8 Example: Expand y = ex sin x upto x3 term using Maclaurin’s theorem.Solution:y(0) = 0y1 = ex cos x + ex sin x = ex cos x + y,y1(0) = 1y2 = - ex sin x + ex cos x + y1 = 2ex cos x,y2(0) = 2y3 = 2ex cos x – 2ex sin x,y3(0) = 2Therefore y = y0 + xy1(0) + (x2/2!) y2(0)+ (x3/3!)y3(0) + ….
9 Example: Expand y = log (1+tan x) up to x3 term using Maclaurin’s theorem.Solution:Given tan x = ey ,y(0) = 0ey. y1 = sec2 x,y1(0) = 1ey y2 + y1 ey = 2 sec x sec x tan x,y2(0) = -1ey(y3 + y2) + (y2 + y1)ey= 2.2 sec x. sec x tan x + 2 sec2 x sec2 xy3(0) = = 4.
10 Example: Obtain Maclaurin’s expression for y = f(x) = log (1 + ex) up to x4 terms.Solution:y(0) = log (1 + 1) = log 2ey = 1 + exDifferentiating we get ey. y1 = ex
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