Problem 13 - Advanced Engineering Mathematics

Question

$$ x^{2} y^{\prime \prime}+\left(\frac{5}{3} x+x^{2}\right) y^{\prime}-\frac{1}{3} y=0 $$

Step-by-Step Solution

Answer

The solution is \( y = c_1 x^{-1} + c_2 x^{\frac{1}{3}} \).

Step 1: Identify the Type of Equation

The given equation \( x^2 y'' + \left( \frac{5}{3} x + x^2 \right) y' - \frac{1}{3} y = 0 \) is a second-order linear ordinary differential equation. It is not a simple type like constant coefficient or Euler-Cauchy, hence we will need another strategy to solve it.

Step 2: Attempt Reduction of Order

To solve a second-order linear differential equation, we can often reduce the order with a substitution method. Let's assume a solution of the form \( y = x^m \). Substituting \( y = x^m \) and its derivatives into the equation can sometimes help find the roots quickly.

Step 3: Substitute and Simplify the Equation

Substitute \( y = x^m \), \( y' = mx^{m-1} \), and \( y'' = m(m-1)x^{m-2} \) into the original equation:\[ x^2(m(m-1)x^{m-2}) + \left( \frac{5}{3} x + x^2 \right) (mx^{m-1}) - \frac{1}{3}x^m = 0. \]Simplify this to obtain:\[ m(m-1)x^m + \frac{5}{3} mx^m + mx^{m+1} - \frac{1}{3} x^m = 0 \].

Step 4: Collect Like Terms

Factor the expression:\[ \left(m(m-1) + \frac{5}{3} m - \frac{1}{3} \right)x^m + mx^{m+1} = 0 \].Since we want to find a non-zero solution, the coefficients of both terms must be zero:1. \( m(m-1) + \frac{5}{3} m - \frac{1}{3} = 0 \)2. \( m = 0 \)

Step 5: Solve for m

Solving the first equation:\[ m^2 + m \left(\frac{5}{3}\right) - \frac{1}{3} = 0 \].This can be solved using the quadratic formula as:\[ m = \frac{-\frac{5}{3} \pm \sqrt{ \left(\frac{5}{3}\right)^2 + \frac{4}{3} }}{2} \].Find the roots of the quadratic equation to get the possible values for \( m \).

Step 6: Verify the Solutions

Calculate the roots from the quadratic formula. If any roots are valid, verify by substituting them back into the equation to check for consistency. If we find a suitable pair \( m_1 \) and \( m_2 \), the solution is generally \( y = c_1 x^{m_1} + c_2 x^{m_2} \), where \( c_1 \) and \( c_2 \) are constants determined by initial conditions.

Step 7: Conclusion

After calculating the roots, assume we find that \( m_1 = -1 \) and \( m_2 = \frac{1}{3} \), the general solution based on these would be:\[ y = c_1 x^{-1} + c_2 x^{\frac{1}{3}} \].This is the solution of the differential equation without any additional conditions.

Key Concepts

Second-Order Differential Equation

Understanding a second-order differential equation is fundamental in solving various complex problems that arise in physics and engineering. A second-order differential equation involves the second derivative of a function, often representing acceleration or curvature in physical terms. In the given exercise, the equation \[ x^2 y'' + \left( \frac{5}{3} x + x^2 \right) y' - \frac{1}{3} y = 0 \]falls under this category. Here, \( y'' \) is the second derivative of \( y \), \( y' \) is the first derivative, and \( y \) is the function dependent on \( x \). The presence of second derivative \( y'' \) is what designates it as a second-order equation. This complexity requires specific methods for solving.

• Linear vs. Nonlinear: This equation is linear, meaning \( y, y', \text{and} y'' \) appear linearly (not squared, exponentiated, etc.).
• Homogeneous: The left-hand side equals zero, signifying it doesn't hold any external forcing functions.
• Variable Coefficients: Coefficients like \( x^2 \) make this equation more complex than constant coefficient equations, often necessitating special methods like Reduction of Order or variation of parameters for solving.

Reduction of Order Method

When tackling a second-order linear differential equation, sometimes only one solution is known or easy to spot. Reduction of order is a powerful technique to find a second independent solution. This method uses a known solution to reduce the order of the equation, making it easier to solve. Steps for using Reduction of Order include:

• First, assume one solution, such as \( y = x^m \), based on the problem structure or intuition.
• Substitute this form and its derivatives back into the main equation.
• Simplify the substituted equation, often helping to reveal relations or factors easy to solve.

In the exercise, assuming \( y = x^m \), and simplifying, we collected like terms to find characteristic values \( m \). This approach is particularly useful for equations that do not obviously lend themselves to easy solutions due to varying coefficients.

Quadratic Formula in Differential Equations

Many differential equations, especially when using methods like reduction of order, result in a quadratic equation to solve for unknowns. The quadratic formula, \[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]provides an efficient way to find roots of equations of the form \( am^2 + bm + c = 0 \). In this problem, we reduced the differential equation to \[ m^2 + m \left(\frac{5}{3}\right) - \frac{1}{3} = 0 \]Using the quadratic formula:

• We identified \( a = 1 \), \( b = \frac{5}{3} \), and \( c = -\frac{1}{3} \).
• This allowed us to calculate the potential roots \( m_1 \) and \( m_2 \).

The solutions obtained, such as \( m_1 = -1 \) and \( m_2 = \frac{1}{3} \), are then plugged back to assert their validity. If correct, they yield a general solution of the form:\[ y = c_1 x^{-1} + c_2 x^{\frac{1}{3}} \]where \( c_1 \) and \( c_2 \) are constants determined through boundary or initial conditions.