Consider the differential equation x2y'' + xy' + y = 0; cos (ln (x)), sin (ln (x)), (0, [infinity]). Verify that the given functions form a fundamental set of solutions of the differential equation on the indicated interval. The functions satisfy the differential equation and are linearly independent since W (cos (ln (x)), sin (ln (x))) = Correct: Your answer is correct. ≠ 0 for 0 < x < [infinity]. Form the general solution. y =

The Wronskian

W (cos (ln (x)), sin (ln (x))) = 1/x

And

The general solution is

y = C1cos (ln (x)) + C2sin (ln (x)

Where C1 and C2 are constants.

Step-by-step explanation:

To verify if the given functions form a fundamental set of solutions to the given differential equation, we find the Wronskian of the two functions.

The Wronskian of functions y1 and y2 is given as

W (y1, y2) = Det (y1, y2; y1', y2')

Where Det (A) signifies the determinant of matrix A

y1' and y2' are the derivatives of y1 and y2 with respect to x respectively.

y1 = cos (ln (x))

y2 = sin (ln (x))

y1' = - (1/x) sin (ln (x))

y2' = (1/x) cos (ln (x))

W (cos (ln (x)), sin (ln (x))) =

Det (cos (ln (x)), sin (ln (x)); - (1/x) sin (ln (x)), (1/x) cos (ln (x)))

= (1/x) cos² (ln (x)) + (1/x) sin² (ln (x))

= (1/x) [cos² (ln (x)) + sin² (ln (x)) ]

= 1/x (1)

= 1/x

≠ 0

Because the Wronskian is not zero, we say the solutions are linearly independent

The general solution is therefore,

y = C1cos (ln (x)) + C2sin (ln (x)

Where C1 and C2 are constants.