# Particular solution to a second-order differential equation with trigonometry

The Problem $$y'' + 4y = \tan\left(2t\right)\sec\left(2t\right)$$

What I know I know that the general solution for the equation is:

$$y_H = K_1e^{2t} + K_2e^{-2t}$$

The Question what is the format for the guess of the particular solution for this equation?

My Guess Using Variation of Parameters

$$y_p = -e^{2t}\int{\frac{tan(2t)sec(2t)e^{-2t}}{-4}} + e^{-2t}\int{\frac{tan(2t)sec(2t)e^{2t}}{-4}}$$

Is this correct?

• Hint: Variation of Parameters. – Amzoti Dec 10 '13 at 3:25
• See my answer here for an example math.stackexchange.com/questions/474830/… – Amzoti Dec 10 '13 at 4:03
• Were you able to solve this? – Amzoti Dec 11 '13 at 6:30

## 2 Answers

Check your general solution.
If you have a solution to the homogeneous equation, here $e^{2t}$, try $f(t)e^{2t}$. That will reduce your equation to a first-order equation in $g(t)=f\,'(t)$

I think that you have a mistake from the beginning since r^2+4=0 gives r=2i and r=-2i. Then your general equation is
C[1] Cos[2 t] + C[2] Sin[2 t]
and not
C[1] Exp[2 t] + C[2] Exp[-2 t]
I hope I am right since, apparently, nobody noticed that.