Is It Possible to Use the Method of Undetermined Coefficients Here?

[checkmatechamp]

I'm doing a practice problem I found online, and I get a solution, but I think it should have a sine term in it. I looked up the solution, and most sites say to use variation of parameters, but is it possible to use the method of undetermined coefficients?

The problem is as follows: y'' + 4y = 8cos2t

r^2 + 4 = 0 ----> r^2 = -4 -----> r = +/- 2i

yh = c1*sin2t + c2*cos2t

Particular solution has form At*sin2t + Bt*cos2t (because you add a "t" to both terms so they don't clash with the cos2t or sin2t terms)

First derivative is -2At*sin2t + A*cos2t + 2Bt*cos2t + B*sin2t

Second derivative is -4At*cos2t - 2A*sin2t - 2A*sin2t - 4Bt*sin2t + 2B*cos2t + 2B*cos2t

So then you add that term to the particular solution and get:

-4At*cos2t - 2A*sin2t - 2A*sin2t - 4Bt*sin2t + 2B*cos2t + 2B*cos2t + 4At*sin2t + 4Bt*cos2t

Canceling out terms leaves you with -2A*sin2t - 2A*sin2t + 2B*cos2t + 2B*cos2t = 8*cos2t.

Now the problem is that 8*cos2t is really 8*cos2t + 0*sin2t. Solving for B gives you B = 2, but that means A = 0. But the final solution should have a sine term in it. (And I looked up the solutions, and they do have a sine term)

I thought you could use the Method of Undetermined Coefficients when you were dealing with a sine/cosine function, a regular polynomial, or an exponential function. (I think the fourth one was a function of e^t. I forget off the top of my head.)

 

[pasmith]

I'm doing a practice problem I found online, and I get a solution, but I think it should have a sine term in it. I looked up the solution, and most sites say to use variation of parameters, but is it possible to use the method of undetermined coefficients?

The problem is as follows: y'' + 4y = 8cos2t

r^2 + 4 = 0 ----> r^2 = -4 -----> r = +/- 2i

yh = c1*sin2t + c2*cos2t

Particular solution has form At*sin2t + Bt*cos2t (because you add a "t" to both terms so they don't clash with the cos2t or sin2t terms)

First derivative is -2At*sin2t + A*cos2t + 2Bt*cos2t + B*sin2t

You are inconsistent in your definitions of $A$ and $B$. You start by defining
$$y(t) = At\sin(2t) + Bt\cos(2t)$$ but in computing the derivatives you somehow switched to $$y(t) = At\cos(2t) + Bt\sin(2t).$$
Thus when you substituted these into the ODE you obtained

[y'' + 4y =] -4At*cos2t - 2A*sin2t - 2A*sin2t - 4Bt*sin2t + 2B*cos2t + 2B*cos2t + 4At*sin2t + 4Bt*cos2t

where you failed to notice that the terms I've bolded don't actually cancel.

But continuing on, you find that $B = 2$. Because you flipped the definitions of $A$ and $B$ halfway through, this is actually telling you that $y(t) = 2t\sin(2t)$, which is correct.