Question

Write the equation $$ y^{\prime \prime}+p y^{\prime}+q y=0, $$

Short answer

1. (\(D > 0\)): The equation has two distinct real roots \(r_1\) and \(r_2\). The general solution is then \(y(x) = C_1e^{r_1x} + C_2e^{r_2x}\), where \(C_1\) and \(C_2\) are constants to be determined from the initial conditions. 2. (\(D = 0\)): The equation has one real root \(r_1 = r_2\). The general solution becomes \(y(x) = (C_1 + C_2x)e^{r_1x}\). 3. (\(D < 0\)): The equation has complex conjugate roots \(r_1 = \alpha + \beta i\) and \(r_2 = \alpha - \beta i\). The general solution can be written as \(y(x) = e^{\alpha x}(C_1\cos(\beta x) + C_2\sin(\beta x))\).

Step-by-step solution

Write the Equation in Terms of the Guessing Function

We start by substituting our guess, \(y = e^{rx}\), into the given equation. To do this, we first need to find the first and second derivatives of \(y\): \(y' = re^{rx}\), and \(y'' = r^2e^{rx}\). Now, substitute these into the given equation: \(r^2e^{rx} + pre^{rx} + qe^{rx} = 0\).

Factor the Equation

We can now factor out the term \(e^{rx}\) from the equation: \(e^{rx}(r^2 + pr + q) = 0\). Since \(e^{rx}\) is never zero, we can divide both sides of the equation by it: \(r^2 + pr + q = 0\).

Solve the Characteristic Equation

The equation \(r^2 + pr + q = 0\) is called the characteristic equation of the given differential equation. This is a quadratic equation in \(r\) and can be solved using different methods, such as factoring, completing the square, or the quadratic formula: \(r = \frac{-p \pm \sqrt{p^2 - 4q}}{2}\). There are three main cases depending on the discriminant \(D = p^2 - 4q\): 1. \(D > 0\): The equation has two distinct real roots \(r_1\) and \(r_2\). The general solution is then \(y(x) = C_1e^{r_1x} + C_2e^{r_2x}\), where \(C_1\) and \(C_2\) are constants to be determined from the initial conditions. 2. \(D = 0\): The equation has one real root \(r_1 = r_2\). The general solution becomes \(y(x) = (C_1 + C_2x)e^{r_1x}\). 3. \(D < 0\): The equation has complex conjugate roots \(r_1 = \alpha + \beta i\) and \(r_2 = \alpha - \beta i\). The general solution can be written as \(y(x) = e^{\alpha x}(C_1\cos(\beta x) + C_2\sin(\beta x))\).

Obtain the General Solution

Depending on the specific case, as described above, we will end up with one of the three types of general solutions. If any initial conditions are given, we can use them to find the values of the constants \(C_1\) and \(C_2\). Otherwise, the result is the general solution to the given second-order linear differential equation.

Key concepts

Characteristic Equation

In the study of second-order linear differential equations, the characteristic equation plays a crucial role. This equation emerges when we use a common method called the guessing method.
We start by assuming a solution of the form \(y = e^{rx}\), a function noted for its nice differential properties. After substituting this guess into the differential equation, we attain an equation purely involving the variable \(r\).
Specifically, for the equation \(y'' + py' + qy = 0\), after substitution and simplifying, we arrive at a quadratic form \(r^2 + pr + q = 0\). This is the characteristic equation. The roots of this quadratic equation, \(r\), are instrumental in building the general solution of the original differential equation.
Key takeaway: the characteristic equation reduces a differential equation into an algebraic one, which is often simpler to solve.

General Solution

The general solution of a second-order linear differential equation represents all possible solutions. It incorporates constants that can be determined if initial conditions are provided.
After solving the characteristic equation \(r^2 + pr + q = 0\), we have to consider three scenarios based on the nature of the roots.

• If the roots are distinct and real, the general solution is \(y(x) = C_1e^{r_1x} + C_2e^{r_2x}\).
• If the roots are real and equal, the general solution changes to \(y(x) = (C_1 + C_2x)e^{r_1x}\).
• In cases of complex conjugate roots, the solution is expressed in terms of sine and cosine functions: \(y(x) = e^{\alpha x}(C_1\cos(\beta x) + C_2\sin(\beta x))\).

These forms cover every possible solution to the differential equation, and the coefficients \(C_1\) and \(C_2\) add flexibility to fit specific conditions.

Quadratic Formula

To solve the characteristic equation \(r^2 + pr + q = 0\), we frequently use the quadratic formula, an essential tool in algebra. This formula helps us find the roots of any quadratic equation.
The quadratic formula is expressed as:\[ r = \frac{-p \pm \sqrt{p^2 - 4q}}{2} \]Here, the term under the square root, \(p^2 - 4q\), is known as the discriminant \(D\).
The value of the discriminant determines the nature of the roots:

• \(D > 0\) gives two distinct real roots.
• \(D = 0\) results in a repeated real root.
• \(D < 0\) leads to complex conjugate roots.

Utilizing the quadratic formula simplifies finding the roots, allowing us to construct the general solution effectively.

Complex Conjugate Roots

Complex conjugate roots arise in the context of differential equations when the discriminant \(D = p^2 - 4q\) is negative.
This means the square root term in the quadratic formula results in an imaginary number, indicating two complex roots of the form \(r_1 = \alpha + \beta i\) and \(r_2 = \alpha - \beta i\).
With these roots, the general solution of the differential equation is written as:\[ y(x) = e^{\alpha x}(C_1 \cos(\beta x) + C_2 \sin(\beta x)) \]This form arises from Euler's formula linking complex exponentials to trigonometric functions.

• The terms \(\cos(\beta x)\) and \(\sin(\beta x)\) reflect the oscillatory nature of solutions with complex roots.
• The exponential \(e^{\alpha x}\) determines the growth or decay of the solution.

Complex conjugate roots suggest combining both exponential decay/growth and periodic oscillations in solutions.