By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$y^{\prime \prime}-4 y^{\prime}+4 y=6 e^{2 t}, \quad y_{0}=y_{0}^{\prime}=0$$

Initial conditions are values given at the start of a problem that help determine a unique solution to a differential equation. In this exercise, we are given initial conditions: \( y(0) = 0 \) and \( y'(0) = 0 \). These initial conditions are crucial because they allow us to simplify our calculations when applying the Laplace transform.
Initial conditions act like anchor points. Without them, we could end up with many possible solutions. By setting specific initial values, we narrow it down to one unique solution that fits our problem.
When working through the problem's steps, notice how the initial conditions are applied directly:

• \( \mathcal{L}\left\{ y'' \right\} = s^2 Y(s) - sy(0) - y'(0) \) simplifies to \( s^2 Y(s) \) because both \( y(0) \) and \( y'(0) \) are zero.

By understanding and correctly applying initial conditions, solving differential equations becomes much clearer!

The inverse Laplace transform helps us convert back from the frequency domain to the time domain. After solving for \( Y(s) \) in the Laplace-transformed equation, we need to find \( y(t) \) by taking the inverse Laplace transform.
In this exercise, we found that:

• \( Y(s) = \frac{6}{(s-2)^3} \)

To find \( y(t) \), we use the inverse Laplace transform:

• \( \mathcal{L}^{-1}\left\{ \frac{6}{(s-2)^3} \right\} \).

By utilizing the shifting property and recognizing the inverse transform:

• \( \mathcal{L}^{-1}\left\{ \frac{1}{s^3} \right\} = \frac{t^2}{2} \)

We obtain:

• \( \mathcal{L}^{-1}\left\{ \frac{6}{(s-2)^3} \right\} = 6 \cdot \frac{t^2}{2} e^{2t} \)

Thus:

• \( y(t) = 3t^2 e^{2t} \).

Understanding inverse Laplace transforms allows you to convert results back to the original domain, giving you the solution in a more intuitive form.

Solving differential equations using Laplace transforms involves several key steps that simplify the process. Here's an overview of the crucial steps highlighted in this exercise:
First, apply the Laplace transform to convert the differential equation into an algebraic one. This transforms derivatives into polynomial expressions:

• \( \mathcal{L}\left\{ y'' - 4y' + 4y \right\} = \mathcal{L}\left\{ 6e^{2t} \right\} \).

Next, simplify using properties of the Laplace transform and incorporate initial conditions:

• \( (s^2 - 4s + 4)Y(s) = \frac{6}{s-2} \).

This simplifies to:

• \( (s-2)^2Y(s) = \frac{6}{s-2} \).

Solve for \( Y(s) \):

• \( Y(s) = \frac{6}{(s-2)^3} \).

Finally, find the inverse Laplace transform to get the solution in the time domain:

• \( y(t) = 3t^2 e^{2t} \).

By following these steps, you can effectively transform and solve complex differential equations, making them more manageable and understandable.