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

  • #1

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?
 

Answers and Replies

  • #2
[tex]cos^{2}(x)[/tex] 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.
 
  • #3
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:
  • #4
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 [tex]y = \frac{1}{4}[/tex] (for one solution) and that [tex]y = Re(Ae^{2i\theta})[/tex] where A is some constant. Or you can use your method and simply guess [tex]A sin(2\theta)+Bcos(2\theta)[/tex]. Add that guess to your other particular solution ([tex]y = \frac{1}{4}[/tex]), then add your particular solutions to your homogeneous solution. This will result in the general solution of the differential equation.
 
  • #5
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?)
 
  • #6
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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