Consider \(y''+3y' + 2y = \cos(2t) + t.\)
We start by finding the homogeneous solution. That is the solution to \(y'' + 3y' + 2y = 0.\)
The characteristic polynomial is $r^2+3r+2 = (r+2)(r+1) = 0$ and so the homogeneous solution is \(y_h = c_1 e^{-t} + c_2 e^{-2t}.\)
Now the guess for the particular solutions is
\[\displaystyle y_p =\underset{\text{from the cosine term}}{\underbrace{ A\cos(2t)+B\sin(2t)}}+ \underset{\text{from the $t$ term}}{\underbrace{Ct +D}}.\]The derivatives of $y_p$ are \(\begin{split} y_p' &= -2A\sin(2t) +2 B\cos(2t) + C\\ y_p'' &= -4 A\cos(2t) -4 B\sin(2t) \end{split}\)
We plug these into the differential equation: \(\begin{split} &y_p''+3y_p' + 2y_p \\ &= (-4 A\cos(2t) - 4B\sin(2t)) + 3 (-2A\sin(2t) +2B\cos(2t)+C) \\ &\qquad + 2(A\cos(2t)+B\sin(2t)+Ct+D)\\ &= (-2A+6B)\cos(2t) + (-2B -6A)\sin(2t) + 2Ct + (2D+3C)\\ &= \cos(2t)+t. \end{split}\)
So we get this system of equations: \(\begin{split} (-2A+6B)&= 1\\ (-6A-2A)&= 0\\ 2C&=1\\ 2D+3C &= 0. \end{split}\)
To solve for $A$ and $B$ we have
\(\begin{split} -3\big(-2A + 6B &= 1 \big)\\ -6A - 2A &= 0\\ \hline 0A -20 B &= -3 \end{split}\qquad B = 3/20.\) Similarly, we can find that $A = -1/20$.
Solving for $C$ and $D$ is much easier. It is given by
\[C = \frac{1}{2},\qquad D = \frac{-3}{4}.\]Thus the general solution is
\[y = c_1 e^{-t}+c_2e^{-2t} + \frac{1}{20}(3\sin(2t)-\cos(2t)) + \frac{1}{4} (2t - 3).\]Consider \(y'' + 4y = \cos(2t).\)
Again, we first find the homogeneous solution \(y_h = c_1\cos(2t) + c_2 \sin(2t).\)
Normally our guess for the particular solution would be
\[y_p = A \cos(2t) + B\sin(2t).\]Unfortunately, since both of these terms appear in the homogenous solutions, the above guess is wrong. The correct guess involves multiplying the above guess by $t$: \(y_p = t(A\cos(2t)+B\sin(2t)).\)
We compute the first two derivatives:
\[\begin{split} y_p' &= (A\cos(2t)+B\sin(2t)) + t (-2A \sin(2t) + 2B\cos(2t))\\ y_p''&= (-2A\sin(2t) + 2B\sin(2t)) + (-2A\sin(2t) + 2B\cos(2t)) \\ &\qquad + t(-4A\cos(2t) -4B\sin(2t)) \end{split}\]So, plugging this into the differential equation gives
\[\begin{split} &y_p'' + 4y_p \\ &=(-4A\sin(2t) + 4B\cos(2t)) + t(-4A\cos(2t) -4B\sin(2t))\\ &\qquad + 4t(A\cos(2t)+B\sin(2t))\\ &= -4A\sin(2t) + 4B\cos(2t)\\ &= \cos(2t). \end{split}\]This leads to the system \(\begin{split} -4A = 0\\ 4B = 1 \end{split},\qquad B = 1/4, A = 0.\)
So
\[y_p = \frac{1}{4}t\sin(2t).\]And lastly the general solution is
\[y = c_1\cos(2t)+ c_2\sin(2t) + \frac{1}{4}t\sin(2t).\]