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A linear second-order differential equation is one where the highest derivative is of second order, and the equation is linear in nature. This particular type is identified by the form \( ax'' + bx' + cx = f(t) \), where \( a, b, \) and \( c \) are constants, and \( f(t) \) is some function of \( t \).In our example, the equation is \( \ddot{x} + x = 3 \), which clearly fits this form:- The second derivative \( \ddot{x} \) relates linearly.- The term \( x \) is the dependent variable.- It's in second-order without a first derivative term present, making it linear in both variables. Remember that understanding the structure of the equation is key to solving it effectively.

For solving a linear second-order differential equation, it's essential first to handle the associated homogeneous equation. This is done by setting the right-hand side of the differential equation to zero. Thus, for \( \ddot{x} + x = 3 \), the homogeneous part is \( \ddot{x} + x = 0 \).This expressed equation is tackled through its characteristic equation which, in this case, becomes \( r^2 + 1 = 0 \). By solving this, we arrive at the roots \( r = \pm i \), which are complex. For these specific roots, the solution is typically expressed as:

• \( x_h(t) = A\cos(t) + B\sin(t) \)

A and B are constants that will be determined later using initial conditions. The homogeneous solution provides a foundation and understanding of system behaviors.

Finding a particular solution involves identifying a specific solution to the non-homogeneous differential equation, which includes the function on the right side of the equation. In our example, \( \ddot{x} + x = 3 \), this accounts for the constant \( 3 \).Given that the non-homogeneous part is a constant (\( 3 \)), it's logical to try a solution of the constant form \( x_p(t) = C \). Taking the derivatives of this form, the result remains 0, simplifying substitution back into the original equation:

• \( 0 + C = 3 \)

Solving this simple equation yields \( C = 3 \). Consequently, our particular solution becomes \( x_p(t) = 3 \). This solution addresses the external input or force within the system.

The initial conditions in this problem - given as \( x(\pi) = 1 \) and \( \dot{x}(\pi) = 2 \) - are crucial for determining the particular values of constants \( A \) and \( B \) within the general solution. They transform the general solution into a specific one tailored to the problem's context.Starting with the general solution \( x(t) = A\cos(t) + B\sin(t) + 3 \), substitute \( t = \pi \) to adhere to the initial condition:

• \( 1 = -A + 3 \)

From which \( A = 2 \) is derived.Working with the derivative, we extract \( \dot{x}(t) = -A\sin(t) + B\cos(t) \) and likewise apply the initial condition:

• \( 2 = -B \)

Establishing that \( B = -2 \). These initial conditions ensure that the equation models the behavior accurately as prescribed by the starting conditions.