Part (a): Solve the IVP using Laplace Transforms

We are given the second-order linear non-homogeneous differential equation:

$\frac{d^2 y}{dt^2} - 6 \frac{dy}{dt} + 15y = 2 \sin(3t)$ with initial conditions $y(0) = -1$ and $y'(0) = -4$.

Step 1: Take the Laplace Transform of both sides

Recall the Laplace transforms of derivatives:

$\mathcal{L}\left\{ y'(t) \right\} = s Y(s) - y(0)$ $\mathcal{L}\left\{ y''(t) \right\} = s^2 Y(s) - s y(0) - y'(0)$ Also, the Laplace transform of $\sin(3t)$ is:

$\mathcal{L}\left\{ \sin(3t) \right\} = \frac{3}{s^2 + 9}$

Now apply the Laplace Transform to both sides of the equation:

$\mathcal{L}\left\{ \frac{d^2 y}{dt^2} - 6 \frac{dy}{dt} + 15y \right\} = \mathcal{L}\left\{ 2 \sin(3t) \right\}$

Using the properties of Laplace Transforms:

$(s^2 Y(s) - s y(0) - y'(0)) - 6(s Y(s) - y(0)) + 15 Y(s) = 2 \cdot \frac{3}{s^2 + 9}$

Substitute $y(0) = -1$ and $y'(0) = -4$:

$(s^2 Y(s) + s - 4) - 6(s Y(s) + 1) + 15 Y(s) = \frac{6}{s^2 + 9}$

Simplify the equation:

$s^2 Y(s) + s - 4 - 6s Y(s) - 6 + 15 Y(s) = \frac{6}{s^2 + 9}$

Combine like terms:

$(s^2 - 6s + 15) Y(s) + (s - 10) = \frac{6}{s^2 + 9}$

Thus, the equation becomes:

$(s^2 - 6s + 15) Y(s) = \frac{6}{s^2 + 9} - (s - 10)$

Step 2: Solve for $Y(s)$

Rearrange to isolate $Y(s)$:

$Y(s) = \frac{\frac{6}{s^2 + 9} - (s - 10)}{s^2 - 6s + 15}$

Simplify further if necessary, then apply the inverse Laplace Transform to find $y(t)$.

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Part (b): Express the IVP as a vector-matrix transformation and apply Laplace Transforms

We rewrite the second-order differential equation as a system of first-order ODEs by introducing the following variables:

$y_1(t) = y(t), \quad y_2(t) = y'(t)$

Thus, the system becomes:

y_1'(t) &= y_2(t) \\ y_2'(t) &= 6y_2(t) - 15y_1(t) + 2 \sin(3t) \end{aligned}$$ We can write this system in vector-matrix form: $$\frac{d}{dt} \begin{pmatrix} y_1 \\ y_2 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -15 & 6 \end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \end{pmatrix} + \begin{pmatrix} 0 \\ 2 \sin(3t) \end{pmatrix}$$ Let $$\mathbf{Y}(t) = \begin{pmatrix} y_1(t) \\ y_2(t) \end{pmatrix}$$, and we can express the system as: $$\frac{d}{dt} \mathbf{Y}(t) = A \mathbf{Y}(t) + \mathbf{f}(t)$$ where: $$A = \begin{pmatrix} 0 & 1 \\ -15 & 6 \end{pmatrix}, \quad \mathbf{f}(t) = \begin{pmatrix} 0 \\ 2 \sin(3t) \end{pmatrix}$$ #### Step 1: Apply Laplace Transform Taking the Laplace Transform of both sides: $$s \mathbf{Y}(s) - \mathbf{Y}(0) = A \mathbf{Y}(s) + \mathcal{L}\{\mathbf{f}(t)\}$$ Using $$\mathbf{Y}(0) = \begin{pmatrix} -1 \\ -4 \end{pmatrix}$$ and the Laplace Transform of $$\mathbf{f}(t)$$, solve for $$\mathbf{Y}(s)$$ by rearranging: $$(sI - A) \mathbf{Y}(s) = \mathbf{Y}(0) + \mathcal{L}\{\mathbf{f}(t)\}$$ Then, compute the inverse Laplace Transform to find $$\mathbf{Y}(t)$$. --- Would you like more detailed steps on any part, or further clarification on any steps? ### Relative Questions: 1. How do you apply partial fraction decomposition when solving for $$Y(s)$$? 2. What is the significance of the initial conditions in the Laplace Transform method? 3. How does the matrix approach compare to the standard method for solving ODEs? 4. Can you explain how the eigenvalues of matrix $$A$$ affect the solution behavior? 5. How does diagonalization simplify solving systems of differential equations? ### Tip: When solving differential equations using Laplace Transforms, always ensure that you account for initial conditions when transforming and inverting functions.