Respuesta :
Answer:
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 .
