ANSWERED TEST WITH ERRORS: Use your knowledge on Higher-order derivative and Chain Rule to show that z" + 4z' + 8z = 0 if z = e-2*(sin 2x + cos 2x). Z =e-2x(sin 2x + cos 2x) Let u = 2x, then du = 2dx du = 2 dx Z = e-"(sinu + cos u) z' = e-" Dx(sin u + cos u) + (sinu + cos u) Dx(e-") Z' = e-" (cosu - sinu) - e-"(sinu + cosu) = e-" cosu - e"" sinu - e- sinu - e- cosu du ay = -2e- sin u du Substitute - = -2e- sin u, du = 2 using the Chain Rule, dz _ dy du dx au = (-2e-" sin u)2 = -4e-w sinu Substitute U = 2x, Z' = -4e-2x sin 2x To get the second derivative, again let u = 2x then du = 2dx = 2 dx z'= -4e-" sin u z"= -4[e-"D(sinu) + sinu Dx(e-u) = .4e"" cos u - 4e-" sinu dy = -4e-" cosu - 4e- sinu du Substitute -= -4e" cosu - 4e"" sinu, = = 2 using the Chain Rule, dz dy du = (-4e-" cosu - 4e-" sinu)2 = -8e-" cos u - 8e - sinu du du dxSubstitute U = 2x, z" = -8e-2x cos 2x- 8e-2* sin 2x Show z" + 4z' + 8z = 0. (-4e-2* sin 2x ) + 4(-8e-2* cos 2x - 8e-2* sin 2x) + 8[e-2x(sin 2x + cos 2x)] = 0 -4e-2x sin 2x - 32e"2X cos 2x - 32e"2x sin 2x + 8e-2x sin 2x + 8e-2X cos 2x = 0 -24e-2x cos 2x - 28e-2x sin 2x # 0 IDENTIFIED ERROR CORRECTION OF ERROR EXPLANATION OF CORRECTION