Problem 14 - A First Course in Differential Equations with Modeling Applications

Question

Solve the given differential equation. $$x^{2} y^{\prime \prime}-7 x y^{\prime}+41 y=0$$

Step-by-Step Solution

Answer

The solution is \(y(x) = x^4(C_1 \cos(5 \ln x) + C_2 \sin(5 \ln x))\).

Step 1: Recognize the form of the differential equation

The given differential equation is a second-order linear homogeneous differential equation with variable coefficients. It is of the form \(x^2y'' - 7xy' + 41y = 0\). Since the equation fits the standard form, we can solve it using a method suitable for Cauchy-Euler equations.

Step 2: Assume a solution of the form

For Cauchy-Euler equations, we assume a solution in the form \(y = x^m\). We find the derivatives \(y' = mx^{m-1}\) and \(y'' = m(m-1)x^{m-2}\), and then substitute these expressions back into the differential equation.

Step 3: Substitute and simplify

Substitute \(y = x^m\), \(y' = mx^{m-1}\), and \(y'' = m(m-1)x^{m-2}\) into the equation:\[x^2(m(m-1)x^{m-2}) - 7x(mx^{m-1}) + 41x^m = 0\]This simplifies to:\[m(m-1)x^m - 7mx^m + 41x^m = 0\]Combine like terms:\[(m(m-1) - 7m + 41)x^m = 0\]

Step 4: Factor and solve the characteristic equation

Since \(x^m eq 0\), we can solve the algebraic equation:\[m^2 - m - 7m + 41 = 0\]which simplifies to:\[m^2 - 8m + 41 = 0\]Solve the quadratic equation using the quadratic formula: \(m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -8\), and \(c = 41\).

Step 5: Calculate the roots of the characteristic equation

Using the quadratic formula:\[m = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot 41}}{2 \cdot 1}\]\[m = \frac{8 \pm \sqrt{64 - 164}}{2}\]\[m = \frac{8 \pm \sqrt{-100}}{2}\]\[m = \frac{8 \pm 10i}{2}\]\[m = 4 \pm 5i\]

Step 6: Write the general solution

Since the roots \(m = 4 \pm 5i\) are complex, we write the general solution as:\[y(x) = x^4(C_1 \cos(5 \ln x) + C_2 \sin(5 \ln x))\]where \(C_1\) and \(C_2\) are arbitrary constants.

Key Concepts

Cauchy-Euler Equation

The Cauchy-Euler equation is a special type of differential equation often found in mathematical and engineering problems. It is characterized by its variable coefficients that are powers of the independent variable, typically expressed in the form:

• \(a_n x^n y^{(n)} + a_{n-1} x^{n-1} y^{(n-1)} + \cdots + a_1 x y' + a_0 y = 0\)

In this problem, the equation is \(x^2 y'' - 7x y' + 41y = 0\), which matches the description of a second-order Cauchy-Euler equation. Recognizing this form is crucial because it guides the method of solution. The beauty of Cauchy-Euler equations is that they can be transformed into simpler algebraic problems using assumed solutions and characteristic equations.

Second-order Linear Differential Equations

Second-order linear differential equations are equations involving a second derivative, where the solution is often a function of one independent variable. The given equation \(x^2 y'' - 7x y' + 41y = 0\) is a prime example. It's second-order because the highest derivative is \(y''\), and it's linear since each term is proportional to either \(y\), \(y'\), or \(y''\) without powers or products of these terms.

Understanding this structure is vital because it tells us we can apply linear methods to find solutions. Such equations form the backbone of many physical phenomena, from electrical circuits to mechanical vibrations, and often involve characteristic equations to solve.

Homogeneous Differential Equations

A homogeneous differential equation is characterized by having the right-hand side equal to zero, representing a "balance" within the system described by the equation. In the context of the exercise, we have \(x^2 y'' - 7x y' + 41y = 0\). Here, 'homogeneity' implies all terms are set to zero, which signifies the solution can be expressed as combinations of simpler functions without external forcing terms.

The significance lies in the ability to assume a solution, often involving powers of variables or exponential terms, leading to simplified calculations. Recognizing homogeneity assists in determining the types of solutions possible and the nature of further calculations.

Complex Roots

Complex roots occur when solving the characteristic equation derived from a differential equation, often indicating oscillatory solutions. In this exercise, solving the characteristic equation \(m^2 - 8m + 41 = 0\) resulted in complex roots \(m = 4 \pm 5i\). These roots suggest the solution will involve trigonometric functions.

• Real part: indicates an exponential term in the solution
• Imaginary part: corresponds to oscillations, reflected as sine and cosine terms

When you encounter complex roots, the solution typically takes the form involving exponential, sine, and cosine functions, combined to handle the imaginary part mathematically, i.e., \(x^m (C_1 \cos(k \ln x) + C_2 \sin(k \ln x))\). Here, \(C_1\) and \(C_2\) are constants determined by initial conditions.

General Solution

The general solution of a differential equation provides a form that encompasses all possible specific solutions, typically involving arbitrary constants whose values depend on boundary conditions or initial values. For our problem, the general solution derived is:

• \(y(x) = x^4(C_1 \cos(5 \ln x) + C_2 \sin(5 \ln x))\)

This format reflects our analysis with complex roots where\

• \(x^m\): indicates the power term from the real part of roots
• \(\cos(5 \ln x)\) and \(\sin(5 \ln x)\): capture the oscillatory behavior
• \(C_1, C_2\): arbitrary constants that can be adjusted based on conditions

Grasping the general solution is essential as it allows for adaptability across different scenarios, where specific values replace the constants to match precise conditions.