2. By studying the form of $$ y_{c}(t) $$ that you found in the process of solving question 1 , give the general solution of $$ \frac{d^{2} y}{d t^{2}}+4 \frac{d y}{d t}+5 y=2 \sin t $$

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The homogeneous part of the differential equation is

$$
\frac{d^{2}y}{dt^{2}} + 4\frac{dy}{dt} + 5y = 0.
$$

**Step 1. Find the Complementary (Homogeneous) Solution**

The characteristic equation is

$$
r^2 + 4r + 5 = 0.
$$

Solve for $$ r $$ using the quadratic formula:

$$
r = \frac{-4 \pm \sqrt{(4)^2 - 4(1)(5)}}{2} = \frac{-4 \pm \sqrt{16-20}}{2} = \frac{-4 \pm \sqrt{-4}}{2}.
$$

Since $$\sqrt{-4} = 2i$$, the roots are

$$
r = -2 \pm i.
$$

Thus, the complementary solution is

$$
y_{c}(t) = e^{-2t}\left( C_{1}\cos t + C_{2}\sin t \right).
$$

**Step 2. Find a Particular Solution**

We now seek a particular solution to

$$
\frac{d^{2}y}{dt^{2}} + 4\frac{dy}{dt} + 5y = 2\sin t.
$$

Since the non-homogeneous term is $$ 2\sin t $$ and there is no conflict with the terms in the complementary solution (which involve $$ e^{-2t} $$), we look for a particular solution of the form

$$
y_{p}(t) = A\cos t + B\sin t.
$$

Differentiate:

$$
y_{p}'(t) = -A\sin t + B\cos t,
$$

$$
y_{p}''(t) = -A\cos t - B\sin t.
$$

Substitute $$ y_{p}(t) $$, $$ y_{p}'(t) $$, and $$ y_{p}''(t) $$ into the differential equation:

$$
\begin{aligned}
y_{p}''(t) + 4y_{p}'(t) + 5y_{p}(t) &= \left(-A\cos t - B\sin t\right) + 4\left(-A\sin t + B\cos t\right) + 5\left(A\cos t + B\sin t\right)\[1mm]
&= (-A + 4B + 5A)\cos t + (-B - 4A + 5B)\sin t\[1mm]
&= (4A + 4B)\cos t + (-4A + 4B)\sin t.
\end{aligned}
$$

This expression must equal the right-hand side $$2\sin t$$. Equate the coefficients on both sides:

For the coefficient of $$\cos t$$:

$$
4A + 4B = 0,
$$

and for $$\sin t$$:

$$
-4A + 4B = 2.
$$

From the first equation,

$$
A + B = 0 \quad \Longrightarrow \quad B = -A.
$$

Substitute $$B = -A$$ in the second equation:

$$
-4A + 4(-A) = 2 \quad \Longrightarrow \quad -4A - 4A = 2 \quad \Longrightarrow \quad -8A = 2.
$$

Solve for $$A$$:

$$
A = -\frac{2}{8} = -\frac{1}{4}.
$$

Then,

$$
B = -A = \frac{1}{4}.
$$

Thus, the particular solution is

$$
y_{p}(t) = -\frac{1}{4}\cos t + \frac{1}{4}\sin t.
$$

**Step 3. Write the General Solution**

The general solution is the sum of the complementary and particular solutions:

$$
y(t) = y_{c}(t) + y_{p}(t) = e^{-2t}\left( C_{1}\cos t + C_{2}\sin t \right) - \frac{1}{4}\cos t + \frac{1}{4}\sin t.
$$
The general solution is: $$ y(t) = e^{-2t}\left( C_{1}\cos t + C_{2}\sin t \right) - \frac{1}{4}\cos t + \frac{1}{4}\sin t. $$

By studying the form of that you found in the process of solving question 1 , give
the general solution of

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