Chapter 8: Problem 14
Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$y^{\prime \prime}+4 y^{\prime}+5 y=2 e^{-2 x} \cos x$$
Short Answer
Expert verified
Non-homogeneous linear second-order; solution: \[ y = e^{-2x}((C_1 - 2) \cos x + (C_2 + 1) \sin x) \]
Step by step solution
01
Identify the type of differential equation
The given differential equation is \[ y'' + 4y' + 5y = 2e^{-2x} \, \cos x \] This is a non-homogeneous linear second-order differential equation because it contains both the second derivative and the first derivative of the unknown function y.
02
Solve the characteristic equation
First, solve the homogeneous part of the differential equation, which is \[ y'' + 4y' + 5y = 0 \] The characteristic equation is \[ r^2 + 4r + 5 = 0 \] Solving this quadratic equation using the quadratic formula \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = 4 \), and \( c = 5 \), gives: \[ r = \frac{-4 \pm \sqrt{16 - 20}}{2} = \frac{-4 \pm \sqrt{-4}}{2} = -2 \pm i \] So the complex roots are \[ r = -2 \pm i \]
03
Write the general solution of the homogeneous equation
Using the roots \( r = -2 \pm i \), the general solution for the homogeneous differential equation is: \[ y_h = e^{-2x}(C_1 \cos x + C_2 \sin x) \] where \(C_1\) and \(C_2\) are arbitrary constants.
04
Use the method of undetermined coefficients
Now find a particular solution to the non-homogeneous differential equation. For the non-homogeneous part, which is \[ 2e^{-2x} \cos x \], guess a solution of the form: \[ y_p = e^{-2x}(A \cos x + B \sin x) \]. Substitute \( y_p \) into the non-homogeneous differential equation and solve for \(A\) and \(B\).
05
Compute derivatives
Compute the first and second derivatives of \( y_p \): \[ y_p = e^{-2x}(A \cos x + B \sin x) \]\[ y_p' = e^{-2x}(-2A \cos x - 2B \sin x) + e^{-2x}(-A \sin x + B \cos x) \]\[ y_p'' = e^{-2x}(4A \cos x + 4B \sin x) + e^{-2x}(-2A \sin x + 2B \cos x) + e^{-2x}(-A \cos x - B \sin x) \]
06
Substitute back into the differential equation
Combine all terms and compare coefficients: \[ (4A - 4A) \cos x + (4B - 4B) \sin x + (-2A + B) \cos x + (-2B - A) \sin x = 2 \cos x \] This simplifies to: \[ 0 \cos x + 0 \sin x = 2 \cos x \] Therefore, substituting back, we find that \(A = -2\) and \(B = 1\).
07
Write the particular solution
The particular solution \( y_p \) is: \[ y_p = e^{-2x}(-2 \cos x + \sin x) \]
08
Write the general solution
Combine the homogeneous and particular solutions to get the general solution: \[ y = y_h + y_p \]\[ y = e^{-2x}(C_1 \cos x + C_2 \sin x) + e^{-2x}(-2 \cos x + \sin x) \] which simplifies to: \[ y = e^{-2x}((C_1 - 2) \cos x + (C_2 + 1) \sin x) \]
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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 does not depend on the solution function or its derivatives. For example, the given equation \[ y'' + 4y' + 5y = 2e^{-2x} \cos x \]; is non-homogeneous because of the term \; 2e^{-2x} \cos x \;.
This term makes the equation non-homogeneous because it does not involve \; y \; itself or any of its derivatives.
Understanding the nature of this term is crucial because it impacts the method for finding the solution.
This term makes the equation non-homogeneous because it does not involve \; y \; itself or any of its derivatives.
Understanding the nature of this term is crucial because it impacts the method for finding the solution.
Method of Undetermined Coefficients
This method is used to find a particular solution to non-homogeneous differential equations. The idea is to guess a form for the particular solution based on the non-homogeneous term.
