# Particular solution of $x'' + x' + x = \sin2t$.

I'm struggling to figure out a differential equation problem. I'm trying to solve

$$x'' + x' + x = \sin2t$$

where I'm supposed to find a particular solution. I'm not completely sure how to break it up into $x(t) = A\cos(t) + B\sin(t)$

I substitute in for x(t), taking the derivative for $x''$ and $x'$, and I get: $$(-Acos(t) -Bsin(t)) + (-Asin(t) + Bcos(t)) + (Acos(t) + Bsin(t)) = sin(2t)$$ which simplifies to: $$-Asin(t) + Bcos(t) = sin(2t)$$

At this point I get a bit lost. In other solutions, I'd find the values of A and B at $x=0$ and $x= pi/2$, but if I do that here I find the solutions: A= 0, B=0 This in turn, when substituted into the original $Acos(t) + Bsin(t) =$ particular solution just yields 0, which is not the answer.

What did I do wrong? How can I correctly solve this question? Thank you

For your problem, the r.h.s. is $f(t)=\sin 2t$ so you should write $x(t)=A\sin 2t +B\cos 2t$

An easy way is by using the complex representation, noting that $\sin 2t=\Im(e^{i2t})$, and trying a solution of the form $z=Ae^{i2t}$.

Now,

$$z''+z'+z=A((2i)^2+2i+1)e^{i2t}=e^{i2t},$$

giving

$$A=\frac1{-3+2i}=-\frac{3+2i}{13}.$$

Then

$$\Im(Ae^{i2t})=-\frac{3\sin2t-2\cos2t}{13}.$$

Your mistake was to use an argument $t$ instead of $2t$.