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

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SUMMARY

The discussion focuses on finding a particular solution, yp, for the differential equation 2y'' + 9y' + 2y = (cos(x))^2. Participants confirm that the half-angle formula can simplify cos²(x) to (1 + cos(2x))/2, leading to a polynomial and a trigonometric function. The suggested particular solutions include y = 1/4 and y = A sin(2x) + B cos(2x), where A and B are constants to be determined. The final step involves combining these particular solutions with the homogeneous solution to achieve the general solution.

PREREQUISITES
  • Understanding of differential equations, specifically second-order linear equations.
  • Familiarity with the half-angle formula in trigonometry.
  • Knowledge of homogeneous and particular solutions in differential equations.
  • Basic skills in solving for constants in trigonometric functions.
NEXT STEPS
  • Study the half-angle formulas in trigonometry for simplification techniques.
  • Learn about the method of undetermined coefficients for finding particular solutions.
  • Explore the theory behind homogeneous and particular solutions in second-order differential equations.
  • Practice solving differential equations with trigonometric functions as non-homogeneous terms.
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Students studying differential equations, mathematicians, and educators looking to deepen their understanding of solving non-homogeneous differential equations involving trigonometric functions.

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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