1. (2 points) Find the general solution of $$ \frac{d^{2} y}{d t^{2}}+4 \frac{d y}{d t}+5 y=2 \cos t $$

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$$
\textbf{Step 1. Solve the Homogeneous Equation:}
$$

The homogeneous part of the differential equation is
$$
\frac{d^{2} y}{d t^{2}} + 4 \frac{d y}{d t} + 5 y = 0.
$$
Assume a solution of the form $$ y = e^{\lambda t} $$. Substituting into the homogeneous equation gives
$$
\lambda^{2} e^{\lambda t} + 4 \lambda e^{\lambda t} + 5 e^{\lambda t} = 0.
$$
Since $$ e^{\lambda t} \neq 0 $$, we have the characteristic equation:
$$
\lambda^{2} + 4\lambda + 5 = 0.
$$
Solving using the quadratic formula,
$$
\lambda = \frac{-4 \pm \sqrt{4^{2}-4\cdot1\cdot5}}{2} = \frac{-4 \pm \sqrt{16-20}}{2} = \frac{-4 \pm \sqrt{-4}}{2} = -2 \pm i.
$$
Thus, the general solution of the homogeneous equation is
$$
y_{h}(t) = e^{-2t}\left(C_{1} \cos t + C_{2} \sin t\right),
$$
where $$ C_{1} $$ and $$ C_{2} $$ are arbitrary constants.

$$
\textbf{Step 2. Find a Particular Solution:}
$$

Consider the non-homogeneous term $$ 2\cos t $$. We use the method of undetermined coefficients and propose a particular solution of the form
$$
y_{p}(t) = A\cos t + B\sin t,
$$
where $$ A $$ and $$ B $$ are constants to be determined.

Compute the derivatives:
$$
y_{p}'(t) = -A\sin t + B\cos t,
$$
$$
y_{p}''(t) = -A\cos t - B\sin t.
$$

Substitute $$ y_{p} $$, $$ y_{p}' $$, and $$ y_{p}'' $$ into the differential equation:
$$
(-A\cos t - B\sin t) + 4(-A\sin t + B\cos t) + 5(A\cos t + B\sin t) = 2\cos t.
$$

Group the terms for $$ \cos t $$ and $$ \sin t $$:
$$
\cos t: \quad -A + 4B + 5A = (4A + 4B),
$$
$$
\sin t: \quad -B - 4A + 5B = (4B - 4A).
$$

Thus, the equation becomes:
$$
(4A + 4B)\cos t + (4B - 4A)\sin t = 2\cos t.
$$

For the equality to hold for all $$ t $$, the coefficients of $$ \cos t $$ and $$ \sin t $$ must match:
$$
4A + 4B = 2, \quad 4B - 4A = 0.
$$

From the second equation, we have:
$$
4B - 4A = 0 \quad \Rightarrow \quad B = A.
$$

Substitute $$ B = A $$ into the first equation:
$$
4A + 4A = 8A = 2 \quad \Rightarrow \quad A = \frac{1}{4}.
$$
Thus, $$ B = \frac{1}{4} $$ as well.

The particular solution is then:
$$
y_{p}(t) = \frac{1}{4}\cos t + \frac{1}{4}\sin t.
$$

$$
\textbf{Step 3. Write the General Solution:}
$$

The general solution to the non-homogeneous differential equation is the sum of the homogeneous and particular solutions:
$$
y(t) = y_{h}(t) + y_{p}(t) = e^{-2t}\left(C_{1} \cos t + C_{2} \sin t\right) + \frac{1}{4}\cos t + \frac{1}{4}\sin t.
$$
The general solution is: $$ y(t) = e^{-2t}(C_{1} \cos t + C_{2} \sin t) + \frac{1}{4} \cos t + \frac{1}{4} \sin t $$ where $$ C_{1} $$ and $$ C_{2} $$ are arbitrary constants.

(2 points) Find the general solution of

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