Problem 10 - Advanced Engineering Mathematics

Question

In Problems \(1-16\), the indicated function \(y_{1}(x)\) is a solution of the given equation. Use reduction of order or formula (5), as instructed, to find a second solution \(y_{2}(x)\). $$ x^{2} y^{\prime \prime}+2 x y^{\prime}-6 y=0 ; \quad y_{1}=x^{2} $$

Step-by-Step Solution

Answer

The second solution is \(y_2(x) = x^{-3}\).

Step 1: Identify the Differential Equation

We are given the differential equation \(x^2 y'' + 2xy' - 6y = 0\). The function \(y_{1}(x) = x^2\) is a known solution. Our task is to find a second linearly independent solution \(y_{2}(x)\).

Step 2: Reduction of Order Formula

Since we know \(y_1(x) = x^2\) is a solution to the differential equation, we can use the reduction of order technique. Assume the second solution is of the form \(y_{2}(x) = v(x) imes y_{1}(x) = v(x) imes x^2\).

Step 3: Set Up the Second Solution

Substitute \(y = v(x) imes x^2\) into the original differential equation and compute its derivatives: \(y' = v' x^2 + 2vx\) and \(y'' = v''x^2 + 4v'x + 2v\).

Step 4: Substitute and Simplify

Substitute \(y\), \(y'\), and \(y''\) into the differential equation: \[ x^2(v''x^2 + 4v'x + 2v) + 2x(v'x^2 + 2vx) - 6(vx^2) = 0.\]Simplify the expression.

Step 5: Cancel Common Factors and Solve for v(x)

Remove common factors and simplify further, reducing terms to \[ v''x^2 + 6v'x = 0.\]Divide by \(x\) (assuming \(x eq 0\)), obtaining \[ x v'' + 6 v' = 0.\]

Step 6: Solve the New Differential Equation

We have a first-order differential equation: \[ x v'' + 6v' = 0.\]Let \(u = v'\), which gives us \(u' = v''\). Substitute into the equation to get \[ x u' + 6u = 0.\]This is separable, so \[ \frac{du}{dx} = -\frac{6u}{x}.\]

Step 7: Perform Separation of Variables

Separate variables as \[ \frac{du}{u} = -\frac{6}{x} dx.\]Integrate both sides to find \[ \ln |u| = -6 \ln |x| + C, \]where \(C\) is a constant of integration.

Step 8: Solve for u(x) and v(x)

Exponentiate both sides to solve for \(u\): \[ u = C x^{-6}.\]Since \(u = v'\), integrate again: \[ v(x) = \int C x^{-6} \, dx = -\frac{C}{5} x^{-5} + C_1.\]

Step 9: Construct the Second Solution y_2(x)

The solution \(v(x)\) gives the second linearly independent solution as:\[ y_2(x) = v(x) \times x^2 = \left(-\frac{C}{5} x^{-5} + C_1\right) x^2.\] Simplifying gives:\[ y_2(x) = -\frac{C}{5} x^{-3} + C_1 x^2.\]

Step 10: Determine y_2(x) without Overlapping y_1(x)

To ensure \(y_2(x)\) is linearly independent, choose constants to avoid overlap with \(y_1(x) = x^2\). Set \( C_1 = 0 \) leading to: \[ y_2(x) = -\frac{C}{5} x^{-3}.\]

Key Concepts

Second-Order Differential Equations

A second-order differential equation is an equation involving the unknown function, its derivatives, and is characterized by the second derivative as the highest order derivative. In the given exercise, our equation is \( x^2 y'' + 2xy' - 6y = 0 \). This differential equation is linear and homogeneous because it equates to zero and does not include terms that are functions of \( x \) or constants alone.

In mathematics, second-order differential equations often appear in modeling physical phenomena such as mechanical vibrations, electrical circuits, and other dynamic systems. Learning how to solve these equations is essential for understanding complex systems.

The prescribed solution \( y_1 = x^2 \) is one such function that satisfies this differential equation when substituted back into it. The question then aims to find another linearly independent solution using a method called Reduction of Order. This technique helps simplify finding a second solution when one solution is already known.

Linearly Independent Solutions

Linearly independent solutions are critical for solving second-order differential equations. When you have two linearly independent solutions, you can form the general solution of the equation, which encompasses all possible solutions. Simply, two functions \( y_1(x) \) and \( y_2(x) \) are linearly independent if there are no constants \( a \) and \( b \), not both zero, such that \( ay_1(x) + by_2(x) = 0 \) for all \( x \) values.

In this problem, since \( y_1 = x^2 \) is a known solution, the second solution \( y_2(x) \) found using the Reduction of Order method becomes linearly independent because it cannot be expressed as a multiple of \( y_1(x) \). This is ensured by selecting specific constants during its derivation, as shown in the solution step where \( C_1 = 0 \) was chosen, thereby making \( y_2(x) = -\frac{C}{5} x^{-3} \) linearly independent from \( y_1(x) \).

Having two such independent solutions allows you to express the general solution of the differential equation in the form \( c_1 y_1(x) + c_2 y_2(x) \), capturing all solutions.

Homogeneous Differential Equations

A homogeneous differential equation is one where all terms involve the function \( y \) and its derivatives, and it equals zero. These equations are of the form \( ay'' + by' + cy = 0 \).

The exercise presents a homogeneous equation \( x^2 y'' + 2xy' - 6y = 0 \). This form is significant because finding solutions to these equations is sometimes simpler compared to non-homogeneous ones, and it allows the Reduction of Order method to be applied effectively.

The approach involves assuming a second solution based on the given one and resolving new simplified equations. Homogeneous equations not only simplify modeling bulk phenomena but also lay foundational concepts for tackling more complex non-homogeneous equations in advanced studies.

In summary, recognizing a differential equation's type - whether homogeneous or not - aids in applying the right methods for obtaining solutions efficiently.