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I have the following differential equation: $$y''+y=\cos(t)\cos(2t)$$

Maybe something can be done to $\cos(t)\cos(2t)$ to make it easier to solve. Any ideas?

Thanks in advance.

  • 1
    Would a sum of cosines be easier than a product of cosines?2011-02-28
  • 0
    Sure! What are you thinking of?2011-02-28
  • 3
    $\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b)$; $\cos(a-b) = \cos(a)\cos(b) + \sin(a)\sin(b)$. So $\cos(a)\cos(b) = \frac{1}{2}(\cos(a+b) + \cos(a-b))$. In particular, $\cos(t)\cos(2t) = \frac{1}{2}(\cos(3t) + \cos(t))$.2011-02-28
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    Do you now manage to solve it?2011-02-28
  • 0
    Thanks, Arturo! That was really helpful.2011-03-01

2 Answers 2