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

#### CaptainBlack

##### Well-known member
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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