By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$y^{\prime \prime}-6 y^{\prime}+9 y=t e^{3 t}, \quad y_{0}=0, y_{0}^{\prime}=5$$

Differential equations involve functions and their derivatives. They help us model phenomena in engineering, physics, economics, and more.
In this exercise, we used a **second-order linear differential equation**.
This means there is a term involving the second derivative of the function y(t).
In simpler terms, it's like calculating the rate of change of a rate of change.
Here, the given differential equation is:
\[y'' - 6y' + 9y = t e^{3t}\]
We also had conditions to satisfy at the start (initial conditions).
Most real-world problems come with these conditions. They ensure our solution is unique and follows a particular path from the beginning.
So, differential equations are all about finding a function (solution) that satisfies both an equation and specific conditions.

Initial conditions are values provided at the start of a problem. They give us specific information about the function and its derivatives at a point.
For our problem, we had:

• y(0) = 0 (initial value of the function)
• y'(0) = 5 (initial value of the first derivative)

These conditions helped us adjust the transformed differential equation correctly.
For instance, when we took the Laplace transform of the given differential equation, we needed to apply y(0) and y'(0):\
\[L\big\{y'' - 6y' + 9y\big\} = L\big\{te^{3t}\big\} \rightarrow s^2Y(s) - sy(0)-y'(0) -6(sY(s)-y(0)) + 9Y(s)\]
This spells out the importance of initial conditions. They shape our unique solution through specific starting points and rates.

The inverse Laplace transform reverts a function back from the Laplace domain to the time domain.
In our case, once we had Y(s), we needed to transform it back to y(t):
\[Y(s) = \frac{1}{(s-3)^4} + \frac{5}{(s-3)^2}\]
We used known inverse Laplace transforms for specific forms:

• \(L^{-1}\big\{\frac{1}{(s-3)^{4}} = \frac{t^{3}}{3!}e^{3t}\big\}\)
• \(L^{-1}\big\{\frac{5}{(s-3)^{2}}\}= 5te^{3t}\)

Combining these, we got our solution:
\[y(t) = \frac{t^3}{6}e^{3t} + 5te^{3t}\]
Understanding how to revert back is key. It lets us relate the transformed function back to our real-world context.

Laplace transforms have unique properties that make solving differential equations easier.
Some critical properties include:

• **Linearity:** \(L\big\{af(t) + bg(t)\big\} = aL\big\{f(t)\big\} + bL\big\{g(t)\big\}\)
• **Differentiation:** \(L\big\{f'(t)\big\} = sF(s) - f(0)\)
• **Shifting:** \(L\big\{e^{at}f(t)\}= F(s-a)\)

These properties helped us convert derivatives into algebraic terms in our problem:
\(L\big\{y''\big\} = s^2Y(s) - sy(0) - y'(0)\)
\(L\big\{y'\big\} = sY(s) - y(0)\)
Then we simply substituted the initial conditions and transformed terms. Knowing the properties of Laplace transforms allows for a systematic and simplified approach to solving complex differential equations.