Setting up a particular equation with (cos(x))^2

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Homework Help Overview

The discussion revolves around finding a particular solution to the differential equation 2y'' + 9y' + 2y = (cos(x))^2. Participants are exploring methods to approach this problem, particularly focusing on the use of trigonometric identities and the structure of the solution.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the initial steps of solving the homogeneous part of the equation and express uncertainty about how to proceed with the particular solution. There are suggestions to use the half-angle formula for cos(x)^2 and to consider the resulting terms in the context of particular solutions.

Discussion Status

There is an ongoing exploration of different approaches to formulating the particular solution. Some participants suggest combining terms derived from the half-angle formula, while others propose specific forms for the particular solution. The conversation indicates a productive exchange of ideas without a clear consensus yet.

Contextual Notes

Participants are working under the constraint of needing to find a particular solution rather than the general solution. There is some confusion regarding the formulation of terms and the appropriate method to combine them.

batmankiller
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Homework Statement


Find a particular solution yp of the differential equation

2y''+9y'+2y=(cos(x))^2

Homework Equations


The Attempt at a Solution


I'm not really sure where to start? I solve for the homogeneous solution and get C1e^ -4.26556+C2e^-.2344355629=0. I'm just not sure where to go from there. Should I do
y(x)=(Acos(x)+Bsin(x))^2 since it's just cos (x)*cos(x) or is not that right approach?
 
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cos^{2}(x) can be broken up using the half angle formula. You'd get a polynomial (degree one) and a trig function that is expandable under Euler's Formula. Add the particular solutions together to get the final particular solution.
 
so (cos(x)^2) is ((1+cos (2x))/2.
Which is 1/2 + cos(2x)/2. Am I missing something to get it into a degree one term and a trig function?
For that I get A+ B*cos(2x)+C*sin(2x)?
 
Last edited:
batmankiller said:
so (cos(x)^2) is ((1+cos (2x))/2.
Which is 1/2 + cos(2x)/2. Am I missing something to get it into a degree one term and a trig function?
For that I get A+ B*cos(2x)+C*sin(2x)?

Sorry, not degree one. I meant it has a constant added to a trig function. But you can solve the equation and get particular solutions by guessing that y = \frac{1}{4} (for one solution) and that y = Re(Ae^{2i\theta}) where A is some constant. Or you can use your method and simply guess A sin(2\theta)+Bcos(2\theta). Add that guess to your other particular solution (y = \frac{1}{4}), then add your particular solutions to your homogeneous solution. This will result in the general solution of the differential equation.
 
Ah ok, I don't even need the general solution, just the particular solution, so should I just consider the two particular solutions of Asin(2x)+Bcos(2x) and y=1/4 (Solving the constants A and B?)
 
batmankiller said:
Ah ok, I don't even need the general solution, just the particular solution, so should I just consider the two particular solutions of Asin(2x)+Bcos(2x) and y=1/4 (Solving the constants A and B?)

Yup. Add the two particular's together and you're done!
 

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