Question

Consider the differential equation \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=g_{1}(t)+i g_{2}(t)\), where \(p(t)\), \(q(t), g_{1}(t)\), and \(g_{2}(t)\) are all real-valued functions continuous on some \(t\)-interval of interest. Assume that \(y_{P}(t)\) is a particular solution of this equation. Generally, \(y_{P}(t)\)

Short answer

Answer: To find the particular solution, \(y_P(t)\), we should follow these steps: 1. Analyze the given differential equation and separate the real and imaginary parts. 2. Assume a particular solution that also has real and imaginary parts, i.e., \(y_P(t) = u(t) + v(t)i\), where \(u\) and \(v\) are real-valued functions. 3. Substitute \(y_P(t)\), \(y_P'(t)\), and \(y_P''(t)\) into the given differential equation. 4. Equate the real and imaginary parts separately, forming two second-order linear inhomogeneous differential equations for \(u(t)\) and \(v(t)\). 5. Solve these two real-valued differential equations for \(u(t)\) and \(v(t)\) separately. 6. Finally, combine the solutions for \(u(t)\) and \(v(t)\) to form the particular solution \(y_P(t) = u(t) + v(t)i\).

Step-by-step solution

Analyze the given differential equation

We have the given differential equation, $$y^{\prime \prime}+p(t) y^{\prime}+q(t) y=g_{1}(t)+i g_{2}(t),$$ where \(p(t)\), \(q(t)\), \(g_1(t)\), and \(g_2(t)\) are all real-valued continuous functions with respect to \(t\). The equation is a second-order linear inhomogeneous differential equation with complex-valued non-homogeneous term. Our goal is to find a particular solution, \(y_P(t)\), for this equation. Note: Remember that \(i\sqrt{-1}\) is the imaginary unit.

Separate the real and imaginary parts of the equation

We can rewrite the given differential equation by separating the real and imaginary parts as follows: $$y^{\prime \prime}+p(t) y^{\prime}+q(t) y=g_1(t) + (g_2(t)i).$$ Then, we will assume that the particular solution, \(y_P(t)\), also has both real and imaginary parts: $$y_P(t) = u(t) + v(t)i,$$ where \(u(t)\) and \(v(t)\) are real-valued functions representing the real and imaginary parts of the particular solution.

Substitute \(y_P(t)\) into the differential equation

Next, we will substitute our assumed form of \(y_P(t)\) into the differential equation. To do this, we first need to find \(y_P'(t)\) and \(y_P''(t)\): $$y_P'(t) = u'(t) + v'(t)i$$ $$y_P''(t) = u''(t) + v''(t)i$$ Now, substitute \(y_P(t)\), \(y_P'(t)\), and \(y_P''(t)\) into the given differential equation: $$u''(t) + v''(t)i + p(t)(u'(t) + v'(t)i) + q(t)(u(t) + v(t)i) = g_1(t) + g_2(t)i$$

Equate real and imaginary parts

Equate the real and imaginary parts of the equation separately: Real part: $$u''(t) + p(t)u'(t) + q(t)u(t) = g_1(t)$$ Imaginary part: $$v''(t) + p(t)v'(t) + q(t)v(t) = g_2(t)$$ Both equations are second-order linear inhomogeneous differential equations with real-valued functions.

Solve the real and imaginary parts separately

Solve the real-valued differential equations for \(u(t)\) and \(v(t)\): 1. Solve for \(u(t)\): $$u''(t) + p(t)u'(t) + q(t)u(t) = g_1(t)$$ 2. Solve for \(v(t)\): $$v''(t) + p(t)v'(t) + q(t)v(t) = g_2(t)$$

Combine the real and imaginary parts

Once we have found the solutions for \(u(t)\) and \(v(t)\) from Step 5, we can combine them to obtain the particular solution \(y_P(t)\) for the original complex-valued differential equation: $$y_P(t) = u(t) + v(t)i$$ With the particular solution found, we have successfully broken down the exercise into steps for finding a particular solution of a given second-order linear inhomogeneous differential equation with complex-valued non-homogeneous term.

Key concepts

Second-order Linear Differential Equations

Second-order linear differential equations involve a second derivative of a function and are expressed in the form \(y'' + p(t)y' + q(t)y = g(t)\). Here, \(p(t)\) and \(q(t)\) are functions of \(t\), while \(g(t)\) is any function that forces the system (called a non-homogeneous term).
Second-order refers to the highest derivative involved in the equation.
This type of equation is commonly used in various scientific and engineering fields, such as physics and control systems.

• Linear: This implies linearity in terms of the unknown function and its derivatives, with no products or powers involved.
• Homogeneous vs. Non-Homogeneous: If \(g(t)\) equals zero, the equation is homogeneous. Otherwise, it is non-homogeneous.
• Complexity: When these equations include complex terms, they require specific solutions that account for real and imaginary components.

Understanding how to work with these equations can help solve various problems related to natural phenomena and engineering constructs.

Particular Solution

Finding a particular solution to a second-order linear differential equation is key to solving the equation fully. A particular solution \(y_P(t)\) satisfies the equation \(y'' + p(t)y' + q(t)y = g(t)\) for the given non-homogeneous term \(g(t)\).
A particular solution is not concerned with the homogeneous part, which is composed of solutions to the corresponding homogeneous equation. Instead, it focuses on providing a specific function that satisfies the entire equation with the non-homogeneous component.

• Process: The objective is to find a function that, when substituted into the differential equation, balances both sides.
• Special Techniques: Various methods such as undetermined coefficients and variation of parameters can be employed to identify this specific solution.
• Combination: The general solution includes both the particular solution and the homogeneous solution.

Understanding and finding a particular solution can provide insights into the specific effects that external forces or inputs have on a system.

Imaginary Unit in Differential Equations

When dealing with differential equations, the imaginary unit \(i\) (where \(i = \sqrt{-1}\)) can appear, especially in equations modeling oscillatory behaviors such as AC circuits or wave functions.
In the given problem, the imaginary unit is used in the non-homogeneous term \(g_1(t) + i g_2(t)\). This kind of term introduces a complex component that requires special attention when solving the equation.

• Meaning: The imaginary unit suggests that part of the input or effect on the system has a phase difference or oscillates in some manner.
• Handling: Solutions involve treating the imaginary part separately from the real part, using methods that maintain the integrity of complex numbers.
• Applications: Often appears in engineering and physics problems involving alternating phenomena or outputs dependent on sinusoidal inputs.

Mastery in comfortably using the imaginary unit within differential equations allows for sophisticated modeling of real-world physical behaviors.

Real and Imaginary Parts Separation

In the context of complex differential equations, effectively separating the real and imaginary parts is a common strategy to simplify the problem. This separation process turns a complex problem into two simpler, independent, real-valued equations.
For our initial differential equation \(y'' + p(t)y' + q(t)y = g_1(t) + i g_2(t)\), integrating the complex function \(y_P(t) = u(t) + v(t)i\) allows for the division into two distinct equations based on real and imaginary components:

• Real Equation: \(u''(t) + p(t)u'(t) + q(t)u(t) = g_1(t)\)
• Imaginary Equation: \(v''(t) + p(t)v'(t) + q(t)v(t) = g_2(t)\)

This separation ensures solving straightforward differential equations that are more manageable, as they involve only real-valued functions.
This method is crucial for efficiently finding solutions in systems influenced by complex inputs, thereby breaking down intricate problems into simple, solvable components.