Second-Order Homogeneous Differential Equation Calculator

A second-order linear homogeneous differential equation with constant coefficients is a cornerstone in the study of differential equations. It takes the general form:

Alternatively, using prime notation ($y' = \frac{dy}{dx}$, $y'' = \frac{d^2y}{dx^2}$):

Key characteristics of this equation type:

• 2nd Order (Second Order): The highest derivative present is the second derivative ($y''$).
• Linear: The dependent variable $y$ and its derivatives ($y', y''$) appear only to the first power and are not multiplied together.
• Homogeneous: The right-hand side of the equation is zero. If it were a non-zero function of $x$, say $f(x)$, the equation would be non-homogeneous.
• Constant Coefficients: The coefficients $a, b,$ and $c$ are real numbers (constants), with the crucial condition that $a \neq 0$ (otherwise, it wouldn't be a second-order equation).

These equations are incredibly powerful for modeling a wide array of physical phenomena. For instance, they describe the motion of a mass attached to a spring (simple harmonic motion, damped oscillations), the flow of charge in an RLC electrical circuit, and many other systems that exhibit oscillatory or exponential decay/growth behavior.

The standard approach to solving $ay'' + by' + cy = 0$ involves a clever substitution. We propose a solution of the exponential form $y = e^{mx}$, where $m$ is a constant to be determined.

Step 1: Assume a Solution Form

Let $y = e^{mx}$. Then, the derivatives are:

• $y' = \frac{dy}{dx} = me^{mx}$
• $y'' = \frac{d^2y}{dx^2} = m^2e^{mx}$

Step 2: Substitute into the DE

Plugging these into the original differential equation $ay'' + by' + cy = 0$ gives:

We can factor out $e^{mx}$ (which is never zero for any real $m$ and $x$):

Step 3: Form the Auxiliary Equation

Since $e^{mx} \neq 0$, for the equation to hold, the term in the parenthesis must be zero. This gives us the auxiliary equation (also known as the characteristic equation):

This is a simple quadratic equation in $m$. The roots of this auxiliary equation dictate the form of the general solution to the differential equation.

Step 4: Find the Roots of the Auxiliary Equation

We use the quadratic formula to find the roots $m_1$ and $m_2$:

The nature of these roots depends on the discriminant, $\Delta = b^2 - 4ac$. There are three distinct cases:

Case 1: Real and Distinct Roots ($\Delta > 0$)

If $b^2 - 4ac > 0$, the auxiliary equation has two distinct real roots, $m_1$ and $m_2$. In this scenario, $e^{m_1x}$ and $e^{m_2x}$ are two linearly independent solutions. The general solution is a linear combination of these:

Where $A$ and $B$ are arbitrary constants. These constants are typically determined if initial conditions (e.g., $y(x_0) = y_0$ and $y'(x_0) = y'_0$) are provided.

Case 2: Real and Equal Roots ($\Delta = 0$)

If $b^2 - 4ac = 0$, the auxiliary equation has one real root (a repeated root), $m_1 = m_2 = m = -\frac{b}{2a}$. One solution is $y_1 = e^{mx}$. A second, linearly independent solution can be found (e.g., using the method of reduction of order) to be $y_2 = xe^{mx}$. The general solution is then:

$A$ and $B$ are arbitrary constants.

Case 3: Complex Conjugate Roots ($\Delta < 0$)

If $b^2 - 4ac < 0$, the auxiliary equation has two complex conjugate roots. These roots can be written as $m_1 = \alpha + i\beta$ and $m_2 = \alpha - i\beta$, where:

• $\alpha = -\frac{b}{2a}$ (the real part)
• $\beta = \frac{\sqrt{4ac - b^2}}{2a}$ (the imaginary part, $\beta > 0$)

The two linearly independent solutions would initially appear as $e^{(\alpha+i\beta)x}$ and $e^{(\alpha-i\beta)x}$. Using Euler's formula ($e^{i\theta} = \cos\theta + i\sin\theta$), these can be rewritten in terms of real-valued functions. The general solution becomes:

$A$ and $B$ are arbitrary constants. This form of the solution often represents damped oscillations in physical systems.

Step 5: Apply Initial Conditions (if given)

If initial conditions such as $y(x_0) = y_0$ and $y'(x_0) = y'_0$ are provided, they are used to solve for the constants $A$ and $B$ in the general solution, yielding a unique particular solution.