Part II: Fill the Blank Space with the
4. Let , then
5.
6.
7. The particular solution of
8. The indicated is a solu
, then

Question

Part II: Fill the Blank Space with the 4. Let $$ f(t)=\left\{\begin{array}{rr}0, & 0 \leq t<3 \ 2, & t \geq 3\end{array}ight. $$, then $$ L $$ 5. $$ L^{-1}\left\{\frac{-2 s+6}{s^{2}+4}ight\}= $$ 6. $$ L\left\{t^{n} f(t)ight\}= $$ 7. The particular solution of $$ y^{\prime \prime}+2 y= $$ 8. The indicated $$ y_{1}(x)=x e^{-x} $$ is a solu $$ y^{\prime \prime}+2 y^{\prime}+y=0 $$, then $$ y_{2}(x)= $$

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**4. Let $$ f(t) = \begin{cases} 0, & 0 \leq t < 3 \ 2, & t \geq 3 \end{cases} $$, then $$ \mathcal{L}\{f(t)\} $$ is:**

The function $$ f(t) $$ is a step function that activates at $$ t = 3 $$. It can be represented using the Heaviside (step) function $$ u(t - 3) $$:

$$
f(t) = 2 \cdot u(t - 3)
$$

The Laplace transform of $$ u(t - a) $$ is $$ \frac{e^{-as}}{s} $$. Therefore:

$$
\mathcal{L}\{f(t)\} = 2 \cdot \frac{e^{-3s}}{s} = \frac{2 e^{-3s}}{s}
$$

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**5. $$ \mathcal{L}^{-1}\left\{\frac{-2s + 6}{s^{2} + 4}\right\} = $$ ?**

First, decompose the expression:

$$
\frac{-2s + 6}{s^{2} + 4} = -2 \cdot \frac{s}{s^{2} + 4} + 6 \cdot \frac{1}{s^{2} + 4}
$$

Using standard inverse Laplace transforms:

$$
\mathcal{L}^{-1}\left\{\frac{s}{s^{2} + a^{2}}\right\} = \cos(at)
$$
$$
\mathcal{L}^{-1}\left\{\frac{1}{s^{2} + a^{2}}\right\} = \frac{\sin(at)}{a}
$$

Here, $$ a = 2 $$:

$$
\mathcal{L}^{-1}\left\{-2 \cdot \frac{s}{s^{2} + 4}\right\} = -2 \cos(2t)
$$
$$
\mathcal{L}^{-1}\left\{6 \cdot \frac{1}{s^{2} + 4}\right\} = 6 \cdot \frac{\sin(2t)}{2} = 3 \sin(2t)
$$

Combining both parts:

$$
\mathcal{L}^{-1}\left\{\frac{-2s + 6}{s^{2} + 4}\right\} = -2 \cos(2t) + 3 \sin(2t)
$$

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**6. $$ \mathcal{L}\left\{t^{n} f(t)\right\} = $$ ?**

The Laplace transform of $$ t^n f(t) $$ can be expressed using the differentiation property of Laplace transforms:

$$
\mathcal{L}\{t^{n} f(t)\} = (-1)^n \frac{d^n}{ds^n} \mathcal{L}\{f(t)\}
$$

Where $$ \mathcal{L}\{f(t)\} = F(s) $$. Therefore:

$$
\mathcal{L}\{t^{n} f(t)\} = (-1)^n \frac{d^n}{ds^n} F(s)
$$

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**7. The particular solution of $$ y'' + 2y = $$ ?**

*Note: It appears that the right-hand side of the differential equation is missing. To determine the particular solution, the form of the nonhomogeneous term (right-hand side) is required. Common forms include polynomials, exponentials, sines, cosines, or combinations thereof. Please provide the complete equation for a specific particular solution.*

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**8. The indicated $$ y_{1}(x) = x e^{-x} $$ is a solution of $$ y'' + 2 y' + y = 0 $$, then $$ y_{2}(x) = $$ ?**

Given the differential equation:

$$
y'' + 2 y' + y = 0
$$

The characteristic equation is:

$$
r^2 + 2r + 1 = 0 \quad \Rightarrow \quad (r + 1)^2 = 0 \quad \Rightarrow \quad r = -1 \text{ (double root)}
$$

For a double root $$ r = -1 $$, the general solution to the homogeneous equation is:

$$
y(x) = (C_1 + C_2 x) e^{-x}
$$

Given that $$ y_1(x) = x e^{-x} $$ is one solution, the second linearly independent solution is:

$$
y_{2}(x) = e^{-x}
$$

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If you have additional information for problem **7** or need further assistance with any other questions, feel free to ask!
**4.** The Laplace transform of $$ f(t) $$ is $$ \frac{2 e^{-3s}}{s} $$. **5.** The inverse Laplace transform is $$ -2 \cos(2t) + 3 \sin(2t) $$. **6.** The Laplace transform of $$ t^{n} f(t) $$ is $$ (-1)^n \frac{d^n}{ds^n} F(s) $$. **7.** Cannot determine the particular solution without the right-hand side of the differential equation. **8.** The second solution is $$ y_{2}(x) = e^{-x} $$.

Part II: Fill the Blank Space with the
4. Let , then
5.
6.
7. The particular solution of
8. The indicated is a solu
, then

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Part II: Fill the Blank Space with the 4. Let , then 5. 6. 7. The particular solution of 8. The indicated is a solu , then

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Part II: Fill the Blank Space with the
4. Let , then
5.
6.
7. The particular solution of
8. The indicated is a solu
, then