Problem 27 - Advanced Engineering Mathematics

Question

Verify that the given functions form a fundamental set of solutions of the differential equation on the indicated interval. Form the general solution. $$ x^{2} y^{\prime \prime}-6 x y^{\prime}+12 y=0 ; x^{3}, x^{4},(0, \infty) $$

Step-by-Step Solution

Answer

The general solution is \(y(x) = C_1 x^3 + C_2 x^4\).

Step 1: Check if the Functions are Solutions

To check if \(x^3\) and \(x^4\) are solutions, we differentiate them and substitute back into the differential equation. For \(y = x^3\), the derivatives are \(y' = 3x^2\) and \(y'' = 6x\). Substitute into the equation: \[x^2(6x) - 6x(3x^2) + 12x^3 = 6x^3 - 18x^3 + 12x^3 = 0.\] This satisfies the equation. Similarly, for \(y = x^4\), the derivatives are \(y' = 4x^3\) and \(y'' = 12x^2\). Substitute into the equation: \[x^2(12x^2) - 6x(4x^3) + 12x^4 = 12x^4 - 24x^4 + 12x^4 = 0.\] This also satisfies the equation.

Step 2: Check Linear Independence

We verify linear independence using the Wronskian. The Wronskian of functions \(f(x) = x^3\) and \(g(x) = x^4\) is computed as:\[W(f,g) = \begin{vmatrix} x^3 & x^4 \ 3x^2 & 4x^3 \end{vmatrix} = x^3 \cdot 4x^3 - x^4 \cdot 3x^2 = 4x^6 - 3x^6 = x^6.\]Since the Wronskian \(W(f,g) = x^6\) is non-zero for \(x > 0\), the functions are linearly independent.

Step 3: Form the General Solution

Since \(x^3\) and \(x^4\) form a fundamental set of solutions and are linearly independent, the general solution to the differential equation is a linear combination of these solutions: \[ y(x) = C_1 x^3 + C_2 x^4, \]where \(C_1\) and \(C_2\) are arbitrary constants.

Key Concepts

Fundamental Set of Solutions

In the world of differential equations, the term 'fundamental set of solutions' refers to a group of solutions that can be used to express the general solution of a differential equation. When solving a linear differential equation, finding particular solutions is not enough.
We need to find a set of solutions that are linearly independent. These solutions form a basis for all possible solutions of the differential equation.Simply put, if we have a differential equation of order 'n', we look for 'n' solutions that are independent of one another. In this exercise, the functions given are \(x^3\) and \(x^4\).
When checked, both functions satisfy the original differential equation, making them valid candidates for the fundamental set of solutions. Hence, any solution for the differential equation is a linear combination of these functions.

Linear Independence

Linear independence is a crucial concept when dealing with differential equations. For functions to be considered a fundamental set of solutions, they must be linearly independent.
This means that no one function in the set can be written as a linear combination of the others.To check if functions \(x^3\) and \(x^4\) are linearly independent, we can use the Wronskian determinant. If the Wronskian is non-zero over the interval of interest, the functions are linearly independent.
In this exercise, the Wronskian of \(x^3\) and \(x^4\) is computed and found to be \(x^6\). Since \(x^6\) is not zero for \(x > 0\), \(x^3\) and \(x^4\) are indeed linearly independent.
This property ensures that our solutions cover all possibilities for the differential equation, enabling the construction of the general solution.

Wronskian Determinant

The Wronskian determinant is a powerful tool in determining the linear independence of functions. Named after Józef Hoene-Wroński, it is a function of the solutions and their derivatives.
By computing this determinant, we can verify if a set of functions are linearly independent.Given two functions, \(f(x)\) and \(g(x)\), the Wronskian \(W(f,g)\) is calculated as the determinant of a matrix with the functions and their first derivatives:

• The matrix looks like this:\[\begin{vmatrix} f(x) & g(x) \ f'(x) & g'(x) \end{vmatrix}\]
• If the Wronskian \(W(f,g)\) is not zero, the functions are independent.

In this problem, the determinant calculation of \(f(x) = x^3\) and \(g(x) = x^4\) yields \(W = x^6\).
This outcome indicates linear independence, confirming that \(x^3\) and \(x^4\) form a basis for the solutions to the differential equation over the specified interval \((0, \infty)\).