Problem 27 - Calculus Volume 3

Question

Find the general solution to the linear differential equation. $$ 36 \frac{d^{2} y}{d x^{2}}+12 \frac{d y}{d x}+y=0 $$

Step-by-Step Solution

Answer

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

Step 1: Write the Differential Equation in Standard Form

The given differential equation is \( 36 \frac{d^2 y}{dx^2} + 12 \frac{dy}{dx} + y = 0 \). The standard form for a second-order linear homogeneous differential equation is \( a \frac{d^2 y}{dx^2} + b \frac{dy}{dx} + c y = 0 \), where \( a = 36 \), \( b = 12 \), and \( c = 1 \).

Step 2: Write the Characteristic Equation

To solve the differential equation, we must find its characteristic equation, which is derived from the differential equation by replacing \( \frac{d^2 y}{dx^2} \) with \( r^2 \), \( \frac{dy}{dx} \) with \( r \), and \( y \) with 1. This gives us the characteristic equation: \[ 36r^2 + 12r + 1 = 0 \]

Step 3: Solve the Characteristic Equation

Solve the quadratic characteristic equation \( 36r^2 + 12r + 1 = 0 \) using the quadratic formula, \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 36 \), \( b = 12 \), and \( c = 1 \). Calculate the discriminant: \( b^2 - 4ac = 12^2 - 4 \cdot 36 \cdot 1 = 144 - 144 = 0 \). Since the discriminant is zero, there is a repeated real root.

Step 4: Calculate the Repeated Root

Using the quadratic formula, find the repeated root: \[ r = \frac{-12 \pm \sqrt{0}}{2 \times 36} = \frac{-12}{72} = -\frac{1}{6} \]. The root is \( r = -\frac{1}{6} \).

Step 5: Write the General Solution

For a repeated root \( r \), the general solution of the differential equation is given by \[ y(x) = C_1 e^{rx} + C_2 x e^{rx} \]. Substitute \( r = -\frac{1}{6} \) into the solution form: \[ y(x) = C_1 e^{-\frac{1}{6}x} + C_2 x e^{-\frac{1}{6}x} \], where \( C_1 \) and \( C_2 \) are arbitrary constants.

Key Concepts

Characteristic Equation

Linear differential equations can be quite mysterious, but the characteristic equation helps us unveil their solutions. It is formed from the given differential equation by substituting the derivatives with powers of a new variable, typically represented as \( r \). By doing this, we transform the differential equation into an algebraic equation. For instance, in our equation, \( 36 \frac{d^2 y}{dx^2} + 12 \frac{dy}{dx} + y = 0 \), we substitute as follows:

• \( \frac{d^2 y}{dx^2} \rightarrow r^2 \)
• \( \frac{dy}{dx} \rightarrow r \)
• \( y \rightarrow 1 \)

Thus, we obtain the characteristic equation \( 36r^2 + 12r + 1 = 0 \). Solving this quadratic equation will give us the roots, indicating the behavior of the system described by the differential equation. The roots can vary:

• Distinct real roots
• Complex roots
• Repeated real roots

Each type of root forms a different pattern in the solution. In our example, however, we find a repeated real root, which directly influences the form of the general solution.

General Solution

The general solution of a linear homogeneous differential equation gives us a family of solutions which can describe any potential state of the system. Once we have the roots of the characteristic equation, we can construct this solution.
For example, when the characteristic equation has repeated roots, as it does in our problem \( r = -\frac{1}{6} \), the general solution takes on a special form:

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

This equation captures all possible solutions by using two arbitrary constants, \( C_1 \) and \( C_2 \), which will be determined by initial conditions or additional constraints, if provided. Substituting our repeated root into this solution format results in:

• \( y(x) = C_1 e^{-\frac{1}{6}x} + C_2 x e^{-\frac{1}{6}x} \)

This shows how each individual solution component depends on the root, providing exponential terms that diminish over time in our case.

Homogeneous Differential Equation

Homogeneous differential equations are special because they set the scene for the characteristic equation in a straightforward manner. They are equations where every term involves the function \( y \), its derivatives, or is zero; there are no isolated constants or external inputs.
In the case of our equation:

• \( 36 \frac{d^2 y}{dx^2} + 12 \frac{dy}{dx} + y = 0 \)

It is homogeneous since all the terms consist of derivatives of \( y \). Such equations always equate to zero.
This core property means that all possible solutions form a vector space, allowing us to use methods like the characteristic equation to find solutions easily.
Homogeneous equations often model systems in equilibrium, where solutions describe stable or unstable states depending on the nature of the roots from the characteristic equation. In our example, the repeated root \( r = -\frac{1}{6} \) suggests an exponential decay, a typical feature in systems approaching stability.