Problem 16 - Calculus Volume 3

Question

Find the general solution to the linear differential equation. $$ 3 y^{\prime \prime}-14 y^{\prime}+8 y=0 $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Differential Equation

The given differential equation is a second-order linear homogeneous differential equation with constant coefficients, expressed as \(3y^{\prime \prime} - 14y^{\prime} + 8y = 0\).

Step 2: Write the Characteristic Equation

To solve the differential equation, we write the characteristic equation, which is obtained by assuming a solution of the form \(y = e^{rx}\). This leads to the characteristic equation: \(3r^2 - 14r + 8 = 0\).

Step 3: Solve the Characteristic Equation

We solve the quadratic characteristic equation \(3r^2 - 14r + 8 = 0\) using the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), with \(a = 3\), \(b = -14\), and \(c = 8\).

Step 4: Calculate the Discriminant

First, calculate the discriminant: \(b^2 - 4ac = (-14)^2 - 4 \times 3 \times 8 = 196 - 96 = 100\).

Step 5: Find the Roots

Since the discriminant is positive, there are two distinct real roots. Substitute into the quadratic formula: \[r = \frac{14 \pm \sqrt{100}}{6}\] which simplifies to \(r = \frac{14 \pm 10}{6}\).

Step 6: Simplify the Roots

Calculate the roots: \(r_1 = \frac{24}{6} = 4\) and \(r_2 = \frac{4}{6} = \frac{2}{3}\). Thus, \(r_1 = 4\) and \(r_2 = \frac{2}{3}\).

Step 7: Write the General Solution

The general solution of the differential equation, based on the roots, is given by \(y(x) = C_1 e^{4x} + C_2 e^{\frac{2}{3}x}\), where \(C_1\) and \(C_2\) are arbitrary constants.

Key Concepts

Characteristic Equation

A characteristic equation is a method used to solve linear differential equations. In our example, the differential equation is \(3y'' - 14y' + 8y = 0\). This is a second-order linear homogeneous differential equation.
To transition from the differential equation to the characteristic equation, we assume a solution of the form \(y = e^{rx}\). This simplifies the process because when you take derivatives, terms involving \(e^{rx}\) will reappear.
Replacing into the differential equation,

• the second derivative \(y''\) becomes \(r^2 e^{rx}\),
• the first derivative \(y'\) becomes \(r e^{rx}\),
• and \(y\) itself remains as \(e^{rx}\).

Thus, the equation \(3r^2 e^{rx} - 14r e^{rx} + 8 e^{rx} = 0\) simplifies to the characteristic equation: \(3r^2 - 14r + 8 = 0\). Notice how the exponential factor \(e^{rx}\) can be factored out because it never equals zero, simplifying our task to finding the roots of the quadratic equation.

Quadratic Formula

The quadratic formula is crucial for solving quadratic equations such as \(3r^2 - 14r + 8 = 0\). The formula is given by:\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \(a\), \(b\), and \(c\) are the coefficients from the quadratic equation \(ax^2 + bx + c = 0\). For our characteristic equation, \(a = 3\), \(b = -14\), and \(c = 8\).
Calculating the discriminant \(b^2 - 4ac\) gives insight into the nature of the roots. Here, \((-14)^2 - 4 \times 3 \times 8 = 100\). Since 100 is positive, there are two distinct real roots.
Substituting back into the quadratic formula, we calculate \(r = \frac{14 \pm 10}{6}\).
The solutions are \(r_1 = \frac{24}{6} = 4\) and \(r_2 = \frac{4}{6} = \frac{2}{3}\). These roots are essential for constructing the general solution of the differential equation.

General Solution of Differential Equations

The general solution of a differential equation combines all possible solutions. For second-order linear differential equations like ours, the general solution takes a specific form depending on the roots of the characteristic equation. Here, the roots \(r_1 = 4\) and \(r_2 = \frac{2}{3}\) are distinct and real, so the solution is given by:

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

Thus, substituting our roots, we get:\[ y(x) = C_1 e^{4x} + C_2 e^{\frac{2}{3}x} \] where \(C_1\) and \(C_2\) are arbitrary constants determined by initial conditions.
This solution represents a family of curves, each curve describing a possible solution for different initial conditions. This flexibility is why we use arbitrary constants.
In practice, determining specific values for these constants would require additional information, such as initial position or velocity.