(0.2) = 0.80227 from = 2 − 1 , (0) = 1 by Taylor series method.
Answers
Answer:
Solution:
Given y′=x2y-1,y(0)=1,h=0.1,y(0.2)=?
Here, x0=0,y0=1,h=0.1
Differentiating successively, we get
y′=x2y-1
y′′=2xy+x2y′
y′′′=2y+4xy′+x2y′′
y′′′′=6y′+6xy′′+x2y′′′
Now substituting, we get
y0′=x
2
0
y0-1=-1
y0′′=2x0y0+x
2
0
y0′=0
y0′′′=2y0+4x0y0′+x
2
0
y0′′=2
y0′′′′=6y0′+6x0y0′′+x
2
0
y0′′′=-6
Putting these values in Taylor's Series, we have
y1=y0+hy0′+
h2
2!
y0′′+
h3
3!
y0′′′+
h4
4!
y0′′′′+...
=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 (x1,y1) in place of (x0,y0) and repeat the process
Now substituting, we get
y1′=x
2
1
y1-1=-0.991
y1′′=2x1y1+x
2
1
y1′=0.17015
y1′′′=2y1+4x1y1′+x
2
1
y1′′=1.40592
y1′′′′=6y1′+6x1y1′′+x
2
1
y1′′′=-5.82983
Putting these values in Taylor's Series, we have
y2=y1+hy1′+
h2
2!
y1′′+
h3
3!
y1′′′+
h4
4!
y1′′′′+...
=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
Explanation: