Question

In Problem 17 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''=x+y², y(0)=1, y'(0)=1

Short answer

The solution of the above question is y(x)=1+x+(1/2)x²+(1/2)x³+(1/6)x⁴+(1/10)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''=x+y²

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

=1

y'''(x) =1+2yy'

y'''(0)=1+ 2× 1× 1

=3

y⁴(x)= 2yy''+ 2(y')²

y⁴(0)= 2× 1× 1+2× (1)²

=4

y⁵(x)=2y'y''+2yy'''+4y'y''

=2yy'''+6y'y''

y⁵(0)=2× 1× 3+6× 1× 1

=6+6

=12

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

y(x)=1+ (1/1)x [(1/2× 1)x²+ [3/(3× 2× 1)]x³+[4/(4× 3× 2× 1)]x⁴+[5/(5× 4× 3× 2× 1)] x⁵

=1+x+(1/2)x²+ (1/2)x³+(1/6)x⁴+(1/10)x⁵

Graph

Graph is included in the following question;

Hence, the final answer is y(x)=1+x+(1/2)x²+ (1/2)x³+(1/6)x⁴+(1/10)x⁵