Consider the differential equation x3y ''' + 12x2y '' + 25xy ' − 25y = 0; x, x−5, x−5 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(x, x−5, x−5 ln(x))

The set of solutions are independent, and therefore form a fundamental set of solution.

Step-by-step explanation:

We want to verify that the functions x, x - 5, (x - 5)lnx for a fundamental set of solutions to the differential equation

x³y''' + 12x²y'' + 25xy' - 25y = 0

On the interval (0, infinity).

By definition, the Wronskian of y_1, y_2, y_3 is given as the determinant:

W(y_1, y_2, y_3) =

|y_1y_2y_3|

|y'_1y'_2y'_3|

|y''_1y''_2y''_3|

If W(y_1, y_2, y_3) ≠ 0, the functions are independent.

If the funtions form a fundamental set of solutions to the differential equation, then they are independent, otherwise, the are dependent, and hence, do not form a set of fundamental solution.

Now, let us check if the given functions form a set of fundamental solution.

W(x, x - 5, x - 5 ln(x)) =

|x(x - 5)(x - 5)lnx|

|11(x - 5)/x + lnx|

|00(5 + x)/x²|

= x((5 + x)/x² - 0) - (x - 5)((x + 5)/x² - 0) + (x - 5)lnx(0 - 0)

= (x + 5)/x - (x - 5)(x + 5)/x²

= (x + 5)/x - (x² - 25)/x²

= (5x + x² - x² - 25)/x²

= (5x - 25)/x²

= 5(x - 5)/x²

≠0

Hence, the set of solutions are independent, and therefore form a fundamental set of solution.

answer is 144

step-by-step explanation:

step-by-step explanation:

the true ones are a b and d