Problem 2 - Technical Mathematics with Calculus

Question

Find the general solution to each differential equation. $$2 y^{*}-5 y^{\prime}-3 y=0$$

Step-by-Step Solution

Answer

The general solution to the differential equation \(2y^{''}-5y'-3y=0\) is \(y = C_1 e^{3x} + C_2 e^{-\frac{1}{2}x}\) where \(C_1\) and \(C_2\) are arbitrary constants.

Step 1: Identify the Type of Differential Equation

The given differential equation \(2y^{''}-5y'-3y=0\) is a homogeneous linear second-order differential equation with constant coefficients. The standard form is \(a y^{''} + b y^{'} + c y = 0\).

Step 2: Write Down the Characteristic Equation

To find the solution, we start by forming the characteristic equation of the differential equation, which is obtained by replacing \(y^{''}\) with \(m^2\), \(y'\) with \(m\), and \(y\) with 1. The characteristic equation is \(2m^2 - 5m - 3 = 0\).

Step 3: Solve the Characteristic Equation

We need to find the roots of the characteristic equation. This can be done either by factoring, completing the square, or using the quadratic formula. The discriminant \(D\) is \(b^2 - 4ac = (-5)^2 - 4(2)(-3) = 25 + 24 = 49\), so there are two real distinct roots. Using the quadratic formula, \(m = \frac{-b \pm \sqrt{D}}{2a}\), we find the roots to be \(m_1 = \frac{5 + 7}{4} = 3\) and \(m_2 = \frac{5 - 7}{4} = -1/2\).

Step 4: Write Down the General Solution

Since the characteristic equation has two distinct real roots \(m_1\) and \(m_2\), the general solution to the differential equation is \(y = C_1 e^{m_1 x} + C_2 e^{m_2 x}\), where \(C_1\) and \(C_2\) are constants. Thus, the general solution is \(y = C_1 e^{3x} + C_2 e^{-\frac{1}{2}x}\).

Key Concepts

Homogeneous Linear Second-Order Differential Equation

Understanding a homogeneous linear second-order differential equation is vital to solving many problems in physics and engineering. It is characterized by an equation in the form of \(a y^{''} + b y^{'} + c y = 0\), where \(a\), \(b\), and \(c\) are constants, and the function \(y\) and its derivatives \(y'\) (first derivative) and \(y''\) (second derivative) depend on a variable, usually time or space.

In the homogeneous linear form, 'homogeneous' means that there is no term without \(y\) or its derivatives—all terms involve the unknown function. In the exercise given, \(2y^{''}-5y'-3y=0\) fits this form perfectly, with \(a=2\), \(b=-5\), and \(c=-3\). The solution to such equations typically involves exponentials with arguments that depend on the roots of the characteristic equation, which leads us to our next section.

Characteristic Equation

When faced with a homogeneous linear second-order differential equation, the characteristic equation plays a crucial role in finding the solution. By transforming the original equation, substituting \(y''\) with \(m^2\), \(y'\) with \(m\), and \(y\) with 1, you get a quadratic equation in the form \(am^2+bm+c=0\), which retains the coefficients of the original differential equation.

For the exercise in question, \(2y^{''}-5y'-3y=0\) leads to the characteristic equation \(2m^2 - 5m - 3 = 0\). Solving this is the gateway to finding the general solution of the differential equation. The roots of the characteristic equation represent the exponents in the final solution of the differential equation and determine the behavior of the system described by the differential equation, such as damping or oscillation patterns.

Quadratic Formula

The quadratic formula is a powerful tool for solving equations of the form \(ax^2+bx+c=0\), where \(a\), \(b\), and \(c\) are known values and \(x\) represents the unknown variable. It's given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) and offers a systematic way to find the roots of any quadratic equation, providing that \(a \eq 0\).

The discriminant \(D = b^2 - 4ac\) crucially determines the nature of the roots—when \(D > 0\), there are two distinct real roots; if \(D = 0\), one repeated real root; and for \(D < 0\), two complex roots. In our exercise, the calculated discriminant is \(49\), which means we have two real distinct roots. These roots are crucial for constructing the general solution to the differential equation.