Problem 7 - A First Course in Differential Equations with Modeling Applications

Question

Find the general solution of the given second-order differential equation. $$12 y^{\prime \prime}-5 y^{\prime}-2 y=0$$

Step-by-Step Solution

Answer

The general solution is: \(y(t) = C_1 e^{\frac{2}{3}t} + C_2 e^{-\frac{1}{4}t}\).

Step 1: Identify the type of differential equation

The given equation is a linear homogeneous second-order differential equation with constant coefficients.

Step 2: Write the characteristic equation

For a second-order differential equation of the form \(ay'' + by' + cy = 0\), the characteristic equation is \(ar^2 + br + c = 0\). Here, it becomes \(12r^2 - 5r - 2 = 0\).

Step 3: Solve the characteristic quadratic equation

The characteristic equation \(12r^2 - 5r - 2 = 0\) is quadratic, so we solve it using the quadratic formula: \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute \(a = 12\), \(b = -5\), and \(c = -2\).

Step 4: Calculate the discriminant

The discriminant of the characteristic equation is given by \(b^2 - 4ac = (-5)^2 - 4(12)(-2) = 25 + 96 = 121\).

Step 5: Find the roots of the characteristic equation

Applying the quadratic formula, \[ r = \frac{-(-5) \pm \sqrt{121}}{2 \times 12} = \frac{5 \pm 11}{24} \]. So the roots are \(r_1 = \frac{16}{24} = \frac{2}{3}\) and \(r_2 = \frac{-6}{24} = -\frac{1}{4}\).

Step 6: Write the general solution

Since the roots of the characteristic equation, \(r_1 = \frac{2}{3}\) and \(r_2 = -\frac{1}{4}\), are real and distinct, the general solution of the differential equation is \(y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}\). Substitute the values of \(r_1\) and \(r_2\) to get: \[ y(t) = C_1 e^{\frac{2}{3}t} + C_2 e^{-\frac{1}{4}t} \].

Key Concepts

Linear Homogeneous Equations

A linear homogeneous equation is a specific type of differential equation where all the terms are proportional to the function or its derivatives. In a second-order differential equation like in the example, each term involves the unknown function, its first derivative, or its second derivative, and there are no free-standing constants. The term 'homogeneous' indicates that the equation equates to zero. This form is represented as:

• \( a y'' + b y' + c y = 0 \)

Here, there are no external functions or constants added, except for equal to zero, making the equation linear and homogeneous. Understanding this classification is crucial as it defines the equation’s characteristic method of solution, where such an equation relies heavily on characteristics of polynomials to solve.

Characteristic Equation

The characteristic equation arises from transforming a second-order linear homogeneous differential equation into a regular algebraic form. This is done by assuming a solution of the exponential form. For our given equation, it's of the type:

• \( 12y'' - 5y' - 2y = 0 \)

To find solutions, we propose \( y = e^{rt} \), where \( r \) is a constant, and derive the characteristic equation as:

• \( ar^2 + br + c = 0 \)

Substituting the values \( a = 12 \), \( b = -5 \), and \( c = -2 \), we formulate \( 12r^2 - 5r - 2 = 0 \). This quadratic form is essential as the roots \( r \) determine the nature and form of the equation's solution.

Quadratic Formula

The quadratic formula is a reliable tool for solving quadratic equations like the characteristic equation. For expressions of the form \( ax^2 + bx + c = 0 \), it finds the roots based on the coefficients \( a \), \( b \), and \( c \). It is expressed as:

• \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula helps in calculating the roots for our characteristic equation \( 12r^2 - 5r - 2 = 0 \). By inputting values:

• \( a = 12 \), \( b = -5 \), and \( c = -2 \),
• calculate the discriminant \( b^2 - 4ac \).

Here, the discriminant, \( 121 \), is positive, implying two distinct real roots exist, which are yielded by the quadratic formula.

General Solution

The general solution of the differential equation is determined from the distinct real roots obtained from the characteristic equation. The roots, \( r_1 = \frac{2}{3} \) and \( r_2 = -\frac{1}{4} \), yield a solution in the form:

• \( y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \)

This formula reflects the general solution for linear homogeneous differential equations with real and distinct roots. The arbitrary constants \( C_1 \) and \( C_2 \) represent the general nature of the solution, indicating the equation possesses a family of potential solutions. Each specific solution is determined by initial or boundary conditions that are typically applied in practical scenarios, ensuring the solution meets particular real-world requirements.