numerically approximate the solution to a first order initial-value problem with the Taylor Method - Maple Help

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Student[NumericalAnalysis][Taylor] - numerically approximate the solution to a first order initial-value problem with the Taylor Method

 Calling Sequence Taylor(ODE, IC, t=b, opts) Taylor(ODE, IC, b, opts)

Parameters

 ODE - equation; first order ordinary differential equation of the form $\frac{ⅆ}{ⅆt}y\left(t\right)=f\left(t,y\right)$ IC - equation; initial condition of the form y(a)=c, where a is the left endpoint of the initial-value problem t - name; the independent variable b - algebraic; the point for which to solve; the right endpoint of this initial-value problem opts - (optional) equations of the form keyword=value, where keyword is one of numsteps, output, comparewith, digits, order, or plotoptions; options for numerically solving the initial-value problem

Description

 • Given an initial-value problem consisting of an ordinary differential equation ODE, a range a <= t <= b, and an initial condition y(a) = c, the Taylor command computes an approximate value of y(b) by expanding y(t) into a Taylor polynomial at each step along the interval.
 • If the second calling sequence is used, the independent variable t will be inferred from ODE.
 • The endpoints a and b must be expressions that can be evaluated to floating-point numbers. The initial condition IC must be of the form y(a)=c, where c can be evaluated to a floating-point number.
 • The Taylor command is a shortcut for calling the InitialValueProblem command with the method = taylor option.

Notes

 • To approximate the solution to an initial-value problem using a method other than the Taylor Method, see InitialValueProblem.

Examples

 > $\mathrm{with}\left(\mathrm{Student}[\mathrm{NumericalAnalysis}]\right):$
 > $\mathrm{Taylor}\left(\frac{ⅆ}{ⅆt}y\left(t\right)=\mathrm{cos}\left(t\right),y\left(0\right)=0.5,t=3\right)$
 ${0.6235}$ (1)
 > $\mathrm{Taylor}\left(\frac{ⅆ}{ⅆt}y\left(t\right)=y\left(t\right)+{t}^{2},y\left(1\right)=3.10,t=4,\mathrm{order}=5,\mathrm{output}=\mathrm{Error}\right)$
 ${0.03161}$ (2)
 > $\mathrm{Taylor}\left(\frac{ⅆ}{ⅆt}y\left(t\right)=\mathrm{cos}\left(t\right),y\left(0\right)=0.5,t=3,\mathrm{output}=\mathrm{plot}\right)$