Verify that the given two-parameter family of functions is the general solution of the nonhomogeneous differential equation on the indicated interval.

2x2y'' + 5xy' + y = x2 − x;

y = c1x−1/2 + c2x−1 + 1/15(x^2)-1/6(x), (0,infinity)

The functions (x^-1/2) and (x^-1) satisfy the differential equation and are linearly independent since W(x^-1/2, x^-1)= ? for 0

The question is :

2x²y'' + 5xy' + y = x² - x;

y = c1x^(1/2) + c2x^(-1) + 1/15(x^2) - 1/6(x), (0,infinity)

The functions (x^-1/2) and (x^-1) satisfy the differential equation and are linearly independent since W(x^-1/2, x^-1)= ? for 0

The functions x^(-1/2) and x^(-1) are linearly independent since their wronskian is (-1/2)x^(-5/2) ≠ 0.

Step-by-step explanation:

Suppose the functions x^(-1/2) and x^(-1) satisfy the differential equation 2x²y'' + 5xy' + y = x² - x;

and are linearly independent, then their wronskian is not zero.

Wronskian of y1 and y2 is given as

W(y1, y2) = y1y2' - y1'y2

Let y1 = x^(-1/2)

y1' = (-1/2)x^(-3/2)

Let y2 = x^(-1)

y2' = -x^(-2)

W(y1, y2) =

x^(-1/2)(-x^(-2)) - (-1/2)x^(-3/2)x^(-1)

= -x^(-5/2) + (1/2)(x^(-5/2)

= (-1/2)x^(-5/2)

So, W(y1, y2) = (-1/2)x^(-5/2) ≠ 0

Which means the functions are linearly independent.

should have been evaluated with one odd value and one even value. the expressions have equivalent values when . when , the first expression has a value of 3 and the second

step-by-step explanation: