Question

In Problem 19 proceed as in Example 3 and obtain the first six non zero terms of a Taylor series solution, centered at 0, of the given initial value problem. Use a numerical solver and a graphing utility to compare the solution curve with the graph of the Taylor polynomial. y^''=x²+y²-2y^', y(0)=1, y^'(0)=1

Short answer

The solution of the above question is y(x)=1+x-(1/2)x²+(2/3)x³-(1/4)x⁴+(7/60)x⁵

Step-by-step solution

Taylor’s theorem

A Taylor series is a way to approximate any function as a polynomial with an infinite number of terms. Each term of the Taylor polynomial comes from the function's derivatives at a single point.

y(x)= y(0)+ [y^'(0)/1! ] x +[y^''(0)/2!] x²+ y^'''[(0)/3!] x³+ y⁴[(0)/4!] x⁴+ y⁵[(0)/5!] x⁵

Taylor’s theorem will be applied;

We will apply the solution form of Taylor’s theorem in the following answer:

y^''= x²+y²-2y^'

y(0)=1

y^'(0)=1

y(x)= y(0)+ [y^'(0)/1!] x+ [y^''(0)/2!] x²+ [y^'''(0)/3!] x³+[y⁴(0)/4!] x⁴ +[y⁵(0)/5!] x⁵

y^''(0)=0+ (1)²-2× 1

= -1

y^'''(x)= 2x+ 2yy^'+2y^''

y^'''(0)= 0+2× 1× 1-2× (-1)

=2+2

=4

y⁴(x)=2+2 yy^''+ (2y^')²- 2y^'''

y⁴(0)=2+2× 1× (-1)+2× 1²- 2× 4

=-6

y⁵(x)=2yy^'''+2y^'y^''+4y^'y^''-2y⁴

= 6y'y^''+2yy^'''-2y⁴

y⁵(0)= 6× 1× (-1)+2× 1× 4-2× (-6)

=14

Finally, substituting with all the initial conditions, then we have:

y(x)= 1+(1/1)x+(-1/2× 1) x²+ (4/3× 2× 1) x³+ (-6/4× 3× 2× 1) x⁴+ (14/5× 4× 3× 2× 1) x⁵

y(x)=1+x- (1/2) x²+ (2/3) x³-(1/4) x⁴+(7/60)x⁵

Graph

Graph is included in the following question;

Hence, the final answer is y(x)=1+x- (1/2) x²+ (2/3) x³-(1/4) x⁴+(7/60)x⁵