Question

In Problem 18 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^''+y²=1, y(0)=2, y^'(0)=3

Short answer

The solution of the above question is y(x)=2+3x-(3/2)x²-2x³-(1/4)x⁴+(17/20)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^''+y²=1

y''= 1-y²

y(0)=2

y'(0)=3

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)=1-(2)²

=-3

y'''(x)= -2yy'

y'''(0)=-2× 2× 3

= -12

y⁴(x) = -2yy''- 2(y')²

y⁴(0) = -2× 2× -3-2× (3)²

=12-18

= -6

y⁵(x)= -2 y'y''-2yy'''-4y'y''

= -2yy'''-6y'y''

y⁵(0)= -2× 2× -12-6× 3× -3

=58+54

=102

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

y(x)=2+(3/1)x+(-3/2× 1)x²- [(-12/3× 2× 1)]x³+[-6/4× 3× 2× 1]x⁴+ [102/5× 4× 3× 2× 1] x⁵

=2+3x(-3/2)x²- 2x³-(1/4) x⁴+ (17/20) x⁵

Graph

Graph is included in the following question;

Hence, he final answer is y=2+3x(-3/2)x²- 2x³-(1/4) x⁴+ (17/20) x⁵