Problem 5 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Solve the differential equations. \(y^{\prime \prime}+5 y^{\prime}=0\)

Step-by-Step Solution

Answer

The general solution to the differential equation \(y^{\prime \prime}+5 y^{\prime}=0\) is \(y = c_1 + c_2 e^{-5x}\), with \(c_1, c_2\) as arbitrary constants.

Step 1: Recognizing the Differential Equation

The given differential equation is of the form \((D^2 + 5D)y = 0\), where \(D\) represents the differential operator, \(d/dx\).

Step 2: Formulate the Characteristic Equation

The characteristic equation associated with this differential equation is \(m^2 + 5m = 0\), where \(m\) represents the roots of the equation.

Step 3: Solve the Characteristic Equation

Solving the characteristic equation gives us the roots \(m = 0\) and \(m = -5\), obtained by factoring the equation as \(m(m + 5) = 0\).

Step 4: Construct the General Solution

With the roots of the characteristic equation, we can construct the general solution to the differential equation. This is of the form \(y = c_1 e^{m1x} + c_2 e^{m2x}\), where \(c_1, c_2\) are arbitrary constants, and \(m1, m2\) are roots of the characteristic equation.

Step 5: Substitute for the Roots

Substituting \(m1 = 0\) and \(m2 = -5\), we have \(y = c_1 e^{0x} + c_2 e^{-5x} = c_1 + c_2 e^{-5x}\). This is the general solution to the given differential equation.

Key Concepts

Characteristic Equation

The characteristic equation plays a crucial role in solving linear homogeneous differential equations, such as the one given in our exercise, \(y''+5y'=0\). This algebraic equation is created by assuming a solution of the form \(e^{mx}\) for the differential equation, where \(m\) is an unknown that needs to be solved for. The coefficients of \(m\) are derived from the coefficients of the differential equation.

When we replace the derivatives in the differential equation with powers of \(m\), what results is the characteristic equation, \(m^2+5m=0\). It is essentially an eigenvalue problem, where we are looking for values of \(m\) that make the equation true, leading us to the roots of the equation. By solving the characteristic equation, we gain the necessary information to outline the general structure of the solution to the differential equation.

Differential Operator

In the context of our exercise, the differential operator, denoted by \(D\), is a shorthand notation representing the derivative with respect to \(x\), specifically \(\frac{d}{dx}\). When applied to a function, it yields the derivative of that function. For example, \(Dy\) is equivalent to \(y'\), and \(D^2y\) to \(y''\), the second derivative of \(y\).

The use of a differential operator simplifies the process of working with differential equations. It allows us to manipulate derivatives algebraically, as demonstrated in the solution procedure for this problem. By utilizing the differential operator, the original differential equation is transformed into a more manageable form, paving the way toward finding the characteristic equation.

General Solution of Differential Equation

The general solution of a differential equation encompasses all possible solutions for the given equation. It is represented in a form that contains arbitrary constants, which can later be determined given initial conditions or boundary values.

In the example exercise, once the characteristic equation has been solved, and the roots \(m=0\) and \(m=-5\) have been found, these are used to construct the general solution, \(y = c_1 e^{m1x} + c_2 e^{m2x}\). This structure demonstrates that the general solution to a linear homogeneous second-order differential equation is a linear combination of exponential functions raised to the power of the roots of the characteristic equation, multiplied by arbitrary constants \(c_1\) and \(c_2\). These constants are placeholders that allow the solution to be tailored to specific initial conditions or other constraints.

Roots of the Equation

The roots of the characteristic equation are the values of \(m\) that satisfy the equation. In our exercise, the roots can be found by setting \(m^2+5m=0\) and solving for \(m\). In this case, we factor out an \(m\) to get \(m(m+5)=0\), indicating that \(m=0\) and \(m=-5\) are the solutions to the characteristic equation.

The significance of finding the roots lies in their direct impact on the form of the general solution. Each root translates to a component of the solution, \(e^{m1x}\) or \(e^{m2x}\), that corresponds to a particular behavior of the system described by the differential equation. The roots can be real or complex, and their nature (distinct, repeated, or complex conjugates) affects the structure of the general solution. In our example, the real roots lead to exponential functions that outline the dynamic response of the system over time.