For each initial-value problem given below, three choices for a possible solution y = y(x) are given. determine whether each choice is or is not a solution to the given initial-value problem. = dx i ii. y(x) = 24x + 1 a..= 4y i. y(x) = €4x b. x y = 2y i. y(x) = x2 with y(o) = 5 ii. y(x) = 5e4x with y(2) = 20 ii. y(x) = 10x iii. y(x) = 5x2 n. 2 19) on
c.- 9y = 0 with y(0) = 1 and y'(o) = 9 i. y(x) = 2e3x – 2-3x ii. y(x) = 237 iii. y(x) = 23x +1. @242 - 4x + 6y = 36x6 with y(1) = 1 and y'(1) = 12 i. y(x) = 10r3 - 9x2 ii. y(x) = 3x - 2x2 ü. y(x) = 3x6 – 2x3 1.5. for the following, let dr.