Question

verify that the given functions are solutions of the differential equation, and determine their Wronskian. $$ x^{3} y^{\prime \prime \prime}+x^{2} y^{\prime \prime}-2 x y^{\prime}+2 y=0 ; \quad x, \quad x^{2}, \quad 1 / x $$

Short answer

Answer: Yes, the functions \(x\), \(x^2\), and \(\frac{1}{x}\) are linearly independent solutions to the given third-order differential equation.

Step-by-step solution

Verify that the functions are solutions of the differential equation.

We will insert the functions \(x\), \(x^2\), and \(\frac{1}{x}\) into the given differential equation one by one. 1. Function: \(y = x\) First and second derivatives: \(y' = 1\) and \(y'' = 0\), and \(y''' = 0\) Insert derivatives into the equation: $$x^3(0) + x^2(0) - 2x(1) + 2(x) = 0 \implies 0$$ Therefore, \(y = x\) is a solution. 2. Function: \(y = x^2\) First and second derivatives: \(y' = 2x\) and \(y'' = 2\), and \(y''' = 0\) Insert derivatives into the equation: $$x^3(0) + x^2(2) - 2x(2x) + 2(x^2) = 2x^2 - 4x^2 + 2x^2 = 0$$ Therefore, \(y = x^2\) is a solution. 3. Function: \(y = \frac{1}{x}\) First and second derivatives: \(y' = -\frac{1}{x^2}\) and \(y'' = \frac{2}{x^3}\), and \(y''' = -\frac{6}{x^4}\) Insert derivatives into the equation: $$x^3\left(-\frac{6}{x^4}\right) + x^2\left(\frac{2}{x^3}\right) - 2x\left(-\frac{1}{x^2}\right) + 2\left(\frac{1}{x}\right) = -6 + 2 + 2 + 2 = 0$$ Therefore, \(y = \frac{1}{x}\) is a solution.

Calculate the Wronskian.

Now, we will calculate the Wronskian for the functions \(x\), \(x^2\), and \(\frac{1}{x}\) to verify their linear independence. The definition of the Wronskian for three functions \(f(x)\), \(g(x)\), and \(h(x)\) is: $$W(x) = \det \begin{bmatrix} f(x) & g(x) & h(x) \\ f'(x) & g'(x) & h'(x) \\ f''(x) & g''(x) & h''(x) \end{bmatrix}$$ For our functions, the Wronskian is: $$W(x) = \det \begin{bmatrix} x & x^2 & \frac{1}{x} \\ 1 & 2x & -\frac{1}{x^2} \\ 0 & 2 & \frac{2}{x^3} \end{bmatrix}$$ Expanding the determinant, we get: $$W(x) = x\det \begin{bmatrix} 2x & -\frac{1}{x^2} \\ 2 & \frac{2}{x^3} \end{bmatrix} - x^2\det \begin{bmatrix} 1 & -\frac{1}{x^2} \\ 0 & \frac{2}{x^3} \end{bmatrix} + \frac{1}{x}\det \begin{bmatrix} 1 & 2x \\ 0 & 2 \end{bmatrix}$$ Which simplifies down to: $$W(x) = x(4 - 2) - x^2(2) + \frac{1}{x}(2) = 2x - 2x^2 + \frac{2}{x}$$ The Wronskian, \(W(x)\) is a non-zero function, which means the functions \(x\), \(x^2\), and \(\frac{1}{x}\) are linearly independent. In conclusion, the given functions are indeed solutions of the differential equation, and their Wronskian shows that they are linearly independent.

Key concepts

Wronskian

When dealing with systems of differential equations, the Wronskian is an important concept. It helps determine whether a set of functions is linearly independent. Here, we focus on three functions:

• \( x \)
• \( x^2 \)
• \( \frac{1}{x} \)

To compute the Wronskian for these functions, we construct a determinant using their derivatives. The Wronskian matrix is given by:\[W(x) = \det \begin{bmatrix} f(x) & g(x) & h(x) \ f'(x) & g'(x) & h'(x) \ f''(x) & g''(x) & h''(x) \end{bmatrix}\]For our functions, plugging in the derivatives, the matrix becomes:\[W(x) = \det \begin{bmatrix} x & x^2 & \frac{1}{x} \ 1 & 2x & -\frac{1}{x^2} \ 0 & 2 & \frac{2}{x^3} \end{bmatrix}\]Solving the determinant, we find that \( W(x) = 2x - 2x^2 + \frac{2}{x} \). If the Wronskian is not equal to zero for any value in its domain, it proves the functions are linearly independent.

Linear Independence

Linear independence is a key concept when working with solutions to differential equations. A set of functions is linearly independent if no function in the set can be written as a linear combination of the others.
For the functions \( x \), \( x^2 \), and \( \frac{1}{x} \), checking linear independence involves computing the Wronskian.

• If the Wronskian is zero everywhere, the functions are linearly dependent.
• If it is non-zero at any point, the functions are linearly independent.

In our example, the Wronskian is \( W(x) = 2x - 2x^2 + \frac{2}{x} \). Since this expression is not zero across its domain, these functions are confirmed to be linearly independent. This means that no one function can be recreated from a combination of the others.

Solution Verification

Verifying that a function is a solution to a differential equation involves checking if, when substituted into the equation, it holds true.
To verify solutions for the given differential equation \( x^{3} y^{\prime \prime \prime}+x^{2} y^{\prime \prime}-2 x y^{\prime}+2 y=0 \):1. **Substitute the function and its derivatives** - Input the function \( y = x \) and its derivatives into the equation. Check if the equation equals zero.2. **Check each term separately** - Do the same for the function \( y = x^2 \), ensuring each derivative term is correct, and verify it results in zero.3. **Repeat for all given functions** - For \( y = \frac{1}{x} \), insert into the differential equation and confirm it solves the equation.All three functions—\( x \), \( x^2 \), and \( \frac{1}{x} \)—solve the equation, thus verifying they are indeed solutions. This process ensures the correctness of solutions related to differential equations, a vital part of understanding and application in calculus.