Green's Function Method

The problem asks for the solution of a linear second-order non-homogeneous differential equation with given initial conditions, using the Green's function method. The differential equation is:

\(\frac{d^2x}{dt^2} + x = f(t)\)

The initial conditions are \(x(0) = 0\) and \(\frac{dx}{dt}(0) = 0\). The source function is given as \(f(t) = \theta(t)\), which is the Heaviside step function.

The Green's function \(G(t, \tau)\) for this differential equation satisfying homogeneous initial conditions is provided as:

\(G(t, \tau) = \begin{cases}0 & \text { for } \quad 0<\rm t<\tau \\ \sin (\rm t−\tau) & \text { for } \quad \rm t>\tau\end{cases}\)

Solution using Green's Function

For a linear non-homogeneous differential equation of the form \(L[x(t)] = f(t)\) with homogeneous initial conditions, the solution \(x(t)\) can be found using the convolution integral involving the Green's function and the source function:

\(x(t) = \int_0^t G(t, \tau) f(\tau) d\tau\)

In this problem, \(f(\tau) = \theta(\tau)\). The Heaviside step function \(\theta(\tau)\) is defined as:

• \(\theta(\tau) = 0\) for \(\tau < 0\)
• \(\theta(\tau) = 1\) for \(\tau > 0\)

Since the integration limit is from \(0\) to \(t\), we are considering values of \(\tau \ge 0\). For \(\tau\) in the range \(0 < \tau \le t\), \(\theta(\tau) = 1\). Thus, \(f(\tau) = 1\) for the relevant integration range.

The Green's function \(G(t, \tau)\) depends on the relationship between \(t\) and \(\tau\). In the integral \(\int_0^t G(t, \tau) f(\tau) d\tau\), the variable \(\tau\) goes from \(0\) up to \(t\). This means for any \(\tau\) within the integration range \(0 \le \tau \le t\), we have \(t \ge \tau\). The Green's function definition relevant for this range is \(G(t, \tau) = \sin(t-\tau)\) for \(t > \tau\). If \(t=\tau\), \(\sin(t-\tau)=0\), and the integral includes this boundary naturally.

Substituting \(G(t, \tau) = \sin(t-\tau)\) and \(f(\tau) = 1\) into the integral formula for \(x(t)\):

\(x(t) = \int_0^t \sin(t-\tau) \cdot 1 \, d\tau\)

\(x(t) = \int_0^t \sin(t-\tau) \, d\tau\)

Evaluating the Integral

To evaluate the definite integral, we can use a substitution. Let \(u = t-\tau\). Then, the differential \(du = -d\tau\).

We need to change the integration limits according to the substitution:

• When \(\tau = 0\), \(u = t - 0 = t\).
• When \(\tau = t\), \(u = t - t = 0\).

The integral becomes:

\(x(t) = \int_t^0 \sin(u) (-du)\)

We can reverse the limits of integration by changing the sign:

\(x(t) = -\int_t^0 \sin(u) \, du = \int_0^t \sin(u) \, du\)

Now, we integrate \(\sin(u)\):

\(\int \sin(u) \, du = -\cos(u)\)

Applying the limits of integration:

\(x(t) = [-\cos(u)]_0^t\)

\(x(t) = -\cos(t) - (-\cos(0))\)

\(x(t) = -\cos(t) + \cos(0)\)

Since \(\cos(0) = 1\):

\(x(t) = -\cos(t) + 1\)

So, the solution is \(x(t) = 1 - \cos(t)\).

Comparing with Options

The derived solution \(x(t) = 1 - \cos(t)\) matches one of the given options.

• Option 1: \(\sin t\)
• Option 2: \(1 - \sin t\)
• Option 3: \(1 - \cos t\)
• Option 4: \(\cos 2t - 1\)

Our solution \(1 - \cos t\) corresponds to Option 3.