Using Taylor's series method find y at x =1.1 and
1.2 by solving dy/dx = x2 + y2 given y(1)=2.3.
Answers
Answer:
Taylor Series method
h=x-x0
y1=y0+hy0′+
h2
2!
y0′′+
h3
3!
y0′′′+
h4
4!
y0′′′′+...
Examples
1. Find y(0.2) for y′=x2y-1, y(0) = 1, with step length 0.1 using Taylor Series method
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
Step-by-step explanation:
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Answer:
Step-by-step explanation: