Question

Use this to find a particular solution of the equation. Then find the general solution and, where indicated, solve the initial value problem and graph the solution. $$ y^{\prime \prime}+y^{\prime}=-8 \cos 2 x+6 \sin 2 x $$

Short answer

There are two arbitrary constants present in the general solution: \(C_1\) and \(C_2\).

Step-by-step solution

Write the given equation

We have the equation: $$ y^{\prime \prime}+y^{\prime}=-8 \cos 2 x+6 \sin 2 x $$

Choose trial functions

Choose the trial functions corresponding to the sine and cosine terms: $$ y_p=A\cos 2x + B\sin 2x $$ where A and B are constants to be determined.

Differentiate trial functions twice

Differentiate \(y_p\) and \(y_p'\) with respect to x to find \(y_p''\): $$ y_p'= -2A\sin 2x + 2B\cos 2x $$ $$ y_p''= -4A\cos 2x -4B\sin 2x $$

Substitute trial derivatives into the equation

Substitute the differentiated trial functions into the equation: $$ (-4A\cos 2x -4B\sin 2x) + (-2A\sin 2x + 2B\cos 2x) = -8 \cos 2 x+6 \sin 2 x $$

Equate coefficients for sine and cosine terms

Equate the coefficients for the sine and cosine terms: $$ -4A + 2B = -8 $$ $$ -4B - 2A = 6 $$

Solve the system of equations

Solve the system of equations for A and B: We have A = 2 and B = -1. Therefore, our particular solution is: $$ y_p=2\cos 2x -\sin 2x $$

Find the general solution of the homogeneous equation

Solve the associated homogeneous equation, which is: $$ y^{\prime \prime}+y^{\prime}=0 $$ The characteristic equation is: $$ r^2 + r = 0 $$ Factoring yields: $$ r(r+1) = 0 $$ The roots are r = 0 and r = -1. Therefore, the complementary function is: $$ y_c=C_1e^{0x} + C_2e^{-x} = C_1 + C_2e^{-x} $$

Find the general solution of the given equation

Add the complementary function to the particular solution: $$ y(x)=y_c+y_p=C_1 + C_2e^{-x} + 2\cos 2x -\sin 2x $$ Here, \(C_1\) and \(C_2\) are arbitrary constants. Since no initial value problem is given, we can stop here, and our final answer is the general solution: $$ y(x)=C_1 + C_2e^{-x} + 2\cos 2x -\sin 2x $$

Key concepts

Particular Solution

When faced with a non-homogeneous differential equation, finding a particular solution means determining one specific solution that fits the equation. This doesn't need to satisfy every possible condition or boundary. It's simply a solution that works for the given equation's non-homogeneous part.
In our exercise, the goal was to find the particular solution for the equation:

• \(y'' + y' = -8\cos 2x + 6\sin 2x\)
• We chose trial functions \(y_p = A\cos 2x + B\sin 2x\), where \(A\) and \(B\) are constants representing arbitrary values we needed to determine.

Through differentiation and substitution, we got \(A = 2\) and \(B = -1\). This gave us the particular solution:

• \(y_p = 2\cos 2x - \sin 2x\)

General Solution

A general solution of a differential equation encompasses all possible particular solutions by incorporating arbitrary constants. These constants can be adjusted to meet different initial conditions or boundary values.
For our particular problem, after finding the homogeneous solution \(y_c\) and the particular solution \(y_p\), we combined these to form the general solution of the equation.
Here, the homogeneous part was derived from solving the associated homogeneous equation:

• \(y'' + y' = 0\)
• This gave us the characteristic equation \(r^2 + r = 0\), with roots \(r = 0\) and \(r = -1\).
• The complementary function became \(y_c = C_1 + C_2e^{-x}\).

Adding this to our particular solution, the general solution was:

• \(y(x) = C_1 + C_2e^{-x} + 2\cos 2x - \sin 2x\)

Here, \(C_1\) and \(C_2\) are arbitrary constants.

Initial Value Problem

An initial value problem (IVP) attaches specific conditions, usually given as initial values of the function or its derivatives, to guide us to one particular solution from the general solution. By solving an IVP, we can pinpoint the exact values of arbitrary constants in the general solution.
In our current problem, an IVP wasn't provided, but typically, you might see an initial condition like this:

• \(y(0) = y_0\)
• \(y'(0) = y_0'\)

From there, you can substitute these conditions into the general solution and its derivative to solve for \(C_1\) and \(C_2\). This results in a specific solution that matches the initial conditions precisely.
So even though we didn't use it here, understanding how an IVP narrows down possible solutions is vital.

Homogeneous Equation

A homogeneous equation in differential equations is one where every term is dependent solely on the function and its derivatives, excluding any standalone constant or function of the independent variable. Often, it appears as the base case in determining the general solution.
In our problem, the associated homogeneous equation was:

• \(y'' + y' = 0\)

Here, the right side is zero, indicating it's a homogeneous form.
The general approach to solving this type is to find roots of the characteristic equation. For example, here:

• Characteristic: \(r^2 + r = 0\), which gave roots \(r = 0\) and \(r = -1\).
• This led to the homogeneous solution: \(y_c = C_1 + C_2e^{-x}\).

This solution forms a crucial part of the general solution, helping us cater to various forms by adjusting the constants.