Homogeneous Linear Second-Order Differential Equations

A homogeneous linear second-order differential equation with constant coefficients has the standard form: $$ ay'' + by' + cy = 0 $$ where \( a \), \( b \), and \( c \) are real constants.

To solve such equations, we examine their associated characteristic equation—a quadratic equation in the variable \( z \):

$$ az^2 + bz + c = 0 $$

The nature of the roots of this equation depends on the discriminant \( \Delta = b^2 - 4ac \), leading to three distinct cases:

• Two distinct real roots, \( z_1 \ne z_2 \), when \( \Delta > 0 \)
  The general solution is: $$ y = c_1 e^{z_1 x} + c_2 e^{z_2 x} $$
• A repeated real root, \( z_1 = z_2 \), when \( \Delta = 0 \)
  The general solution is: $$ y = c_1 e^{z_1 x} + x c_2 e^{z_1 x} $$
• A pair of complex conjugate roots, \( \alpha \pm i\beta \), when \( \Delta < 0 \)
  The general solution is: $$ y = e^{\alpha x} \left( c_1 \cos(\beta x) + c_2 \sin(\beta x) \right) $$

Here, \( c_1 \) and \( c_2 \) are arbitrary real constants determined by initial or boundary conditions.

A Practical Example

Consider the equation:

$$ y'' - 3y' - 4y = 0 $$

This is a second-order linear equation with constant coefficients, where \( a = 1 \), \( b = -3 \), and \( c = -4 \).

We begin by solving the characteristic equation:

$$ z^2 - 3z - 4 = 0 $$

The solutions are:

$$ z = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-4)}}{2} = \frac{3 \pm \sqrt{25}}{2} = \begin{cases} z_1 = 4 \\ z_2 = -1 \end{cases} $$

Since the roots are real and distinct, the general solution is:

$$ y = c_1 e^{4x} + c_2 e^{-x} $$

Example 2

Now consider:

$$ y'' + 6y' + 9y = 0 $$

The characteristic equation is:

$$ z^2 + 6z + 9 = 0 $$

Its solution is a repeated root:

$$ z = \frac{-6 \pm \sqrt{36 - 36}}{2} = -3 $$

So the general solution becomes:

$$ y = c_1 e^{-3x} + x c_2 e^{-3x} $$

Example 3

Finally, consider:

$$ y'' + 2y' + 2y = 0 $$

The characteristic equation is:

$$ z^2 + 2z + 2 = 0 $$

Solving this gives complex conjugate roots:

$$ z = \frac{-2 \pm \sqrt{4 - 8}}{2} = \frac{-2 \pm \sqrt{-4}}{2} = \begin{cases} z_1 = -1 + i \\ z_2 = -1 - i \end{cases} $$

Thus \( \alpha = -1 \) and \( \beta = 1 \), and the general solution is:

$$ y = e^{-x} \left( c_1 \cos x + c_2 \sin x \right) $$

Justification of the Method

A homogeneous linear second-order differential equation with constant coefficients can be written as:

$$ a y'' + b y' + c y = 0 $$

Assuming a trial solution of the form \( y = e^{zx} \), where \( z \) is a constant, we obtain:

$$ y = e^{zx} $$

$$ y' = z e^{zx} $$

$$ y'' = z^2 e^{zx} $$

Substituting into the original equation:

$$ a z^2 e^{zx} + b z e^{zx} + c e^{zx} = 0 $$

Factoring out the exponential term:

$$ (a z^2 + b z + c) e^{zx} = 0 $$

Since \( e^{zx} \) is never zero, the equation reduces to the characteristic equation:

$$ a z^2 + b z + c = 0 $$

This confirms the validity of the method for solving homogeneous linear second-order equations with constant coefficients.

And so forth.