Suppose you solved a second-order equation by rewriting it as a system and found two scalar solutions: y = e^5x and z = e^2x. Think of the corresponding vector solutions y1 and y2 and use the Wronskian to show that the solutions are linearly independent Wronskian = det [ ] = These solutions are linearly independent because the Wronskian is [ ] Choose for all x.

We can calculate the Wronskian using the fundamental solutions that we are provided and their corresponding the derivatives, since the Wroskian is defined as the following determinant.

[tex]W = \left|\begin{array}{cc}y&z\\y'&z'\end{array}\right|[/tex]

Thus replacing the functions of the exercise we get:

[tex]W = \left|\begin{array}{cc}e^{5x}&e^{2x}\\5e^{5x}&2e^{2x}\end{array}\right|[/tex]

Working with the determinant we get

[tex]W = 2e^{7x}-5e^{7x}\\W=-3e^{7x}[/tex]

Thus we have found that the Wronskian is not 0, so the solutions are linearly independent.