The function f(t) satisfies the differential equation $\frac{d^{2}f}{d{t}^2}+f=0$ and the auxiliary conditions, $f\left(0\right)=0,\frac{df}{dt}\left(0\right)=4$ . The Laplace transform of f(t) is given by

Laplace Transform of a Differential Equation

The problem asks us to find the Laplace transform of a function $f(t)$ that satisfies a given differential equation along with specific auxiliary conditions. This involves using the properties of the Laplace transform, especially those related to derivatives.

Understanding the Given Information

- The differential equation is: $\frac{d^{2}f}{d{t}^2}+f=0$
- The auxiliary conditions (also known as initial conditions) are: - $f\left(0\right)=0$ - $\frac{df}{dt}\left(0\right)=4$ (which can also be written as $f^{\prime }\left(0\right)=4$)
- We need to find the Laplace transform of $f(t)$, denoted as $F(s)$ or $\mathcal{L}\left\{f(t)\right\}$.

Applying Laplace Transform to the Differential Equation

To find $F(s)$, we take the Laplace transform of each term in the given differential equation:
$$\mathcal{L}\left\{\frac{d^{2}f}{d{t}^2}+f\right\}=\mathcal{L}\left\{0\right\}$$
Using the linearity property of the Laplace transform, we can separate the terms:
$$\mathcal{L}\left\{\frac{d^{2}f}{d{t}^2}\right\}+\mathcal{L}\left\{f\right\}=\mathcal{L}\left\{0\right\}$$

Laplace Transform of Derivatives

The general formula for the Laplace transform of the first derivative is:
$$\mathcal{L}\left\{\frac{df}{dt}\right\}=sF(s)-f(0)$$
And for the second derivative:
$$\mathcal{L}\left\{\frac{d^{2}f}{d{t}^2}\right\}=s^{2}F(s)-sf(0)-f^{\prime }(0)$$

Substituting Initial Conditions

Now, we substitute the given auxiliary conditions $f(0)=0$ and $f^{\prime }(0)=4$ into the formula for the second derivative's Laplace transform:
$$\mathcal{L}\left\{\frac{d^{2}f}{d{t}^2}\right\}=s^{2}F(s)-s(0)-4$$

$$\mathcal{L}\left\{\frac{d^{2}f}{d{t}^2}\right\}=s^{2}F(s)-4$$

Solving for F(s)

Substitute this back into our transformed differential equation:
$$\left(s^{2}F(s)-4\right)+F(s)=0$$
Now, group the terms containing $F(s)$:
$$s^{2}F(s)+F(s)-4=0$$
Factor out $F(s)$:
$$F(s)(s^2+1)-4=0$$
Add 4 to both sides:
$$F(s)(s^2+1)=4$$
Finally, divide by $(s^2+1)$ to isolate $F(s)$:
$$F(s)=\frac{4}{s^2+1}$$

Comparing with Options

The calculated Laplace transform of $f(t)$ is $\frac{4}{s^2+1}$. Let's compare this with the given options:Our result matches Option 3. This solution demonstrates a common application of Laplace transforms in solving initial value problems for differential equations.