Problem 8 - 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 \(y_{2}(x)\). $$ 6 y^{\prime \prime}+y^{\prime}-y=0 ; \quad y_{1}=e^{x / 3} $$

Step-by-Step Solution

Answer

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

Step 1: Write Down the Given Differential Equation

The differential equation given in the problem is \( 6y'' + y' - y = 0 \). The function \( y_1(x) = e^{x/3} \) is provided as a known solution.

Step 2: Verify the Solution

To verify that \( y_1(x) = e^{x/3} \) is a solution, substitute \( y_1(x) \) into the differential equation. Calculate the first and second derivatives: \( y_1'(x) = \frac{1}{3}e^{x/3} \) and \( y_1''(x) = \frac{1}{9}e^{x/3} \). Substitute these back into the equation to confirm it satisfies the equation.

Step 3: Use Reduction of Order

To find a second solution \( y_2(x) \), assume \( y_2(x) = v(x)e^{x/3} \), where \( v(x) \) is a function to be determined. Thus, take the derivatives: \( y_2' = v'e^{x/3} + \frac{1}{3}ve^{x/3} \) and \( y_2'' = v''e^{x/3} + \frac{2}{3}v'e^{x/3} + \frac{1}{9}ve^{x/3} \).

Step 4: Substitute into the Differential Equation

Substitute \( y_2, y_2', \) and \( y_2'' \) into the original differential equation: \( 6(v''e^{x/3} + \frac{2}{3}v'e^{x/3} + \frac{1}{9}ve^{x/3}) + (v'e^{x/3} + \frac{1}{3}ve^{x/3}) - ve^{x/3} = 0 \).

Step 5: Simplify and Solve for \( v(x) \)

Simplify the equation by collecting terms and factoring out \( e^{x/3} \). This simplifies to: \( 6v'' + 3v' = 0 \). Divide through by 3 to get \( 2v'' + v' = 0 \) and solve this first-order differential equation for \( v' \).

Step 6: Solve the Auxiliary Equation

Solve \( v' = C \) from \( 2v'' + v' = 0 \), which indicates \( v(x) = C_1x + C_2 \). Since \( y_2 \) should be linearly independent of \( y_1 \), use \( v(x) = x \), thus \( y_2(x) = xe^{x/3} \).

Step 7: Verify the Independence of \( y_2 \)

Confirm that \( y_2(x) = xe^{x/3} \) is independent of \( y_1(x) \) by checking that the Wronskian \( W(y_1, y_2) = y_1y_2' - y_2y_1' \) is non-zero. Calculate it to verify linear independence.

Key Concepts

Understanding Differential Equations

Differential equations are equations that involve functions and their derivatives. They are fundamental in expressing physical laws and many dynamic systems.
In our exercise, the given differential equation is a second-order linear equation:

• Second-order means it involves the second derivative, as seen in the term \( y'' \).
• Linear indicates each term is linear in terms of either \( y \) or its derivatives.

The importance of solving differential equations lies in their ability to describe real-world phenomena and predict future states of systems. To solve them, mathematicians and engineers often look for solutions that satisfy the equation under given conditions. In this context, having one known solution helps us find a second solution, which can give more insight into the system's behavior.

Finding a Second Solution

When tackling a differential equation, having one solution can be a stepping stone to finding a complete solution. The method known as "reduction of order" helps us derive a second solution when one is already known.
For our differential equation, we knew that one solution is \( y_1(x) = e^{x/3} \). To find another solution, we hypothesize that the second solution, \( y_2(x) \), has a form that incorporates the initial solution multiplied by an unknown function \( v(x) \):

• This takes the form \( y_2(x) = v(x)e^{x/3} \).
• Through substitution, this method transforms the problem into finding the exact function \( v(x) \).

By solving for \( v(x) \), we eventually find that \( y_2(x) = xe^{x/3} \) serves as another valid solution.

Linear Independence and Its Verification

To ensure the usefulness of our second solution, it must be linearly independent from the first one. Two functions are linearly independent if one is not a constant multiple of the other.
Mathematically, we use the Wronskian determinant to check independence:

• The Wronskian for functions \( y_1 \) and \( y_2 \) is calculated as \( W(y_1, y_2) = y_1y_2' - y_2y_1' \).
• If the Wronskian is non-zero, the functions are linearly independent.

In our case, using \( y_1(x) = e^{x/3} \) and \( y_2(x) = xe^{x/3} \), the Wronskian turns out to be non-zero, confirming their linear independence. This signifies that together, these solutions provide the most general form of the solution for the differential equation in question.