For the equation $$y''-3y'+2y=5\cos(2t),$$ the "guess" for the homogeneous trig function on the right using undetermined coefficients would be (as I am taught): $$ A\cos(2t)+B\sin(2t)=5cos(2t).$$
Then I proceeded to solve for the $ A$ and $B$.
My question is:
If it's $\cos(t)$, does the "guess" have to be in this exact form $(A\cos(x)+B\sin(x))$ ? Meaning does the cosine have to get the "$A$" constant?
I tried this same equation using $A\sin(2t)+B\cos(2t)$ and did not get the same answer. The correct coefficients are; "$A$" is $(\frac{-1}{4})$ and "$B$" is $(\frac{3}{4})$. What I got when I switched was $A=(\frac{-5}{4})$ and $B=(\frac{15}{4})$.
Can you simply not switch the order? Thanks.
It does not matter if you solve for $y= A\cos(2t)+B\sin(2t)$ or $y= B\cos(2t)+A\sin(2t)$ your final solution should be the same.
$$ y= A\cos(2t)+B\sin(2t) \implies y'= -2A\sin(2t)+2B\cos(2t)$$
$$ \implies y''=-4A\cos(2t)-4B\sin(2t)$$
plugging in the equation, $$ y''-3y'+2y=5\cos (2t)$$
We have to solve for $A$ and $B$.
I came up with $A=-1/4$ and $B=-3/4$
Thus my solution for the particular part is $$y=-1/4\cos(2t)-3/4 \sin(2t)$$