Find laplace transform
y''+16y= {1, 0$$ \le t<\pi $$
{0, $$ \pi\le t<\infty $$
y(0)=6
y'(0)=3

To solve the differential equation using the Laplace transform, we begin by applying the Laplace transform to both sides of the given equation.

### Given problem:
\[
y'' + 16y = f(t)
\]
where:
\[
f(t) =
\begin{cases}
1, & 0 \le t < \pi \\
0, & \pi \le t < \infty
\end{cases}
\]
and the initial conditions are:
\[
y(0) = 6, \quad y'(0) = 3
\]

### Step 1: Apply the Laplace Transform
We apply the Laplace transform to both sides of the differential equation. Using standard Laplace transform properties, we have:

\[
\mathcal{L}\{y''(t)\} = s^2 Y(s) - s y(0) - y'(0) = s^2 Y(s) - 6s - 3
\]
\[
\mathcal{L}\{16y(t)\} = 16Y(s)
\]
\[
\mathcal{L}\{y''(t) + 16y(t)\} = s^2 Y(s) - 6s - 3 + 16 Y(s) = f(s)
\]

### Step 2: Laplace Transform of $ f(t) $
The function $ f(t) $ is a piecewise function, which we can write using the Heaviside step function $ u_\pi(t) $:

\[
f(t) = 1 - u_\pi(t)
\]

The Laplace transform of $ f(t) $ is:

\[
\mathcal{L}\{1\} = \frac{1}{s}
\]
\[
\mathcal{L}\{u_\pi(t)\} = \frac{e^{-\pi s}}{s}
\]

Thus:
\[
\mathcal{L}\{f(t)\} = \frac{1}{s} - \frac{e^{-\pi s}}{s}
\]

### Step 3: Solve for $ Y(s) $
Now we substitute everything into the Laplace-transformed equation:

\[
(s^2 + 16) Y(s) - 6s - 3 = \frac{1}{s} - \frac{e^{-\pi s}}{s}
\]

Solve for $ Y(s) $:

\[
Y(s) = \frac{1}{s(s^2 + 16)} - \frac{e^{-\pi s}}{s(s^2 + 16)} + \frac{6s + 3}{s^2 + 16}
\]

### Step 4: Inverse Laplace Transform
We now compute the inverse Laplace transform of each term separately.

1. First term:
\[
\mathcal{L}^{-1}\left\{ \frac{1}{s(s^2 + 16)} \right\}
\]
This is a standard Laplace transform, and its inverse is:
\[
\mathcal{L}^{-1}\left\{ \frac{1}{s(s^2 + 16)} \right\} = 1 - \cos(4t)
\]

2. Second term:
\[
\mathcal{L}^{-1}\left\{ \frac{e^{-\pi s}}{s(s^2 + 16)} \right\}
\]
This is a shifted function using the second shifting theorem (also known as the Heaviside shifting property):
\[
\mathcal{L}^{-1}\left\{ \frac{e^{-\pi s}}{s(s^2 + 16)} \right\} = u_\pi(t) \left[1 - \cos(4(t-\pi))\right]
\]

3. Third term:
\[
\mathcal{L}^{-1}\left\{ \frac{6s + 3}{s^2 + 16} \right\}
\]
This can be split into two standard transforms:
\[
\mathcal{L}^{-1}\left\{ \frac{6s}{s^2 + 16} \right\} = 6 \cos(4t)
\]
\[
\mathcal{L}^{-1}\left\{ \frac{3}{s^2 + 16} \right\} = \frac{3}{4} \sin(4t)
\]

### Final Solution:
The inverse Laplace transform gives the solution for $ y(t) $:

\[
y(t) = 1 - \cos(4t) + 6 \cos(4t) + \frac{3}{4} \sin(4t) - u_\pi(t) \left[ 1 - \cos(4(t-\pi)) \right]
\]

Let me know if you'd like further details, or have any questions!

Here are 5 related questions:
1. How does the second shifting theorem work in Laplace transforms?
2. What is the significance of Heaviside step functions in differential equations?
3. How can Laplace transforms be used to solve non-homogeneous equations?
4. What are the main properties of the Laplace transform?
5. How do initial conditions affect the solution to a differential equation using Laplace transforms?

**Tip:** When solving piecewise functions using Laplace transforms, it's useful to express them in terms of Heaviside functions for easier transformation.

Find the Laplace transform of y'' + 16y = {1, 0 ≤ t < π, 0, π ≤ t < ∞}, y(0) = 6, y'(0) = 3

This detailed solution covers the Laplace transform of a second-order differential equation with a piecewise function, initial conditions y(0) = 6, and y'(0) = 3. Key mathematical concepts such as the Heaviside step function and the second shifting theorem are used to solve the problem.

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