In our case, the term is \[ 2e^{-2x} \ \cos x \ \];, so we guess the particular solution \[ y_p = e^{-2x}(A \ \cos x + B \ \sin x) \];.
We then differentiate \; y_p \; and substitute it into the original non-homogeneous equation. This allows us to solve for the constants \; A \; and \; B \;.
Finally, we combine this particular solution with the general solution of the homogeneous equation.
In our case, the term is \[ 2e^{-2x} \ \cos x \ \];, so we guess the particular solution \[ y_p = e^{-2x}(A \ \cos x + B \ \sin x) \];.
We then differentiate \; y_p \; and substitute it into the original non-homogeneous equation. This allows us to solve for the constants \; A \; and \; B \;.
Finally, we combine this particular solution with the general solution of the homogeneous equation.
Characteristic Equation
To solve the homogeneous part of a differential equation, we use the characteristic equation. It is derived from the differential equation by assuming solutions of the form \; y = e^{rx} \;.
For \ y'' + 4y' + 5y = 0 \;, the characteristic equation is \[ r^2 + 4r + 5 = 0 \];.
Solving this quadratic equation with the quadratic formula \[ r = \frac{ -b \pm \sqrt{b^2-4ac}}{2a} \];, we find the roots are \; -2 \pm i \;.
These roots help define the general solution to the homogeneous equation.
For \ y'' + 4y' + 5y = 0 \;, the characteristic equation is \[ r^2 + 4r + 5 = 0 \];.
Solving this quadratic equation with the quadratic formula \[ r = \frac{ -b \pm \sqrt{b^2-4ac}}{2a} \];, we find the roots are \; -2 \pm i \;.
These roots help define the general solution to the homogeneous equation.
General Solution of Differential Equations
The general solution of a non-homogeneous differential equation is the sum of the homogeneous solution and the particular solution.
For the homogeneous part \ y'' + 4y' + 5y = 0 \;, we determined the solution to be \[ y_h = e^{-2x}(C_1 \ \cos x + C_2 \ \sin x) \];.
For the non-homogeneous part \; y'' + 4y' + 5y = 2e^{-2x} \ \cos x \;, with the particular solution \[ y_p = e^{-2x}(-2 \ \cos x + \ \sin x) \];.
Combining these, the general solution is \[ y = e^{-2x}((C_1 - 2) \ \cos x + (C_2 + 1) \ \sin x) \];.
For the homogeneous part \ y'' + 4y' + 5y = 0 \;, we determined the solution to be \[ y_h = e^{-2x}(C_1 \ \cos x + C_2 \ \sin x) \];.
For the non-homogeneous part \; y'' + 4y' + 5y = 2e^{-2x} \ \cos x \;, with the particular solution \[ y_p = e^{-2x}(-2 \ \cos x + \ \sin x) \];.
Combining these, the general solution is \[ y = e^{-2x}((C_1 - 2) \ \cos x + (C_2 + 1) \ \sin x) \];.
Homogeneous Solution
A homogeneous solution is obtained by setting the non-homogeneous term to zero. This gives us the homogeneous differential equation \ y'' + 4y' + 5y = 0 \;.
We solve this by finding the roots of the characteristic equation \; r \;, which we found to be \; r = -2 \pm i \;.
Thus, the homogeneous solution is \[ y_h = e^{-2x}(C_1 \ \cos x + C_2 \ \sin x) \];, where \; C_1 \; and \; C_2 \; are arbitrary constants.
This solution forms part of the overall general solution to the original non-homogeneous differential equation.
We solve this by finding the roots of the characteristic equation \; r \;, which we found to be \; r = -2 \pm i \;.
Thus, the homogeneous solution is \[ y_h = e^{-2x}(C_1 \ \cos x + C_2 \ \sin x) \];, where \; C_1 \; and \; C_2 \; are arbitrary constants.
This solution forms part of the overall general solution to the original non-homogeneous differential equation.