Problem 12 - A First Course in Differential Equations with Modeling Applications

Question

Solve the given differential equation. $$x^{2} y^{\prime \prime}+8 x y^{\prime}+6 y=0$$

Step-by-Step Solution

Answer

The general solution is \(y(x) = C_1 x^{-1} + C_2 x^{-6}\).

Step 1: Identify the type of differential equation

The given differential equation is \(x^{2} y^{\prime \prime} + 8x y^{\prime} + 6y = 0\). This is a second-order linear homogeneous differential equation with variable coefficients.

Step 2: Convert to standard form using Cauchy-Euler substitution

We recognize this as a Cauchy-Euler equation and use the substitution \(y = x^m\). Therefore, \(y' = mx^{m-1}\) and \(y'' = m(m-1)x^{m-2}\).

Step 3: Substitute and simplify

Substitute \(y = x^m\), \(y' = mx^{m-1}\), and \(y'' = m(m-1)x^{m-2}\) into the differential equation:\[ x^2(m(m-1)x^{m-2}) + 8x(mx^{m-1}) + 6x^m = 0 \]Simplify to:\[ m(m-1)x^m + 8mx^m + 6x^m = 0 \]

Step 4: Factor out common terms

Factor the expression:\[ x^m (m(m-1) + 8m + 6) = 0 \]Since \(x^m eq 0\), the equation simplifies to:\[ m(m-1) + 8m + 6 = 0 \]

Step 5: Solve the characteristic equation

Set the characteristic equation:\[ m^2 + 7m + 6 = 0 \]Solve using the quadratic formula \(m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = 7\), and \(c = 6\).

Step 6: Solve for roots using quadratic formula

Compute the roots:\(b^2 - 4ac = 7^2 - 4 \cdot 1 \cdot 6 = 49 - 24 = 25\)The roots are:\[ m = \frac{-7 \pm 5}{2}\]Therefore, the roots are \(m = -1\) and \(m = -6\).

Step 7: Write the general solution

Since the roots \(m_1 = -1\) and \(m_2 = -6\) are real and distinct, the general solution to the differential equation is:\[ y(x) = C_1 x^{-1} + C_2 x^{-6} \], where \(C_1\) and \(C_2\) are arbitrary constants.

Key Concepts

Second-Order Linear Differential Equation

Differential equations can often look complicated, but they help us understand how things change. The equation \(x^2 y'' + 8x y' + 6y = 0\) is a specific type, known as a second-order linear differential equation. Here's what that means:

• Second-order refers to \(y''\), which is the second derivative of \(y\) with respect to \(x\). It's like asking not just how fast something is changing, but how the rate of change is changing.
• Linear means that each term is either a constant or a simple multiple of \(y\) or its derivatives, like \(y'\) or \(y''\).

This type of equation pops up all over the place in the real world, modeling everything from how springs move to how heat flows.

Homogeneous Differential Equations

Homogeneous differential equations are a special category of differential equations where every term is a multiple of the unknown function or its derivatives. Our example, \(x^2 y'' + 8x y' + 6y = 0\), is homogeneous because there are no free terms or functions of \(x\) alone, without multiplying by \(y\), \(y'\), or \(y''\). Such equations are fundamentally about balance. Think about it like a perfectly balanced see-saw; every side needs a matching counterpart, ensuring the equation equals zero throughout. Homogeneous equations are particularly powerful because they often have elegant solutions, especially when they have constant coefficients or can be transformed into such a form, like with the Cauchy-Euler method.

Characteristic Equation

In solving many second-order linear differential equations, we use a key tool: the characteristic equation. For our problem, after substituting \(y = x^m\), we derive a polynomial equation from the differential equation. This polynomial, \(m^2 + 7m + 6 = 0\), is called the characteristic equation.

• It captures the essence of the differential equation in algebraic form.
• The solutions or roots from this equation, \(m_1 = -1\) and \(m_2 = -6\), tell us important things about the behavior of the solution \(y\).

These roots help structure our general solution, allowing us to piece together solutions that fit the entire differential equation.

General Solution

The concept of a general solution is central to understanding the full scope of possible outcomes for a differential equation. For our example, once we find the roots \(m_1 = -1\) and \(m_2 = -6\) from the characteristic equation, constructing the general solution involves these steps:

• For each real, distinct root like ours, \(y(x) = C_1 x^{-1} + C_2 x^{-6}\).
• \(C_1\) and \(C_2\) are arbitrary constants that can be defined based on initial conditions or specific scenarios.

The beauty of a general solution is its flexibility. It contains an infinite number of specific solutions, each representing different real-world conditions under the umbrella of the same mathematical relationship. This allows mathematicians and scientists to apply the general solution to various contexts simply by tweaking the constants.