Question

\(x_{1}{ }^{\prime}=x_{2}\) and \(x_{2}{ }^{\prime}=-25 x_{1}+2 \sin 4 t\), with \(x_{1}(0)=0\) and \(x_{2}(0)=2\) on \([0,2 \pi]\).

Short answer

The final solutions are functions combining trigonometric homogeneous and particular solutions constrained by initial conditions.

Step-by-step solution

Identify the System of Differential Equations

We have a system of differential equations given by:1. \( x_1' = x_2 \)2. \( x_2' = -25x_1 + 2\sin(4t) \)These equations represent a system where the derivative of \( x_1 \) depends on \( x_2 \), and the derivative of \( x_2 \) is influenced by \( x_1 \) and an external forcing function \( 2\sin(4t) \).

Apply Initial Conditions

The initial conditions provided are \( x_1(0) = 0 \) and \( x_2(0) = 2 \). These are essential for solving the differential equations as they provide the starting values for the system at time \( t = 0 \).

Solve the Homogeneous System

The homogeneous part of the system is derived by setting the external force to zero: \[x_2' = -25x_1\]Using the equation \( x_1' = x_2 \), the system can be rewritten (prioritizing substitution and elimination strategies) to eventually solve the characteristic equation for these linear equations.

Find the Characteristic Equation

Construct the characteristic equation from the homogeneous part: Assuming \( x_1(t) = e^{\lambda t} \), and substituting back into the equations, we find:\[x_2'(t) = -25 e^{\lambda t},x_1'(t) = \lambda e^{\lambda t} = x_2(t),x_2(t) = \lambda x_1(t)\]This leads us to the characteristic equation \( \lambda^2 = -25 \), hence \( \lambda = \pm 5i \).

Solve the Homogeneous Solution

With roots \( \lambda = \pm 5i \), the homogeneous solution is:\[x_1_h(t) = c_1\cos(5t) + c_2\sin(5t)\]\[x_2_h(t) = -5c_1\sin(5t) + 5c_2\cos(5t)\]

Solve the Non-Homogeneous System

Assume a particular solution of the form for \( x_1(t) \) and \( x_2(t) \) due to \( 2\sin(4t) \). Use an ansatz like \( x_1_p(t) = A\cos(4t) + B\sin(4t) \) and find \( A \) and \( B \) by substituting \( x_1(t) + x_2(t) \) according to the given non-homogeneous part.

Combine Solutions

Combine the homogeneous and particular solutions for \( x_1(t) \) and \( x_2(t) \) and apply initial conditions to find constants \( c_1 \) and \( c_2 \).

Apply the Initial Conditions

Use the initial conditions \( x_1(0) = 0 \) and \( x_2(0) = 2 \) to determine \( c_1 \) and \( c_2 \) in the general solution:\[x_1(0) = c_1 = 0\]\[x_2(0) = 5c_2 = 2 \Rightarrow c_2 = \frac{2}{5}\]

Final Solution

Substitute \( c_1 = 0 \) and \( c_2 = \frac{2}{5} \) back into the general solution for the functions:\[x_1(t) = \frac{2}{5}\sin(5t) + A\cos(4t) + B\sin(4t)\]\[x_2(t) = 5\cdot\frac{2}{5}\cos(5t) + A \cdot 4\sin(4t) - B \cdot 4\cos(4t)\]The constants \(A\) and \(B\) will depend on solving the system uniquely.

Key concepts

Initial Conditions

In solving differential equations, initial conditions are critical as they define the state of the system at a specific time point. In this exercise, the initial conditions are: \(x_1(0) = 0\) and \(x_2(0) = 2\). These conditions are applied when \(t = 0\).
These initial values allow us to calculate the constants in our general solution, ensuring it fits the specific solution that matches the behavior at the starting point.

• Without initial conditions, multiple solutions can satisfy the differential equations, making the process ambiguous.
• They provide a unique pathway to finalize the solution.

By applying these conditions, we ensure that the trajectory of the dynamic system is specifically tailored to start at the given point in its state space.

Characteristic Equation

The characteristic equation is fundamental in finding solutions to linear differential equations. It's derived from the homogeneous part of a system. In our case, the system is given by:\[x_1' = x_2, \ x_2' = -25x_1\] Setting the forcing function to zero, we assume solutions of the form \(x_1(t) = e^{\lambda t}\) and substitute back into our equations to find:\[x_2'(t) = -25 e^{\lambda t}, \ x_1'(t) = \lambda e^{\lambda t} = x_2(t), \ x_2(t) = \lambda x_1(t)\]This results in the characteristic equation \(\lambda^2 = -25\), with solutions \(\lambda = \pm 5i\).
These imaginary roots reflect oscillatory behavior, which is typical for systems modeling phenomena like springs or circuits.

• This equation provides the necessary exponents in the solution forms.
• Solving it determines the shape of the homogeneous solution.

Homogeneous Solution

A homogeneous solution is obtained by disregarding any external influences or forcing functions. For our exercise, this involves solving:\[x_1' = x_2, \ x_2' = -25x_1\] Using the characteristic equation roots \(\lambda = \pm 5i\), the homogeneous solutions for \(x_1(t)\) and \(x_2(t)\) are:\[x_1_h(t) = c_1\cos(5t) + c_2\sin(5t)\]\[x_2_h(t) = -5c_1\sin(5t) + 5c_2\cos(5t)\] These solutions represent the natural response of the system without external influence.

• They reflect how the system behaves internally.
• The coefficients \(c_1\) and \(c_2\) are determined by the initial conditions.

The homogeneous solution is a key component in constructing the overall solution, providing the structure for the system's response.

Non-Homogeneous System

A non-homogeneous system includes external influences or forcing functions, often leading to particular solutions that accommodate these forces. For the given system:\[x_1' = x_2, \ x_2' = -25x_1 + 2\sin(4t)\] We have an external forcing term \(2\sin(4t)\).
To tackle this, we propose particular solutions of forms \(x_1_p(t) = A\cos(4t) + B\sin(4t)\) and substitute into the original differential equations to solve for constants \(A\) and \(B\).

• This accounts for the system's externally driven behavior.
• The overall solution is a sum of homogeneous and particular solutions.

Understanding this concept helps interpret how systems react under external conditions, vital in engineering and physics.

System of Equations

A system of equations in the context of differential equations involves solving for multiple functions simultaneously. In this exercise:\[x_1' = x_2, \ x_2' = -25x_1 + 2\sin(4t)\]This is a coupled system where the rate of change of one function depends on another. Systems of equations allow the modeling of complex phenomena where multiple dynamic variables interact.

• The coupled nature means each equation can't be solved in isolation.
• Applying methods for systems, like matrix approaches, can aid in solving larger systems effectively.

By understanding how to analyze and solve these systems, you can model everything from population dynamics to electric circuits efficiently.