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Magazine Chapter 9: Problem 41
Solve the initial-value problem. \(y^{\prime \prime \prime}+y^{\prime}=0, y(\pi)=1, y^{\prime}(\pi)=8, y^{\prime \prime}(\pi)=4\)
Short Answer
Expert verified
The solution is \( y(x) = 5 - 8 \sin x + 4 \cos x \).
Step by step solution
01
Write down the characteristic equation
The given differential equation is a linear homogeneous differential equation. Begin by writing the characteristic equation for the differential equation: \[r^3 + r = 0\]
02
Solve the characteristic equation
Factor the characteristic equation to find the roots. \[r(r^2 + 1) = 0\] The solutions are: \(r = 0\) and \(r^2 = -1\). Solving for \(r\) gives us \(r = 0, r = i, \text{ and } r = -i\).
03
Write down the general solution
The roots (0, i, -i) indicate the solution to the differential equation. The general solution is: \[ y(x) = c_1 + c_2 \sin x + c_3 \cos x \]
04
Apply the initial conditions
Use the initial conditions \(y(\pi) = 1\), \(y'(\pi) = 8\), and \(y''(\pi) = 4\) to find the constants. Start with the general solution: \[ y(\pi) = c_1 + c_2 \sin \pi + c_3 \cos \pi \] Since \(\sin \pi = 0\) and \(\cos \pi = -1\), we get: \[ 1 = c_1 - c_3 \] Next, find \(y'(x)\): \[ y'(x) = c_2 \cos x - c_3 \sin x \] Using \(y'(\pi) = 8\), we get: \[ 8 = c_2 \cos \pi - c_3 \sin \pi \] Since \(\cos \pi = -1\) and \(\sin \pi = 0\), we have: \[ 8 = -c_2 \] Therefore, \(c_2 = -8\). Finally, find \(y''(x)\): \[ y''(x) = -c_2 \sin x - c_3 \cos x \] Using \(y''(\pi) = 4\), we get: \[ 4 = -c_2 \sin \pi - c_3 \cos \pi \] Given \(\sin \pi = 0\) and \(\cos \pi = -1\), we have: \[ 4 = c_3 \]
05
Solve for the constants
From the equations in Step 4, we have: \[ c_1 - c_3 = 1 \] \[ c_2 = -8 \] \[ c_3 = 4 \] Substitute \(c_3 = 4\) into the first equation: \[ c_1 - 4 = 1 \] Solving for \(c_1\) gives: \[ c_1 = 5 \]
06
Write the particular solution
Using the values of the constants \(c_1 = 5\), \(c_2 = -8\), and \(c_3 = 4\), the particular solution is: \[ y(x) = 5 - 8 \sin x + 4 \cos x \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linear Homogeneous Differential Equation
A linear homogeneous differential equation is a type of differential equation where every term is a function of the dependent variable and its derivatives. The term 'homogeneous' means that there is no free term (i.e., no terms without the dependent variable or its derivatives). In simpler terms, all the terms in the equation include the dependent variable or its derivatives. For example, the equation we are dealing with is:
\( y^{\text{{'''}}} + y' = 0 \)
Here, there are no terms that are functions of an independent variable alone. Each term involves \( y \) and its derivatives.
\( y^{\text{{'''}}} + y' = 0 \)
Here, there are no terms that are functions of an independent variable alone. Each term involves \( y \) and its derivatives.
Characteristic Equation
To solve a linear homogeneous differential equation, you first need to find the characteristic equation. The characteristic equation translates the differential equation into a polynomial equation by assuming a solution of the form \( e^{rx} \), where \( r \) is a constant. For the differential equation
\( y''' + y' = 0 \), the characteristic equation is found by replacing \( y''' \) with \( r^3 \) and \( y' \) with \( r \), giving us:
\[ r^3 + r = 0 \]
Solving this characteristic equation allows us to find the roots (values of \( r \)) that relate to the general solution of the differential equation.
\( y''' + y' = 0 \), the characteristic equation is found by replacing \( y''' \) with \( r^3 \) and \( y' \) with \( r \), giving us:
\[ r^3 + r = 0 \]
Solving this characteristic equation allows us to find the roots (values of \( r \)) that relate to the general solution of the differential equation.
General Solution
The general solution of a differential equation is a combination of functions based on the roots of the characteristic equation. For the equation
\( r^3 + r = 0 \), the roots are: \( r = 0, r = i, \text{{and}} r = -i \). These roots give us a set of fundamental solutions, which in this case are: \( e^0 \) (or simply a constant), and \( \cos x \) and \( \sin x \) due to the imaginary roots. Thus, the general solution to our differential equation is:
\[ y(x) = c_1 + c_2 \sin x + c_3 \cos x \]
Here, \( c_1, c_2, \text{{and}} c_3 \) are arbitrary constants that will be determined through initial conditions.
\( r^3 + r = 0 \), the roots are: \( r = 0, r = i, \text{{and}} r = -i \). These roots give us a set of fundamental solutions, which in this case are: \( e^0 \) (or simply a constant), and \( \cos x \) and \( \sin x \) due to the imaginary roots. Thus, the general solution to our differential equation is:
\[ y(x) = c_1 + c_2 \sin x + c_3 \cos x \]
Here, \( c_1, c_2, \text{{and}} c_3 \) are arbitrary constants that will be determined through initial conditions.
Initial Conditions
Initial conditions are values given for the solution and its derivatives at a specific point to determine the constants in the general solution. For our problem, the initial conditions provided are:
\( y(\pi) = 1 \)\( y'(\pi) = 8 \)\( y''(\pi) = 4 \).
We use these conditions to solve for the constants \( c_1, c_2, \text{{and}} c_3 \). By substituting the values and their derivatives at \( x = \pi \) into the general solution and its derivatives, we can form equations to solve for these constants.
\( y(\pi) = 1 \)\( y'(\pi) = 8 \)\( y''(\pi) = 4 \).
We use these conditions to solve for the constants \( c_1, c_2, \text{{and}} c_3 \). By substituting the values and their derivatives at \( x = \pi \) into the general solution and its derivatives, we can form equations to solve for these constants.
Particular Solution
The particular solution is the final solution of the differential equation after applying the initial conditions to the general solution. Here, we substitute the values of the constants back into the general solution. From our initial conditions, we found:
\( c_1 = 5 \)\( c_2 = -8 \)\( c_3 = 4 \).
Substituting these constants into the general solution \( y(x) = c_1 + c_2 \sin x + c_3 \cos x \) gives us the particular solution:
\[ y(x) = 5 - 8 \sin x + 4 \cos x \]
This particular solution satisfies both the differential equation and the initial conditions provided.
\( c_1 = 5 \)\( c_2 = -8 \)\( c_3 = 4 \).
Substituting these constants into the general solution \( y(x) = c_1 + c_2 \sin x + c_3 \cos x \) gives us the particular solution:
\[ y(x) = 5 - 8 \sin x + 4 \cos x \]
This particular solution satisfies both the differential equation and the initial conditions provided.
