Part II: Fill the Blank Space with the 4. Let \( f(t)=\left\{\begin{array}{rr}0, & 0 \leq t<3 \\ 2, & t \geq 3\end{array}\right. \), then \( L \) 5. \( L^{-1}\left\{\frac{-2 s+6}{s^{2}+4}\right\}= \) 6. \( L\left\{t^{n} f(t)\right\}= \) 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 \( y_{2}(x)= \)

Certainly! Here are the filled-in answers for each of the provided problems: --- **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: \[ \mathcal{L}\{f(t)\} = 2 \cdot \frac{e^{-3s}}{s} = \frac{2 e^{-3s}}{s} \] --- **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: \[ \mathcal{L}^{-1}\left\{\frac{-2s + 6}{s^{2} + 4}\right\} = -2 \cos(2t) + 3 \sin(2t) \] --- **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: \[ \mathcal{L}\{t^{n} f(t)\} = (-1)^n \frac{d^n}{ds^n} F(s) \] --- **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.* --- **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: \[ y_{2}(x) = e^{-x} \] --- If you have additional information for problem **7** or need further assistance with any other questions, feel free to ask!