Laplace Transform of a Differential Equation Solution

Problem Analysis

We need to find the Laplace transform, denoted as $Y(s)$, of the solution $y(t)$ to the given second-order linear homogeneous differential equation: $y''(t) + 4y(t) = 0$ with the initial conditions $y(0) = 1$ and $y'(0) = 6$.

Applying Laplace Transform

The Laplace transform is a powerful tool for solving differential equations. We apply the Laplace transform to both sides of the differential equation. We will use the following standard Laplace transform properties:

• $L\{y''(t)\} = s^2 Y(s) - s y(0) - y'(0)$
• $L\{y(t)\} = Y(s)$
• $L\{0\} = 0$

Applying these properties to the equation $y''(t) + 4y(t) = 0$, we get:

$L\{y''(t)\} + 4L\{y(t)\} = L\{0\}$

Substituting the properties:

$(s^2 Y(s) - s y(0) - y'(0)) + 4Y(s) = 0$

Substituting Initial Conditions

Now, we substitute the given initial conditions, $y(0) = 1$ and $y'(0) = 6$, into the transformed equation:

$(s^2 Y(s) - s(1) - 6) + 4Y(s) = 0$

This simplifies to:

$s^2 Y(s) - s - 6 + 4Y(s) = 0$

Solving for Y(s)

Our goal is to find the expression for $Y(s)$. We can rearrange the equation to isolate $Y(s)$. First, group the terms containing $Y(s)$:

$(s^2 + 4)Y(s) - s - 6 = 0$

Next, move the constant and $s$ terms to the right side of the equation:

$(s^2 + 4)Y(s) = s + 6$

Finally, divide by $(s^2 + 4)$ to solve for $Y(s)$:

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

To match the format of the options provided, we can split the fraction:

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

Final Result

The Laplace transformation $Y(s)$ of the solution $y(t)$ for the given differential equation and initial conditions is:

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

This result corresponds to option 3.