91Ó°ÊÓ

Non-homogeneous differential equations are equations that have a non-zero term independent of the function and its derivatives. In simple words, these equations include an external force or input impacting the system. This can be represented as:

• General form: \( a_n x^{(n)} + a_{n-1} x^{(n-1)} + ... + a_1 x' + a_0 x = g(t) \), where \( g(t) e 0 \)

In our problem, the term \( 20 \, \text{δ}(t - \frac{\require{enclose} π}{2}) \) represents an external input at a specific time \( t = \frac{\require{enclose} π}{2} \).

The presence of this delta function makes it necessary to handle the differential equation differently compared to homogeneous equations, where \( g(t) = 0 \).

The delta function, denoted as \( \text{δ}(t - t_0) \), is a mathematical function that is zero everywhere except at \( t = t_0 \). It's also known as the Dirac delta function.

• Properties:
  • \( \text{δ}(t - t_0) = 0 \) for \( t e t_0 \)
  • \( \text{∫_{-∞}^{∞} \text{δ}(t - t_0) \, dt = 1} \)

In our exercise, the delta function \( 20 \, \text{δ}(t - \frac{\require{enclose} π}{2}) \) indicates an impulse of magnitude 20 occurring at time \( t = \frac{\require{enclose} π}{2} \). This causes a sudden change in the system only at that point in time, leading to a discontinuity in the solution.

To solve a homogeneous differential equation, we first need to find the characteristic equation. This is obtained by assuming a solution of the form \( x(t) = e^{rt} \) and substituting it into the homogeneous equation \( x'' + 4x' + 13x = 0 \). This leads to a polynomial equation in terms of \( r \):

• Characteristic polynomial: \( r^2 + 4r + 13 = 0 \)

Solving the characteristic equation, we find the roots \( r = -2 \, \text{±} \, 3i \).

This gives us the general solution for the homogeneous part as:\( x_h(t) = e^{-2t}(C_1 \, \text{cos}(3t) + C_2 \, \text{sin}(3t)) \), where \( C_1 \) and \( C_2 \) are constants determined by initial conditions.

Initial conditions are values given at a specific point (usually \( t = 0 \)) to find the specific solution of the differential equation. They help in determining the constants in the general solution of homogeneous equations.

• Our exercise provides the initial conditions:
• \( x(0) = 0 \)
• \( x'(0) = 10 \)

By substituting these values into our general solution \( x_h(t) \), we can solve for \( C_1 \) and \( C_2 \).

Substituting \( x(0) = 0 \) into the general solution yields:\( C_1 = 0 \).
Substituting \( x'(0) = 10 \) leads to:\( -2C_1 + 3C_2 = 10 \). Solving this, we find \( C_2 = \frac{10}{3} \).

The impulse response method is used to solve differential equations with a delta function. These types of problems often involve a sudden change or 'impulse' at a specific time. The delta function (\( \text{δ}(t - t_0) \)) creates a discontinuity in the derivative.

• Magnitude of impulse: \( \frac{d}{dt}x(t) \big|^{t=t_0^+}_{t=t_0^-} = \text{Magnitude of impulse} \)

In our exercise,\( \frac{d}{dt}x(t) \big|^{t=\frac{\require{enclose} π}{2}^+}_{t=\frac{\require{enclose} π}{2}^-} = 20 \).
This means there's an immediate change in \( x'(t) \) at \( t = \frac{\require{enclose} π}{2} \) worth 20 units.

Combining the homogeneous solution with the impulse response, we adjust the solution to accommodate the jump at \( t = \frac{\require{enclose} π}{2} \). This gives us the complete solution for the original non-homogeneous differential equation.