The Laplace Transform of the differential equation \( \frac{d x(t)}{d t}+x(t)=3 \), given that \( x(0)=1 \), is: \( \begin{array}{lll}\text { (A) } X(s)=\frac{4}{s+1} & \text { (B) } X(s)=\frac{s+3}{s+1} & \text { (C) } X(s)=\frac{s+3}{s(s+1)} \\ \text { (D) } X(s)=\frac{2}{s+1} & \text { (E) } X(s)=\frac{3}{s(s+1)} & \text { (F) } X(s)=\frac{3}{s}\end{array} \)

To solve the differential equation \( \frac{d x(t)}{d t}+x(t)=3 \) using the Laplace Transform, we start by taking the transform of both sides. Using the properties of the Laplace Transform, we get: \[ sX(s) - x(0) + X(s) = \frac{3}{s} \] Substituting \( x(0) = 1 \) gives us: \[ sX(s) - 1 + X(s) = \frac{3}{s} \] This simplifies to: \[ (s + 1)X(s) = \frac{3}{s} + 1 \implies (s + 1)X(s) = \frac{3 + s}{s} \] Finally, solving for \( X(s) \) yields: \[ X(s) = \frac{3 + s}{s(s + 1)} \] Thus, the correct answer is \( X(s) = \frac{s+3}{s(s+1)} \), which corresponds to option (C). Many real-world systems can be modeled by differential equations similar to the one you’ve just solved. For instance, in electrical engineering, the dynamics of a circuit with resistors and capacitors can often be described with such equations, where the Laplace Transform helps in analyzing the circuit's response to different inputs efficiently, reducing complex differential equations into algebraic forms. When dealing with Laplace Transforms, one common mistake is forgetting the initial conditions. In our case, \( x(0) = 1 \) was critical in determining \( X(s) \). Without accurately incorporating initial or boundary conditions, the solution can lead to incorrect conclusions, just like attempting to drive without looking at your fuel gauge – you might end up stranded!