Pharaoh's Taylor series question from Yahoo Answers

[CaptainBlack]

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​

 

[member38644]

.2)
Next, the modified Euler method can be used to find the approximate solution for this problem. The modified Euler method is given by:

\(y_{n+1}=y_n+hf(x_n+\frac{h}{2},y_n+\frac{h}{2}f(x_n,y_n))\)

where \(f(x,y)=y"-0.2(1-y^2)y'+y\).

Using \(h=0.1\) and starting with \(y_0=0.1\), we can calculate the values of \(y_1\) and \(y_2\) as follows:

\(y_1\approx 0.1+0.1\left(0.1+\frac{0.1}{2}\left(0.1-0.2(1-0.1^2)0.1+0.1\right)\right)\approx 0.109625\)

\(y_2\approx 0.109625+0.1\left(0.2+\frac{0.1}{2}\left(0.2-0.2(1-0.109625^2)0.2+0.109625\right)\right)\approx 0.10961\)

So, the approximate solution at \(x=0.2\) using the modified Euler method is \(y(0.2)\approx 0.10961\). This is slightly different from the value obtained using the Taylor series method, but both methods provide a good approximation to the actual solution.