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CaptainBlack

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**Part 1 of Pharaoh's Taylor series and modified Euler question from Yahoo Answers**

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

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

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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