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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.

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marcelvm

Answer:

The solutions are linearly independent because the Wronskian is not equal to 0 for all x.

The value of the Wronskian is [tex]\bold{W=-3e^{7x}}[/tex]

Step-by-step explanation:

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.

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