crucial as they allow us to determine the unique solution from a family of possible solutions.
Initial conditions are values assigned to the function and its derivatives at a specific point, usually \( x = 0 \). They are crucial as they allow us to determine the unique solution from a family of possible solutions.
- They specify the starting behavior of the system described by the equation.
- Help transform a general solution into a particular one that meets the conditions provided.
Homogeneous Equation
A homogeneous equation in the context of differential equations is a form where the equation is set to zero. For the given problem, the homogeneous equation resulting from the given differential equation is:\[y''' - 2y'' - 5y' + 6y = 0\]This equation emphasizes the natural dynamics of the system without any external influences (non-homogeneous terms).
- Solving a homogeneous equation helps find the complementary function, which is a key component of the general solution.
- The concept is crucial to understanding how the natural state or behavior of a system is characterized mathematically.
Non-Homogeneous Equation
A non-homogeneous equation incorporates an external factor or driving function on the right-hand side of the equation. Here, that factor is \( 2e^x(1-6x) \), which requires us to find a particular solution to complete the general solution.
Such equations are expressed like:\[y''' - 2y'' - 5y' + 6y = 2 e^x(1-6x)\]Non-homogeneous terms introduce complexity by affecting the behavior of the solution uniquely, compared to purely homogeneous parts.
Such equations are expressed like:\[y''' - 2y'' - 5y' + 6y = 2 e^x(1-6x)\]Non-homogeneous terms introduce complexity by affecting the behavior of the solution uniquely, compared to purely homogeneous parts.
- Non-homogeneous terms often represent external forces or inputs to a system, like a driving force.
- Finding a particular solution addresses how these additional influences modify the system's behavior.
Particular Solution
The particular solution specifically targets the non-homogeneous part of a differential equation. It represents how external or non-natural terms in the equation influence the system, differing from solutions of the homogeneous part.
In our exercise, to find the particular solution \( y_p(x) \), we need to match it to the non-homogeneous term \( 2 e^x(1-6x) \).
In our exercise, to find the particular solution \( y_p(x) \), we need to match it to the non-homogeneous term \( 2 e^x(1-6x) \).
- Choosing the right form for the particular solution is essential; it often mirrors the form of the non-homogeneous component.
- For instance, here, we try \( y_p(x) = xe^x(Ax + B) \) because of the exponential and polynomial form appearing in the non-homogeneous term.