Problem 11 - 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 \(y_{2}(x)\). $$ x y^{\prime \prime}+y^{\prime}=0 ; \quad y_{1}=\ln x $$

Step-by-Step Solution

Answer

The second solution is \(y_2(x) = \frac{1}{x}\).

Step 1: Write the given differential equation and known solution

The given differential equation is \(x y'' + y' = 0\) and the known solution is \(y_1(x) = \ln x\). Our task is to find a second linearly independent solution \(y_2(x)\) using the reduction of order method.

Step 2: Assume a form for the second solution

According to the reduction of order method, we assume that the second solution has the form \(y_2(x) = v(x) y_1(x) = v(x) \ln x\), where \(v(x)\) is a function to be determined.

Step 3: Derive expressions for \(y'\) and \(y''\)

First, compute \(y' = (v(x) \ln x)' = v'(x) \ln x + \frac{v(x)}{x}\). Compute \(y'' = (v'(x) \ln x + \frac{v(x)}{x})' = v''(x) \ln x + \frac{v'(x)}{x} + \frac{v'(x)}{x} - \frac{v(x)}{x^2}\).

Step 4: Substitute \(y'\) and \(y''\) into the differential equation

Substitute into the differential equation \(x y'' + y' = 0\), giving \(x(v''(x) \ln x + \frac{v'(x)}{x} + \frac{v'(x)}{x} - \frac{v(x)}{x^2}) + v'(x) \ln x + \frac{v(x)}{x} = 0\). Simplifying this equation, we get \(x v''(x) \ln x + 2v'(x) \ln x = 0\).

Step 5: Simplify and solve the equation

Factor out common terms to get \((x v''(x) + 2 v'(x)) \ln x = 0\). Since \(\ln x eq 0\) for \(x > 0\), we have \(x v''(x) + 2v'(x) = 0\). This simplifies to an equation \(v'(x) = \frac{C}{x^2}\) for a constant \(C\).

Step 6: Integrate to find \(v(x)\)

Integrate \(v'(x) = \frac{C}{x^2}\) with respect to \(x\) to get \(v(x) = -\frac{C}{x} + D\), where \(D\) is an integration constant. Without loss of generality, set \(D = 0\) for simplicity, so \(v(x) = -\frac{C}{x}\).

Step 7: Write the second solution

Substitute back into \(y_2(x) = v(x) \ln x = -\frac{C}{x} \ln x\). The second solution \(y_2(x)\) is thus \(y_2(x) = \frac{1}{x}\) by choosing \(C = -1\) for simplification.

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. These equations are essential for modeling real-world phenomena where change occurs. For example, they can describe the growth of populations, the behavior of circuits, or how heat diffuses through a material. A differential equation uses derivatives to show the relationship between a function and the rates at which its outputs change.

In the exercise provided, we deal with a second-order linear differential equation: \( xy'' + y' = 0 \). Here's what this means:

• "Second-order" refers to the highest derivative \( y'' \), which is the second derivative of \( y \).
• "Linear" denotes that the equation is linear in terms of the function \( y \) and its derivatives, without any products or powers besides 1.

Such differential equations often appear in physical problems and theoretical investigations, which is why understanding them is crucial for fields like physics and engineering.

Second Solution

Finding a second solution to a differential equation is about discovering another function that also satisfies the same equation. In the context of the exercise, we already know one solution, \( y_1(x) = \ln x \). However, differential equations like \( xy'' + y' = 0 \) often have more than one solution. Finding a second, linearly independent solution is important for forming a general solution.

The task here is to find another solution \( y_2(x) \), using a method called reduction of order. Here's a quick breakdown:

• Reduction of order is a technique used when one solution is known, and it helps to uncover a second solution.
• We assume the second solution has the form \( y_2(x) = v(x) y_1(x) \), where \( v(x) \) is an undetermined function.
• This method transforms the equation from a second-order differential equation into a simpler form that we can solve for \( v(x) \).

By following this method, we found that \( y_2(x) = \frac{1}{x} \). Together with \( y_1(x) \), these solutions form a complete set for the differential equation.

Linearly Independent Solutions

Linearly independent solutions are critical when dealing with differential equations, especially second-order linear equations. The concept of linear independence ensures that solutions are distinct in form and contribute to the overall structure of the general solution.

In mathematical terms, two functions \( y_1(x) \) and \( y_2(x) \) are linearly independent if there are no constants \( c_1 \) and \( c_2 \), not both zero, such that \( c_1 y_1(x) + c_2 y_2(x) = 0 \) for all \( x \). When dealing with differential equations, establishing linear independence allows us to express the general solution as a combination of these basic solutions.

For the given exercise with the equation \( xy'' + y' = 0 \), we have the solutions \( y_1(x) = \ln x \) and \( y_2(x) = \frac{1}{x} \). To verify linear independence, consider their Wronskian:

• The Wronskian is a determinant that checks if two functions are linearly independent.
• If the Wronskian of these functions is non-zero, they are indeed independent.

Hence, having linearly independent solutions ensures any solution of the differential equation can be expressed as a combination of \( y_1(x) \) and \( y_2(x) \). This completes the method of solving such equations comprehensively.