91Ó°ÊÓ

Differential equations are equations that involve an unknown function and its derivatives. They play a crucial role in various fields like physics, engineering, economics, and biology, modeling everything from the growth rate of populations to the motion of planets.
A differential equation describes a relationship between functions and their rates of change. In simple terms, it tells how a function changes over time or space.
For example, the differential equation given in our exercise is \(x''' = -e^t + 4(t+1)^{-3}\). This is a third-order differential equation because it involves the third derivative \(x'''\). Here, \(-e^t + 4(t+1)^{-3}\) defines how the system changes, driven by specific functions of time \(t\).
Solving such equations reveals the underlying function \(x(t)\) that satisfies the given change criteria.

A homogeneous solution is part of the process of solving a differential equation. It's what you get by solving the equation when you set the non-homogeneous part (the term without the derivative) to zero.
In our case, solving the homogeneous equation means dealing with \(x''' = 0\). This simplifies our equation because we're only considering the derivative terms.
The characteristic equation for a homogeneous differential equation helps find possible solutions. For the equation \(r^3 = 0\), the root is repeated: \(r = 0\).
This gives us the homogeneous solution: \(x_h(t) = C_1 + C_2 t + C_3 t^2\). Here, \(C_1\), \(C_2\), and \(C_3\) are constants that will be determined by further conditions like boundary values.

A particular solution aims to satisfy the entire original non-homogeneous differential equation, including terms without derivatives. This step finds the additional specific functions that make the equation true.
In our exercise, we consider \(-e^t + 4(t+1)^{-3}\) separately.

• For \(-e^t\), a particular solution can be guessed as \(x_{p1} = Ae^t\). Solving yields \(x_{p1} = -e^t\).
• For the term \(4(t+1)^{-3}\), an intelligent guess \((t+1)^{-2}\) fits well with the differential structure. Solving these gives \(x_{p2} = 4(t+1)^{-2}\).

Combining both, the particular solution is \(x_p(t) = -e^t + 4(t+1)^{-2}\). This is specific to how non-homogeneous terms influence the system.

Boundary conditions are additional constraints provided to narrow down the solution to a specific context. They are crucial in determining the constants of integration in differential equations.
For the given problem, the boundary conditions are:

• \(x(0) = -1\)
• \(x(1) = 3-e+2\ln 2\)
• \(x(2) = 6-e^2+2\ln 3\)

To apply these conditions, substitute each into the general solution \(x(t) = C_1 + C_2 t + C_3 t^2 - e^t + 4(t+1)^{-2}\).
For instance, using \(x(0) = -1\), you get \(C_1 - 1 + 4 = -1\), solving for \(C_1\). The process is similar for \(x(1)\) and \(x(2)\), creating a system of equations in terms of \(C_2\) and \(C_3\).
This systematic approach helps pinpoint the right values of \(C_1, C_2, C_3\), uniquely defining the solution.