91Ó°ÊÓ

In Problems 10–13, use the vectorized Euler method with h = 0.25 to find an approximation for the solution to the given initial value problem on the specified interval.

${\mathbf{t}}^{\mathbf{2}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{t}\mathbf{+}\mathbf{2}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{on}\mathbf{[}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{]}$

In Problems 10–13, use the vectorized Euler method with h = 0.25 to find an approximation for the solution to the given initial value problem on the specified interval.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{on}\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{]}$

In Problems 14–24, you will need a computer and a programmed version of the vectorized classical fourth-order Runge–Kutta algorithm. (At the instructor’s discretion, other algorithms may be used.)†

Using the vectorized Runge–Kutta algorithm with h = 0.5, approximate the solution to the initial value$\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{5}\mathbf{ty}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\frac{\mathbf{2}}{\mathbf{3}}$ problemat t = 8.

Compare this approximation to the actual solution $\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{t}}^{\frac{\mathbf{5}}{\mathbf{3}}}\mathbf{-}\mathbf{t}$.

In Problems 14–24, you will need a computer and a programmed version of the vectorized classical fourth-order Runge–Kutta algorithm. (At the instructor’s discretion, other algorithms may be used.)†

Using the vectorized Runge–Kutta algorithm, approximate the solution to the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$at $\mathbf{t}\mathbf{=}\mathbf{1}$. Starting with $\mathbf{h}\mathbf{=}\mathbf{1}$, continue halving the step size until two successive approximations of both $\mathbf{y}\left(\mathbf{1}\right)$and$\mathbf{y}\mathbf{\text{'}}\left(\mathbf{1}\right)$differ by at most 0.1.

In Problems 14–24, you will need a computer and a programmed version of the vectorized classical fourth-order Runge–Kutta algorithm. (At the instructor’s discretion, other algorithms may be used.)†

Using the vectorized Runge–Kutta algorithm for systems with$\mathit{h}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{175}$, approximate the solution to the initial value problem$\mathbf{x}\mathbf{\text{'}}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{;}\mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}$ at$\mathbf{t}\mathbf{=}\mathbf{1}$.

Compare this approximation to the actual solution.

In Problems 14–24, you will need a computer and a programmed version of the vectorized classical fourth-order Runge–Kutta algorithm. (At the instructor’s discretion, other algorithms may be used.)†

Using the vectorized Runge–Kutta algorithm, approximate the solution to the initial value problem

$$\begin{gathered} \frac{\mathbf{du}}{\mathbf{dx}}\mathbf{=}\mathbf{3}\mathbf{u}\mathbf{-}\mathbf{4}\mathbf{v}\mathbf{;}\mathbf{u}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{\text{'}} \\ \frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{=}\mathbf{2}\mathbf{u}\mathbf{-}\mathbf{3}\mathbf{v}\mathbf{;}\mathbf{v}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1} \end{gathered}$$

at x = 1. Starting with h=1, continue halving the step size until two successive approximations of u(1)and v(1) differ by at most 0.001.

Combat Model.A simplified mathematical model for conventional versus guerrilla combat is given by the system$\mathbf{x}{\mathbf{\text{'}}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{)}{\mathbf{x}}_{\mathbf{1}}{\mathbf{x}}_{\mathbf{2}}\mathbf{;}  {\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{10}\mathbf{;}  \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{2}}\mathbf{=}\mathbf{-}{\mathbf{x}}_{\mathbf{1}}\mathbf{;}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{15}$ where${\mathbf{x}}_{\mathbf{1}}$ and${\mathbf{x}}_{\mathbf{2}}$ are the strengths of guerrilla and conventional troops, respectively, and 0.1 and 1 are the combat effectiveness coefficients.Who will win the conflict: the conventional troops or the guerrillas? [Hint:Use the vectorized Runge–Kutta algorithm for systems with h=0.1to approximate the solutions.]

In Project C of Chapter 4, it was shown that the simple pendulum equation$\mathbf{θ}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{²õ¾\pm ²Ôθ}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}$ has periodic solutions when the initial displacement and velocity show that the period of the solution may depend on the initial conditions by using the vectorized Runge–Kutta algorithm with h= 0.02 to approximate the solutions to the simple pendulum problem on

[0, 4] for the initial conditions:

localid="1664100454791" $$\begin{gathered} \mathbf{(}\mathbf{a}\mathbf{)}\mathbf{ }\mathbf{ }\mathbf{θ}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{,}\mathbf{θ}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0} \\ \mathbf{(}\mathbf{b}\mathbf{)}\mathbf{ }\mathbf{ }\mathbf{θ}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{,}\mathbf{θ}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0} \\ \mathbf{(}\mathbf{c}\mathbf{)}\mathbf{ }\mathbf{ }\mathbf{θ}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{,}\mathbf{θ}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0} \end{gathered}$$

[Hint: Approximate the length of time it takes to reach].

Oscillations and Nonlinear Equations. For the initial value problem $\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{\left)}\mathbf{\right(}\mathbf{1}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}\mathbf{)}\mathbf{x}\mathbf{\text{'}}\mathbf{+}\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{x}\mathbf{(}\mathbf{0}\mathbf{)}\mathbf{=}{\mathbf{x}}_{\mathbf{o}}\mathbf{,}\mathbf{x}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$using the vectorized Runge–Kutta algorithm with h = 0.02 to illustrate that as t increases from 0 to 20, the solution x exhibits damped oscillations when ${\mathbf{x}}_{\mathbf{o}}\mathbf{=}\mathbf{1}$, whereas exhibits expanding oscillations when ${\mathbf{x}}_{\mathbf{o}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{1}\mathbf{,}$.

Nonlinear Spring.The Duffing equation$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{+}{\mathbf{ry}}^{\mathbf{3}}\mathbf{=}\mathbf{0}$ where ris a constant is a model for the vibrations of amass attached to a nonlinearspring. For this model, does the period of vibration vary as the parameter ris varied?

Does the period vary as the initial conditions are varied? [Hint:Use the vectorized Runge–Kutta algorithm with h= 0.1 to approximate the solutions for r= 1 and 2,

with initial conditions$\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{a}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$ for a = 1, 2, and 3.]