Formula
7. Taylor Series method
h=x-x_0
y_1 = y_0 + hy_0' + h^2/(2!) y_0'' + h^3/(3!) y_0''' + h^4/(4!) y_0'''' + ...
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Examples
1. Find y(0.2) for y'=x^2y-1, y(0) = 1, with step length 0.1 using Taylor Series method
Solution:
Given y'=x^2y-1, y(0)=1, h=0.1, y(0.2)=?
Here, x_0=0,y_0=1,h=0.1
Differentiating successively, we get
y'=x^2y-1
y''=2xy+x^2y'
y'''=2y+4xy'+x^2y''
y''''=6y'+6xy''+x^2y'''
Now substituting, we get
y_0'=x_0^2y_0-1=-1
y_0''=2x_0y_0+x_0^2y_0'=0
y_0'''=2y_0+4x_0y_0'+x_0^2y_0''=2
y_0''''=6y_0'+6x_0y_0''+x_0^2y_0'''=-6
Putting these values in Taylor's Series, we have
y_1 = y_0 + hy_0' + h^2/(2!) y_0'' + h^3/(3!) y_0''' + h^4/(4!) y_0'''' + ...
=1+0.1*(-1)+(0.1)^2/(2!)*(0)+(0.1)^3/(3!)*(2)+(0.1)^4/(4!)*(-6)+...
=1-0.1+0+0.00033+0+...
=0.90031
:.y(0.1)=0.90031
Again taking (x_1,y_1) in place of (x_0,y_0) and repeat the process
Now substituting, we get
y_1'=x_1^2y_1-1=-0.991
y_1''=2x_1y_1+x_1^2y_1'=0.17015
y_1'''=2y_1+4x_1y_1'+x_1^2y_1''=1.40592
y_1''''=6y_1'+6x_1y_1''+x_1^2y_1'''=-5.82983
Putting these values in Taylor's Series, we have
y_2 = y_1 + hy_1' + h^2/(2!) y_1'' + h^3/(3!) y_1''' + h^4/(4!) y_1'''' + ...
=0.90031+0.1*(-0.991)+(0.1)^2/(2!)*(0.17015)+(0.1)^3/(3!)*(1.40592)+(0.1)^4/(4!)*(-5.82983)+...
=0.90031-0.0991+0.00085+0.00023+0+...
=0.80227
:.y(0.2)=0.80227
This material is intended as a summary. Use your textbook for detail explanation.
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