Problem 7 - 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 \(y_{2}(x)\). $$ 9 y^{\prime \prime}-12 y^{\prime}+4 y=0 ; \quad y_{1}=e^{2 x / 3} $$

Step-by-Step Solution

Answer

The second solution is \( y_2(x) = x e^{2x/3} \).

Step 1: Verify that \( y_1(x) \) is a Solution

First, we start by verifying that the given function \( y_1(x) = e^{2x/3} \) is indeed a solution to the differential equation \( 9 y'' - 12 y' + 4 y = 0 \). We will differentiate \( y_1(x) \) to find \( y'_1 \) and \( y''_1 \).

Step 2: Differentiate \( y_1(x) \) Once

Calculate the first derivative: \[ y'_1(x) = \frac{d}{dx} e^{2x/3} = \frac{2}{3} e^{2x/3} \]

Step 3: Differentiate \( y_1(x) \) a Second Time

Calculate the second derivative: \[ y''_1(x) = \frac{d}{dx} \left( \frac{2}{3} e^{2x/3} \right) = \left( \frac{2}{3} \right)^2 e^{2x/3} = \frac{4}{9} e^{2x/3} \]

Step 4: Substitute Derivatives Into the Differential Equation

Substitute \( y_1(x), y'_1(x), \) and \( y''_1(x) \) into the differential equation:\[ 9 \left( \frac{4}{9} e^{2x/3} \right) - 12 \left( \frac{2}{3} e^{2x/3} \right) + 4 e^{2x/3} = 0 \] Simplifying this results in:\[ 4e^{2x/3} - 8e^{2x/3} + 4e^{2x/3} = 0 \] which confirms that \( y_1(x) \) is a solution.

Step 5: Use Reduction of Order to Find \( y_2(x) \)

Since integration is to be used, look for a solution of the form \( y_2(x) = v(x) e^{2x/3} \) and substitute into the differential equation. Differentiate and substitute to find \( v(x) \).

Step 6: Substitute and Simplify

Setting \( y_2 = v e^{2x/3} \) leads to:\[ y'_2 = v'e^{2x/3} + v \frac{2}{3} e^{2x/3} \]\[ y''_2 = v''e^{2x/3} + 2v' \frac{2}{3} e^{2x/3} + v \left( \frac{2}{3} \right)^2 e^{2x/3} \]Substitute these into the differential equation and simplify using the property that terms need to independently balance out.

Step 7: Solve for \( v(x) \) and Integrate

When simplified, the equation for \( v'' \) becomes:\[ v'' e^{2x/3} = 0 \]This simplifies to \( v'' = 0 \), resulting in \( v' = C_1 \) and \( v = C_1x + C_0 \). Thus, \( y_2 = v \cdot e^{2x/3} = (C_1x + C_0) e^{2x/3} \). Simplify further for \( y_2(x) \), often choosing simple constants like \( C_1=1, C_0=0 \) for clarity.

Step 8: Second Solution \( y_2(x) \)

The non-trivial second solution is \( y_2(x) = x e^{2x/3} \). This is linearly independent of \( y_1(x) \).

Key Concepts

Differential Equation

A differential equation is a mathematical equation that relates a function with its derivatives. In simple terms, it shows how a function changes with respect to one or more variables. In this exercise, we deal with a second order linear differential equation given by: \ \( 9y'' - 12y' + 4y = 0 \). The derivatives \(y''\) and \(y'\) represent the rate at which \(y'\) and \(y\) change, respectively. Solving a differential equation involves finding the function \(y(x)\) that satisfies the equation.

• "Order" specifies the highest derivative present, here it is the second derivative \(y''\).
• This equation is "linear," meaning it involves the first power of \(y\) and its derivatives.

Solutions to such equations often involve discovering functions that fulfill the given relationships between the derivatives and the function itself.

Linearly Independent Solutions

Linearly independent solutions play a crucial role in solving differential equations, especially in the context of linear differential equations. Two functions \(y_1(x)\) and \(y_2(x)\) are considered linearly independent if no constant \(c\) exists such that \(y_2(x) = cy_1(x)\). In differential equations, having two linearly independent solutions is vital to form the general solution.

• For a second order linear differential equation, if you have two linearly independent solutions, \(y_{1}(x)\) and \(y_{2}(x)\), the general solution is \(y(x) = C_{1} y_{1}(x) + C_{2} y_{2}(x)\).
• In this exercise, \(y_{1}(x) = e^{2x/3}\) and \(y_{2}(x) = x e^{2x/3}\) are verified as linearly independent, giving us a complete solution set for the differential equation.

Why is linear independence crucial? It ensures that our solution set covers all possible behaviors modeled by the differential equation.

Second Order Linear Differential Equations

A second order linear differential equation often takes the standard form \(ay'' + by' + cy = 0\). Here, parameters \(a, b,\) and \(c\) are constants. The indices in the derivatives indicate the differential order, and they determine the complexity of the problem.

• Such equations describe a wide range of physical systems, including electrical circuits and harmonic oscillators.
• To simplify finding solutions, mathematicians apply techniques such as the "reduction of order." This reduces the problem into a manageable form and confirms the need for solutions specialized in specific situations.

The goal is to find two linearly independent solutions, as they allow the construction of a general solution via superposition, a principle permitted by linearity. Understanding such equations enhances your ability to interpret and model many scientific phenomena.

Solution Verification

Solution verification involves ensuring that a proposed function truly satisfies the differential equation. For this exercise, it starts with the given solution \(y_1(x) = e^{2x/3}\). Verification requires substituting the function and its derivatives back into the original equation.

• First, calculate the derivatives: \(y'_1\) and \(y''_1\).
• Next, substitute these into the differential equation \(9y'' - 12y' + 4y = 0\).
• The equation should simplify to zero, confirming the solution is correct.

For the second solution, reduction of order is used to create \(y_2(x) = xe^{2x/3}\). By repeating the verification process, you ensure this second solution also holds. Successful verification affirms the accuracy and completeness of the solutions proposed.