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[SOLVED] Pharaoh's Taylor series question from Yahoo Answers

CaptainBlack

Well-known member
Jan 26, 2012
890
Part 1 of Pharaoh's Taylor series and modified Euler question from Yahoo Answers

consider van der pol's equation
y" - 0.2(1-y^2)y' + y = 0 y(0)=0.1 y'(0)=0.1
1)
You are asked to find the approximate solution for this problem using the Taylor series
method. Your expansion should include the first three non-zero terms and you should
work to six decimal places accuracy. First find the approximate solutions for both y (0.1)
and y’(0.1) using the first three non-zero terms of Taylor series expansion for each
function and then use this information to calculate the approximate solution at x = 0.2.

The Taylor series expansion about \(t=0\) is of the form:


\(y(t)=y(0)+y'(0)t+\frac{y''(0)t^2}{2}+.. \)​


We are given \(y(0)\) and \(y'(0)\) in the initial condition, and so from the equation we have:


\(y''(0) = 0.2(1-(y(0))^2)-y(0)=0.2(1-0.1^2)-0.1=-0.0802\)​


So the Taylor series about \(t=0\) is:


\(y(t)=0.1+0.1t-0.0401t^2+... \)​


and using the first three terms of this we have \(y(0.1)\approx 0.109599\),


Also:


\(y'(t)=0.1-0.0802t+...\)​


and so \(y'(0.1) \approx 0.09198\)


Now the Taylor expansion about \(t=0.1\) is:

\(y(t)=y(0.1)+(t-0.1)y'(0.1)+\frac{(t-0.1)^2y''(0.1)}{2}+...\)​


where \(y''(0.1)=0.2\left(1-(y(0.1))^2\right)y'(0.1)-y(0.1)=-0.08178796\).


So:

\(y(0.2)\approx 0.109599+0.1\times 0.0198-0.1^2\times 0.8178796 = 0.108789 \) to 6 six DP​
 
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