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The task requires finding the solution to the given differential equation with initial conditions using the Laplace transform. The equation provided is:

\[
x_1(t) = \frac{d}{dt}(2e^{-t} - 2\cos t),
\]
with the solution \( x_2(t) \) expressed as an integral:

\[
x_2(t) = \int_0^t (2e^{-\tau} - 2 \cos \tau) d\tau.
\]

There is also a table of Laplace transforms for several functions, which can be used to solve this problem by transforming it into the Laplace domain.

### Step-by-Step Solution:

#### 1. Find \( x_1(t) \)

The first function is defined as the derivative of \( 2e^{-t} - 2 \cos t \). Let's differentiate it:

\[
x_1(t) = \frac{d}{dt}(2e^{-t} - 2 \cos t)
= -2e^{-t} + 2 \sin t.
\]

So, \( x_1(t) = -2e^{-t} + 2 \sin t \).

#### 2. Solve for \( x_2(t) \)

To find \( x_2(t) \), we integrate the given function:

\[
x_2(t) = \int_0^t (2e^{-\tau} - 2 \cos \tau) d\tau.
\]

Breaking this into two parts:

\[
x_2(t) = 2 \int_0^t e^{-\tau} d\tau - 2 \int_0^t \cos \tau d\tau.
\]

For the first integral:

\[
\int e^{-\tau} d\tau = -e^{-\tau},
\]

Evaluating from 0 to \( t \):

\[
2 \int_0^t e^{-\tau} d\tau = 2 \left( -e^{-t} + 1 \right) = 2 - 2e^{-t}.
\]

For the second integral:

\[
\int \cos \tau d\tau =

Знайдіть зображення функцій при нульових початкових умовах: x1(t) = d/dt (2e^(-t) - 2cos(t)), x2(t) = ∫(0 to t)(2e^(-τ) - 2cos(τ)) dτ.

Solve a differential equation using Laplace transforms, covering the integral and derivative of exponential and trigonometric functions. Explore step-by-step solutions for x1(t) and x2(t) with zero initial conditions, suitable for undergraduate students.

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