In Problems 41 and 42 solve the given initial-value problem in which the input function g(x) is discontinuous. [Hint: Solve each problem on two intervals, and then find a solution so that y and y are continuous at x p2 (Problem 41) and at x p (Problem 42).] 41. y 4y g(x), y(0) 1, y(0) 2,

hello your question is poorly written attached below is the complete question

attached below

Step-by-step explanation:

using

m^2 - 2m + 10 = 0

m = 2± (√4 - 4(1)(10)) / (2(1)) = 1 ± 3i

Hence the complementary function ; Yc = e^x ( C1 cos3x + C2 sin3x )

attached below is the detailed solution

the least common denominatore of 7 and 2 is 9

step-by-step explanation:

7+2=9

i think da answer is

d.square root of 3