Problem 13 - A First Course in Differential Equations with Modeling Applications

Question

Solve the given differential equation. $$3 x^{2} y^{\prime \prime}+6 x y^{\prime}+y=0$$

Step-by-Step Solution

Answer

The general solution is \(y(x) = x^{-1/2}(C_1 \cos(\frac{\sqrt{3}}{6} \ln x) + C_2 \sin(\frac{\sqrt{3}}{6} \ln x))\).

Step 1: Identify and Simplify the Differential Equation

The given differential equation is second order and linear: \(3x^2y'' + 6xy' + y = 0\). Since this is a Cauchy-Euler differential equation (or an 'equi-potential' differential equation), it can be solved by assuming a solution of the form \(y = x^m\), where \(m\) is a constant.

Step 2: Substitute the Assumed Solution

Substitute \(y = x^m\) into the differential equation. Then, the first derivative \(y' = mx^{m-1}\) and the second derivative \(y'' = m(m-1)x^{m-2}\). This transforms the differential equation to: \[3x^2(m(m-1)x^{m-2}) + 6x(mx^{m-1}) + x^m = 0\]. Simplifying, we get \[3m(m-1)x^m + 6mx^m + x^m = 0\].

Step 3: Factor and Combine Like Terms

Combine like terms in the equation: \[(3m(m-1) + 6m + 1)x^m = 0\]. Since \(x^m eq 0\) for non-zero \(x\), we focus on solving the auxiliary equation: \[3m(m-1) + 6m + 1 = 0\].

Step 4: Solve the Auxiliary Equation

Expand and simplify the auxiliary equation: \[3m^2 - 3m + 6m + 1 = 0\] simplifies to \[3m^2 + 3m + 1 = 0\]. Use the quadratic formula \(m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = 3\), \(c = 1\).

Step 5: Apply the Quadratic Formula

Substitute the values into the quadratic formula: \[m = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3}\]. This simplifies to \[m = \frac{-3 \pm \sqrt{9 - 12}}{6}\], leading to \[m = \frac{-3 \pm \sqrt{-3}}{6}\].

Step 6: Simplify Complex Roots

The discriminant is negative, so the roots are complex. Simplifying further, \[m = \frac{-3 \pm i\sqrt{3}}{6}\] becomes \[m = -\frac{1}{2} \pm i\frac{\sqrt{3}}{6}\].

Step 7: Write the General Solution

Since the roots are \(-\frac{1}{2} \pm i\frac{\sqrt{3}}{6}\), the general solution for real coefficients is given by: \[y(x) = x^{-1/2}(C_1 \cos(\frac{\sqrt{3}}{6} \ln x) + C_2 \sin(\frac{\sqrt{3}}{6} \ln x))\]. Here, \(C_1\) and \(C_2\) are arbitrary constants.

Key Concepts

Second-Order Differential Equations

Second-order differential equations are a type of differential equation involving derivatives up to the second degree. These equations are vital in modeling various physical phenomena, such as oscillatory systems, wave propagation, and heat conduction. A second-order differential equation has the general form:\[A(x)y'' + B(x)y' + C(x)y = F(x)\]where:

• \(y''\) is the second derivative of \(y\) with respect to \(x\), reflecting how the rate of change itself changes.
• \(y'\) is the first derivative.
• \(y\) represents the original function.
• \(A(x)\), \(B(x)\), \(C(x)\), and \(F(x)\) are functions of \(x\) that dictate the behavior and characteristics of the equation.

Second-order equations can be linear or nonlinear, homogeneous or non-homogeneous. In this exercise, we encountered a homogeneous, linear second-order differential equation, specifically a Cauchy-Euler type, which we'll explore further.

Complex Roots

When solving differential equations, the nature of the roots of the auxiliary equation greatly influences the form of the solution. Often, we have to deal with scenarios where the roots are complex. This happens when the discriminant of the quadratic auxiliary equation is negative.

• Complex roots are generally in the form \(a \pm bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit providing \(i^2 = -1\).
• In this specific problem, the auxiliary equation yields complex roots \(-\frac{1}{2} \pm i\frac{\sqrt{3}}{6}\). This indicates an oscillatory solution in terms of \(x\).

Complex roots lead to solutions expressed with sine and cosine functions, capturing the oscillatory nature inherent in complex conjugates. This results in the general solution structure involving trigonometric functions like cosine and sine.

General Solution

The general solution to a differential equation encompasses all possible solutions, usually defined by arbitrary constants. For second-order Cauchy-Euler differential equations with complex roots, the solution takes a specific form:

• \(y(x) = x^a(C_1 \cos(b \ln x) + C_2 \sin(b \ln x))\)
• where \(a\) and \(b\) come from the roots of the auxiliary equation \(a \pm bi\).

In our example, the roots were \(-\frac{1}{2} \pm i\frac{\sqrt{3}}{6}\) leading to the general solution:\[y(x) = x^{-1/2}(C_1 \cos(\frac{\sqrt{3}}{6} \ln x) + C_2 \sin(\frac{\sqrt{3}}{6} \ln x))\]The general solution reflects the behavior of the differential equation, capturing all pathways the system can take, represented by the arbitrary constants \(C_1\) and \(C_2\), which can be determined if initial conditions are provided.

Equi-potential Differential Equation

Equi-potential differential equations, also known as Cauchy-Euler equations, are a special class of linear differential equations characterized by their unique structure:\[a_n x^n y^{(n)} + a_{n-1} x^{n-1} y^{(n-1)} + ... + a_0 y = 0\]This form implies that the order of differentiation corresponds with the degree of the variable \(x\). Due to this relationship, they can often be solved by assuming solutions of the form \(y = x^m\).

• Cauchy-Euler equations are notably useful in addressing problems with logarithmic or power-based systems, commonly found in physics and engineering fields.
• The method of assuming \(y = x^m\) simplifies the problem into finding the correct \(m\)-values that satisfy the auxiliary polynomial equation, thereby determining the roots.

Understanding equi-potential equations is crucial, as they frequently model systems where proportional relationships exist between forces, such as in beam bending or stress distribution computations.