Predictor-corrector, starting values with Taylor method

[Wyrm]

Homework Statement

Hi! I need to solve the following ODE:

$$xy'=1-y+x^2y^2, \qquad y(0)=1$$

using a predictor-corrector method. Starting values need to be found using a Taylor method.
The exact solution is of the form $$\frac{\tan{x}}{x}$$

Homework Equations

Taylor method of third order (for 1st starting value):

$$y_1=y_0+hy_0'+\frac{h^2}{2}y_o''+\frac{h^3}{6}y_o'''$$

The Attempt at a Solution

Right here I bumped into a problem: in order to evaluate the 1st starting value (for example, if I'm using Adams-Bashforth 3rd order as a predictor and Adams-Moulton 3rd order method as a corrector) I get the following:

$$y'=\frac{1}{x}-\frac{y}{x}+xy^2$$

Dividing with $$x=0$$ in order to get yields infinity :(. Any help would be appreciated.

 

[mighty2000]

Hello! Thank you for sharing your problem. Solving ODEs using numerical methods can be challenging, but I'm sure we can figure it out together.

To start, let's review the Taylor method of third order. This method uses the first, second, and third derivatives of the function at the initial point to approximate the solution. In this case, we have y'=1-y+x^2y^2. We can rewrite this as:

y'=1-y+x^2y^2
y'=1-y+x^2(y^2)

Now, let's find the derivatives:

y'= -1 + 2xy^2 + x^2(2yy')
y''= 4xy + 2x^2y' + 2xy(y')
y'''= 4y + 2x + 2x^2(y') + 2xy(y'') + 2xy'(y')

We can substitute these into the Taylor method formula:

y_1=y_0+hy_0'+\frac{h^2}{2}y_o''+\frac{h^3}{6}y_o'''
y_1= y_0 + h(-1 + 2xy_0^2 + x^2(2yy')) + \frac{h^2}{2}(4xy_0 + 2x^2y_0' + 2xy_0(y')) + \frac{h^3}{6}(4y_0 + 2x_0 + 2x^2(y_0') + 2xy_0(y'') + 2xy_0'(y'))

Now, we can plug in our initial condition, y(0)=1, to get our first starting value:

y_1= 1 + h(-1 + 2(0)^2 + (0)^2(2(1)(0))) + \frac{h^2}{2}(4(0)(1) + 2(0)^2(0) + 2(0)(0)) + \frac{h^3}{6}(4(1) + 2(0) + 2(0)^2(0) + 2(0)(0) + 2(0)(0))

Simplifying, we get:

y_1= 1 - h + \frac{h^3