Question

verify that the functions \(y_{1}\) and \(y_{2}\) are solutions of the given differential equation. Do they constitute a fundamental set of solutions? $$ x^{2} y^{\prime \prime}-x(x+2) y^{\prime}+(x+2) y=0, \quad x>0 ; \quad y_{1}(x)=x, \quad y_{2}(x)=x e^{x} $$

Short answer

Answer: Yes, the functions \(y_1(x)=x\) and \(y_2(x)=xe^x\) form a fundamental set of solutions for the given differential equation.

Step-by-step solution

Verify \(y_{1}(x)\) as a solution

To check if \(y_{1}(x) = x\) is a solution, we need to find its first and second derivative and substitute them into the given differential equation. Let's find the required derivatives: $$ y_{1}(x) = x \\ y_{1}^{\prime}(x) = 1 \\ y_{1}^{\prime \prime}(x) = 0 $$ Now substitute the derivatives and the function into the differential equation: $$ x^{2}(0) - x(x+2)(1) + (x+2)(x) = 0 \\ -x(x+2) + x(x+2) = 0 $$ The equation holds true, which means \(y_{1}(x) = x\) is a solution to the given differential equation.

Verify \(y_{2}(x)\) as a solution

Now, we'll check if \(y_{2}(x)=xe^x\) is a solution. We'll find its first and second derivatives and substitute them in the given differential equation: $$ y_{2}(x) = x e^{x} \\ y_{2}^{\prime}(x) = e^{x} + x e^{x} \\ y_{2}^{\prime \prime}(x) = 2e^{x} + x e^{x} $$ Substitute the derivatives and the function into the differential equation: $$ x^{2}(2e^{x} + xe^{x}) - x(x+2)(e^{x} + xe^{x}) + (x+2)(xe^{x}) \\ = x^{2}e^{x}(2 + x) - x(x+2)e^{x}(1+x) + (x+2)xe^{x} $$ The equation holds true, which means \(y_{2}(x) = xe^x\) is a solution to the given differential equation.

Check if the solutions form a fundamental set of solutions

To check if \(y_{1}(x)\) and \(y_{2}(x)\) form a fundamental set of solutions, we will find their Wronskian, which is: $$ W(y_{1}, y_{2}) = \begin{vmatrix} y_{1} & y_{2} \\ y_{1}^{\prime} & y_{2}^{\prime} \end{vmatrix} = y_{1}y_{2}^{\prime} - y_{1}^{\prime}y_{2} $$ Now substitute the functions and their first derivatives: $$ W(y_{1}, y_{2}) = (x)(e^{x} + xe^{x}) - (1)(xe^{x}) \\ = xe^{x} + x^2e^{x} - xe^{x} \\ = x^2e^{x} $$ Since the Wronskian, \(W(y_{1}, y_{2})=x^2e^{x}\), is non-zero for \(x>0\), we can conclude that \(y_{1}(x)=x\) and \(y_{2}(x)=xe^x\) form a fundamental set of solutions.

Key concepts

Fundamental Set of Solutions

When dealing with differential equations, particularly linear homogenous equations, it's important to find a fundamental set of solutions. But what exactly does that mean? A fundamental set of solutions consists of solutions that span the solution space of the differential equation.
If you have two functions that solve the differential equation and their combinations can form any other solution, you have a fundamental set. For our equation, the solutions were given as \(y_1(x) = x\) and \(y_2(x) = xe^x\). These two solutions were verified to satisfy the differential equation individually.
The crucial test for a fundamental set is the Wronskian, which we'll look into next. A non-zero Wronskian on the interval of interest usually indicates that the functions make up a fundamental set. In this case, since \(W(y_1, y_2)\) was determined to be \(x^2e^x\), which is non-zero for \(x>0\), they indeed form a fundamental set.
This means that any solution to the differential equation can be expressed as a linear combination of \(y_1(x)\) and \(y_2(x)\), ensuring the completeness of solutions over the interval.

Wronskian

The Wronskian is an important tool in verifying whether two or more functions form a fundamental set of solutions for a linear differential equation. It’s named after Józef Hoene-Wroński. In essence, the Wronskian is a determinant that helps us determine the linear independence of solutions.
For us, given two functions \(y_1(x) = x\) and \(y_2(x) = xe^x\), the Wronskian is calculated by:

• Constructing a matrix from the functions and their derivatives
• Manually calculating the determinant of that matrix

For this problem, the Wronskian \(W(y_1, y_2)\) is:
\[ W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2'\end{vmatrix} = y_1y_2' - y_1'y_2 \]
Substituting the derivatives and simplifying, we get:
\[ W(y_1, y_2) = xe^x + x^2e^x - xe^x = x^2e^x \]
Since this result is non-zero for \(x > 0\), it confirms that \(y_1(x)\) and \(y_2(x)\) are linearly independent, verifying they form a fundamental set of solutions.

Solution Verification

Verifying if functions are solutions to a given differential equation involves substituting them into the equation and checking if it holds true. Here, we were tasked with verifying \(y_1(x) = x\) and \(y_2(x) = xe^x\) as solutions to the differential equation:
\[ x^2 y'' - x(x+2) y' + (x+2)y = 0 \]
For \(y_1(x)\): We calculated that its first and second derivatives are \(y_1'(x) = 1\) and \(y_1''(x) = 0\). Substituting these into the differential equation, the equation simplifies to zero, confirming \(y_1(x)\) is a solution.
For \(y_2(x)\): We found its first and second derivatives as \(y_2'(x) = e^x + xe^x\) and \(y_2''(x) = 2e^x + xe^x\). Substituting these into the differential equation, it also simplified to zero, confirming \(y_2(x)\) as a solution.
Thus, both functions satisfy the differential equation, and based on the earlier Wronskian check, they form a fundamental set of solutions for the specified range \(x > 0\). This ensures any solution can be expressed as a combination of these two verified functions.