Show that the set of solutions of a second-order linear homogeneous differential equation is linearly independent. {eax, xeax}, a ≠ b
Find the Wronskian for the set of solutions.
{eax, xeax}, a ≠ b

The set of solutions are linearly independent because the wronskian is

e^(2ax) ≠ 0

Step-by-step explanation:

To show that the set of solutions {e^(ax), xe^(ax)} are linearly independent, we find the Wronskian of the set of solutions.

Wronskian of y1 and y2 is given as:

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

y1 = e^(ax)

y1' = ae^(ax)

y2 = xe^(ax)

y2' = axe^(ax) + e^(ax)

W(y1, y2) = e^(ax)[axe^(ax) + e^(ax)] - ae^(ax).xe^(ax)

= (ax + 1)e^(2ax) - axe^(2ax)

= e^(2ax)

The wronskian of the given set of solutions is e^(2ax), which is not zero.

Because it is not zero, the set of solutions are linearly independent.

c

step-by-step explanation:

it isn't d, because the interquartile range is from 27-57. and the interquartile range is basically half the data, it's pretty obvious.

it isn't b, because like i said earlier, the interquartile range is from 27-57. 57-27 is 30. the real range has to be longer

not a either. although it looks like it, 3/4 of the data is not greater than 27. but if c is wrong than this should be correct

the correct answer is c, because 13 is the point in which the plot starts

use formula for the circle a=3.14

step-by-step explanation: