Question

Consider the initial value problem \(u^{\prime \prime}+p(t) u^{\prime}+q(t) u=g(t), u(0)=u_{0}, u^{\prime}(0)=u_{0}^{\prime}\) Transform this problem into an initial value problem for two first order equations.

Short answer

Question: Transform the given initial value problem (IVP) with the equation \(u^{\prime\prime} + p(t)u^{\prime} + q(t)u = g(t)\), and initial conditions \(u(0) = u_0\) and \(u^{\prime}(0) = u_0^{\prime}\), into an initial value problem for two first-order equations. Answer: The transformed initial value problem for two first-order equations is: First-order equation 1: \(u^{\prime}(t) = v(t)\) Initial condition 1: \(u(0) = u_0\) First-order equation 2: \(v^{\prime}(t) = -p(t)v(t) - q(t)u(t) + g(t)\) Initial condition 2: \(v(0) = u_{0}^{\prime}\)

Step-by-step solution

Introduce a new variable v

Let's introduce a new variable \(v(t)\) such that \(v = u^{\prime}(t)\). Now, we can differentiate \(v(t)\) with respect to \(t\) and write \(v^{\prime}(t) = u^{\prime\prime}(t)\)

Write two first-order equations in terms of u and v

Now we rewrite the given equation in terms of the new variable \(v(t)\): 1. \(u^{\prime}(t) = v(t)\) 2. \(v^{\prime}(t) = u^{\prime\prime}(t) = -p(t)u^{\prime}(t) - q(t)u(t) + g(t)\) Substitute \(u^{\prime}(t) = v(t)\) in equation 2: \(v^{\prime}(t) = -p(t)v(t) - q(t)u(t) + g(t)\)

Initial Conditions

Now we rewrite the initial conditions in terms of the new variable \(v(t)\): 1. \(u(0) = u_0\) 2. \(v(0) = u^{\prime}(0) = u_{0}^{\prime}\) Now, we have transformed the initial value problem into an initial value problem for two first-order equations: First-order equation 1: \(u^{\prime}(t) = v(t)\) Initial condition 1: \(u(0) = u_0\) First-order equation 2: \(v^{\prime}(t) = -p(t)v(t) - q(t)u(t) + g(t)\) Initial condition 2: \(v(0) = u_{0}^{\prime}\)

Key concepts

Differential Equations

Differential equations are mathematical constructs that provide a way to describe the relationships between functions and their rates of change. These equations are foundational in expressing natural phenomena in disciplines ranging from physics to economics. A differential equation typically consists of derivatives, which represent rates of change, and one or more functions that are subject to these changes.

For instance, in physics, differential equations model the motion of particles, while in biology they can represent how populations evolve over time. The versatility and applicability of differential equations make them a critical area of study in mathematics and its applications.

First-Order Differential Equations

First-order differential equations are a particular set of differential equations wherein the highest derivative present is of the first order. These equations often take the form \( y' = f(x, y) \), which represents a rate of change of the variable \( y \) with respect to the variable \( x \), and \( f \) is some function in terms of \( x \) and \( y \).

These types of equations are simpler to solve compared to higher-order differential equations and are widely used to model processes that depend on a single rate of change, such as cooling/heating processes, simple harmonic motion, or population growth.

Initial Conditions

Initial conditions are specific values assigned to the function and its derivatives at the start of the problem, which is usually at \( t=0 \). They serve as a set of constraints that make the solution to a differential equation unique. Without initial conditions, a differential equation can have multiple solutions. But with initial conditions, we can pinpoint a particular solution that fits the context of the problem.

For example, knowing the initial position and velocity of a particle allows us to solve the differential equation governing its motion, predicting its future position and velocity at any given time. Initial conditions are critical for accurately modeling and understanding real-world scenarios.

Transforming Higher-Order Equations

Higher-order differential equations involve derivatives of second-order or above. Solving these complex equations directly can often be challenging. A common technique is to transform a higher-order differential equation into a system of first-order differential equations. This process simplifies the original equation because first-order differential equations are easier to manage.

In the context of the initial value problem provided, a second-order differential equation is transformed by introducing a new variable \( v(t) = u'(t) \). The original equation then breaks down into a system of two first-order differential equations, with each equation expressing a different aspect of the problem's dynamics. By doing so, we can apply well-established methods to solve or analyze the first-order system and thereby solve the original higher-order problem.