Problem 31 - Calculus: Early Transcendental Functions

Question

Write the second-order equation as a system of first-order equations. $$y^{\prime \prime}-(\cos x) y^{\prime}+x y^{2}=2 x$$

Step-by-Step Solution

Answer

The second-order equation can be written as a system of first-order equations as follows: \(y' = v; v' = \cos(x)v - xy^{2} + 2x\).

Step 1: Introduce a new variable

Introduce a new variable \(v\) such that \(v = y'\). In this manner, the given second order equation \(y'' - \cos(x)y' + xy^{2}=2x\) can be broken up. Now the revised version of the equation is \(v' - \cos(x)v + xy^{2}=2x\).

Step 2: Rewrite the given second order equation

Rewrite the given equation in terms of \(v\) and \(y'\), replacing \(y''\) with \(v'\). The equation transforms into: \(v' - \cos(x)v + xy^{2}=2x\)

Step 3: Formulate the system of equations

Form a system of first order equations using the equations \(y' = v\) and \(v' - \cos(x)v + xy^{2}=2x\). Therefore, the system of first order equations becomes: \(y' = v; v' = \cos(x)v - xy^{2} + 2x\).

Key Concepts

First-Order System

Understanding how to convert a second-order differential equation into a first-order system is a fundamental skill in the study of differential equations. A first-order system consists of equations that involve only the first derivative of the functions involved. To simplify complex second-order equations or to utilize numerical methods for finding solutions, it's often helpful to express them as a system of first-order equations.

In the given exercise, to convert the second-order equation into a system of first-order equations, we introduced a new variable, which transforms the original equation into two equations that only deal with first derivatives. By setting \(v = y'\), we effectively reduce the original higher-order equation to a system that is easier to analyze or simulate, and this technique is widely used in both theoretical and applied mathematics.

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. In applications, the functions generally represent physical quantities, the derivatives represent their rates of change, and the differential equation defines a relationship between the two. They are pivotal in engineering, physics, economics, and other disciplines.

The exercise we're looking at involves a second-order differential equation, which means it includes a second derivative. These equations can be more challenging to solve because they require knowledge of both calculus and differential equation techniques. By breaking them down into first-order systems, we make the task of finding a solution more manageable, especially since first-order differential equations are often simpler to handle both analytically and numerically.

Cosine Function

The cosine function is one of the basic functions encountered in trigonometry and appears frequently in different areas of mathematics, including calculus. Defined initially for the angle of a right triangle, the cosine function relates the angle to the ratio of the adjacent side over the hypotenuse.

In the context of differential equations, trigonometric functions like cosine often appear as coefficients, making the equations more complex due to their periodic nature. In our exercise, \(\cos x\) is the coefficient of \(y'\), which contributes to the complexity of the solving process. Understanding the properties of cosine, such as its periodicity and limits, can be crucial when seeking solutions to differential equations that contain trigonometric functions.

Calculus

Calculus is a branch of mathematics that studies continuous change and includes the study of limits, functions, derivatives, integrals, and infinite series. It's the language that describes dynamic phenomena and the driving force behind many scientific and engineering applications.

When facing a second-order differential equation, calculus tools become indispensable. They enable us to perform operations such as differentiation, which is the process of finding the derivative of a function, and integration, which, in the context of differential equations, can be used to find particular solutions. In solving our exercise, we use differentiation to introduce a new variable and turn the problem into a first-order system. This manipulation is steeped in the principles of calculus and showcases the power of calculus in solving complex mathematical challenges.