Example 1:
Solve the initial value problem y' = -2xy²,  y(0) = 1 for y at x = 1 with step lengh .2 using Taylor series method of order four.

Solution:
Given y' = f(x,y) = -2xy² Þ y''   = -2y² - 4xyy'
                                            y''' = -8yy' - 4xy'² - 4xyy'' 
                                           y^iv  = -12y'² - 12yy'' - 12xy'y'' - 4xyy'''
                                           y^v   = -48y'y'' - 16yy''' -12xy''² - 16xy'y''' - 4xyy^iv

The forth order Taylor's formula is

y(x_i+h) = y(x_i) + h y'(x_i, y_i) + h²  y''(x_i, y_i)/2! + h³ y'''(x_i, y_i)/3! + h⁴ y^iv(x_i, y_i)/4! + . . .

given at x=0, y=1 and with h = .2 we have

y'   = -2(0)(1)² = 0.0
y''  = -2(1)² - 4(0)(1)(0) = -2
y''' = -8(1)(0) - 4(0)(0)² - 4(0)(1)(-2) = 0.0
y^iv  = -12(0)² - 12(1)(-2) - 12(0)(0)(-2) - 4(0)(1)(0) = 24

y(0.2) = 1 + .2 (0) + .2² (-2)/2! + 0 + .2⁴ (24)/4!  = .9615

now at x = .2 we have y = .9615

y' = -0.3699,  y'' = -1.5648,  y''' =  3.9397 and  y^iv  =  11.9953

then y(0.4) = 1 + .2 (-.3699) + .2² (-1.5648)/2! + .2³(3.9397)/3! + .2⁴ (11.9953)/4!  =  0.8624
       y(0.6) = 1 + .2 (-.5950) + .2² (-0.6665)/2! + .2³ (4.4579)/3! + .2⁴ (-5.4051)/4!   =  0.7356
       y(0.8) = 1 + .2 (-.6494) + .2² (-0.0642)/2! + .2³ (2.6963)/3! + .2⁴ (-10.0879)/4! =  0.6100
       y(1.0) = 1 + .2 (-.5953) + .2² (-0.4178)/2! + .2³ (0.9553)/3! + .2⁴ (-6.7878)/4!   =  0.5001

now at x = 1.0 we have y = .5001
y' = -0.5001,  y'' = 0.5004,  y''' =  -.000525 and  y^iv  =  -3.0005

Error in the approximation

E₄ = h⁴ (y⁴(1.0) - y⁴(0.8))/ 5! = h⁴ (-3.0005-11.9953)/ 5! = -1.9994e-004

Analytical solution y(x) = 1/(1+x²) at x = 1, y = .5

Alternate method:

By Taylor series method
y(x₀ + x - x₀) = y₀ + (x-x₀)y'(x₀) + (x-x₀)²y''(x₀)/2! + (x-x₀)³y'''(x₀)/3! + (x-x₀)⁴y^iv(x₀)/4! + . . . 
Since x₀ = y' = y''' = 0,  y = 1, y'' = -2 and y^iv = 24 at x = 0, 
y( x ) = 1. -  x² + x⁴  + . . .