Questions: Use the Laplace transform to solve the following initial value problem: y'' - 4y' + 20y = 0 y(0) = 0, y'(0) = 4 a. Using Y for the Laplace transform of y(t), i.e., Y = Ly(t), find the equation you get by taking the Laplace transform of the differential equation b. Now solve for Y(s) = 4 / (s-2)^2 + 4^2 c. By completing the square in the denominator and inverting the transform, find y(t) = e^(2t) sin(4t)

Solution Steps

To solve the given initial value problem using the Laplace transform, follow these steps:

a. Take the Laplace transform of the entire differential equation. Use the properties of the Laplace transform to handle derivatives, and apply the initial conditions to simplify the expression.

b. Solve for \( Y(s) \) by isolating it in the transformed equation.

c. Use the inverse Laplace transform to find \( y(t) \). This involves recognizing the form of \( Y(s) \) as a standard Laplace transform pair and applying the inverse transform.

We start with the initial value problem given by the differential equation:

\[ y^{\prime \prime} - 4y^{\prime} + 20y = 0 \]

Applying the Laplace transform, we have:

\[ \mathcal{L}\{y^{\prime \prime}\} - 4\mathcal{L}\{y^{\prime}\} + 20\mathcal{L}\{y\} = 0 \]

Using the properties of the Laplace transform, we substitute:

\[ s^2 Y(s) - sy(0) - y'(0) - 4(s Y(s) - y(0)) + 20 Y(s) = 0 \]

Given the initial conditions \(y(0) = 0\) and \(y'(0) = 4\), we simplify to:

\[ s^2 Y(s) + 4s Y(s) + 20 Y(s) - 4 = 0 \]

Rearranging the equation, we isolate \(Y(s)\):

\[ Y(s)(s^2 + 4s + 20) = 4 \]

Thus, we find:

\[ Y(s) = \frac{4}{s^2 + 4s + 20} \]

Next, we need to perform the inverse Laplace transform to find \(y(t)\). First, we complete the square in the denominator:

\[ s^2 + 4s + 20 = (s + 2)^2 + 16 \]

Now, we can express \(Y(s)\) as:

\[ Y(s) = \frac{4}{(s + 2)^2 + 4^2} \]

Recognizing this as a standard form, we apply the inverse Laplace transform:

\[ y(t) = e^{-2t} \sin(4t) \cdot \text{Heaviside}(t) \]

Final Answer