Sure. Here is the solution:
Let y1(x) = x(1 + e^x) and y2(x) = x(2 − e^x) be solutions of the differential equation y + p(x)y + q (x) y = 0, where the functions p(x) and q(x) are continuous in the open interval I =]0 , [infinity][. Without trying to find the functions p(x) and q(x), show that the functions y3(x) = x and y4(x) = xe^x form a fundamental set of solutions of the differential equation.
To show that y3(x) and y4(x) form a fundamental set of solutions of the differential equation, we need to show that they are linearly independent and that their Wronskian is not equal to zero.
To show that y3(x) and y4(x) are linearly independent, we can use the fact that any linear combination of two linearly independent solutions is also a solution. In this case, if we let y(x) = c1y3(x) + c2y4(x), where c1 and c2 are constants, then y + p(x)y + q (x) y = c1(x + p(x)x + q (x)x) + c2(xe^x + p(x)xe^x + q (x)xe^x) = 0. This shows that y(x) is a solution of the differential equation for any values of c1 and c2. Therefore, y3(x) and y4(x) are linearly independent.
To show that the Wronskian of y3(x) and y4(x) is not equal to zero, we can calculate the Wronskian as follows: W(y3, y4) = y3y4′ − y3′y4 = x(xe^x) − (x + xe^x)(x) = xe^x(x − 1) ≠ 0. This shows that the Wronskian of y3(x) and y4(x) is not equal to zero. Therefore, y3(x) and y4(x) form a fundamental set of solutions of the differential equation.