Mathematical Models and Numerical Methods Involving First-Order Equations

A shell of mass 2 kg is shot upward with an initial velocity of 200 m/sec. The magnitude of the force on the shell due to air resistance is |v|/20. When will the shell reach its maximum height above the ground? What is the maximum height

When the velocity v of an object is very large, the magnitude of the force due to air resistance is proportional to v² with the force acting in opposition to the motion of the object. A shell of mass 3 kg is shot upward from the ground with an initial velocity of 500 m/sec. If the magnitude of the force due to air resistance is 0.1v², when will the shell reach its maximum height above the ground? What is the maximum height?

Escape Velocity. According to Newton’s law of gravitation, the attractive force between two objects varies inversely as the square of the distances between them. That is, ${\mathbf{F}}_{\mathbf{g}}\mathbf{=}{\mathbf{GM}}_{\mathbf{1}}{\mathbf{M}}_{\mathbf{2}}\mathbf{/}{\mathbf{r}}^{\mathbf{2}}$ where ${\mathbf{M}}_{\mathbf{1}}{\mathbf{andM}}_{\mathbf{2}}$ are the masses of the objects, r is the distance between them (center to center), F_g is the attractive force, and G is the constant of proportionality. Consider a projectile of

constant mass m being fired vertically from Earth (see Figure 3.12). Let t represent time and v the velocity of the projectile.

1. Show that the motion of the projectile, under Earth’s gravitational force, is governed by the equation $\frac{\mathbf{dv}}{\mathbf{dt}}\mathbf{=}\mathbf{-}\frac{{\mathbf{gR}}^{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}$, where r is the distance between the projectile and the center of Earth, R is the radius of Earth, M is the mass of Earth, and $\mathbf{g}\mathbf{=}\mathbf{GM}\mathbf{/}{\mathbf{R}}^{\mathbf{2}}$ .
2. Use the fact that dr /dt = v to obtain $\mathbf{v}\frac{\mathbf{dv}}{\mathbf{dt}}\mathbf{=}\mathbf{-}\frac{{\mathbf{gR}}^{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}$
3. If the projectile leaves Earth’s surface with velocity v_o, show that ${\mathbf{v}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{2}{\mathbf{gR}}^{\mathbf{2}}}{\mathbf{r}}\mathbf{+}{{\mathbf{v}}_{\mathbf{o}}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{gR}$
4. Use the result of part (c) to show that the velocity of the projectile remains positive if and only if ${{\mathbf{v}}_{\mathbf{o}}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{gR}\mathbf{>}\mathbf{0}$. The velocity ${\mathbf{v}}_{\mathbf{e}}\mathbf{=}\sqrt{\mathbf{2}\mathbf{gR}}$ is called the escape velocity of Earth.
5. If g = 9.81 m/sec ² and R = 6370 km for Earth, what is Earth’s escape velocity?
6. If the acceleration due to gravity for the moon is g_m = g/6 and the radius of the moon is R_m = 1738 km, what is the escape velocity of the moon?

A sailboat has been running (on a straight course) under a light wind at 1 m/sec. Suddenly the wind picks up, blowing hard enough to apply a constant force of 600 N to the sailboat. The only other force acting on the boat is water resistance that is proportional to the velocity of the boat. If the proportionality constant for water resistance is b = 100 N-sec/m and the mass of the sailboat is 50 kg, find the equation of motion of the sailboat. What is the limiting velocity of the sailboat under this wind?

In Problem 21 it is observed that when the velocity of the sailboat reaches 5 m/sec, the boat begins to rise out of the water and “plane.” When this happens, the proportionality constant for the water resistance drops to b₀ = 60 N-sec/m. Now find the equation of motion of the sailboat. What is the limiting velocity of the sailboat under this wind as it is planning?

Sailboats A and B each have a mass of 60 kg and cross the starting line at the same time on the first leg of a race. Each has an initial velocity of 2 m/sec. The wind applies a constant force of 650 N to each boat, and the force due to water resistance is proportional to the velocity of the boat. For sailboat A the proportionality constants are  before planing when the velocity is less than 5 m/sec and  when the velocity is above 5 m/sec. For sailboat B the proportionality constants are  before planing when the velocity is less than 6 m/sec and  when the velocity is above . If the first leg of the race is 500 m long, which sailboat will be leading at the end of the first leg?

Rocket Flight. A model rocket having initial mass mo kg is launched vertically from the ground. The rocket expels gas at a constant rate of a kg/sec and at a constant velocity of b m/sec relative to the rocket. Assume that the magnitude of the gravitational force is proportional to the mass with proportionality constant g. Because the mass is not constant, Newton’s second law leads to the equation (mo - αt) dv/dt - αβ = -g(m0 – αt), where v = dx/dt is the velocity of the rocket, x is its height above the ground, and m0 - αt is the mass of the rocket at t sec after launch. If the initial velocity is zero, solve the above equation to determine the velocity of the rocket and its height above ground for 0≤t<m0/α.

An RL circuit with a 5  resistor and a 0.05-H inductor carries a current of 1 A at t = 0, at which time a voltage source E(t) = 5 cos 120t V is added. Determine the subsequent inductor current and voltage.

An RC circuit with a $\mathbf{1}\mathbf{\Omega }$ resistor and a $\mathbf{0}\mathbf{.}\mathbf{000001}\mathbf{-}\mathbf{F}$ capacitor is driven by a voltage $\mathbf{E}\left(t\right)\mathbf{=}\mathbf{sin}\mathbf{100}\mathbf{tV}$. If the initial capacitor voltage is zero, determine the subsequent resistor and capacitor voltages and the current.

The pathway for a binary electrical signal between gates in an integrated circuit can be modeled as an RC circuit, as in Figure 3.13(b); the voltage source models the transmitting gate, and the capacitor models the receiving gate. Typically, the resistance is  $\mathbf{100}\mathbf{\Omega }$ and the capacitance is very small, say,  ${\mathbf{10}}^{\mathbf{-}\mathbf{12}} \mathbf{F}$ (1 picofarad, pF). If the capacitor is initially uncharged and the transmitting gate changes instantaneously from 0 to 5 V, how long will it take for the voltage at the receiving gate to reach (say)  ? (This is the time it takes to transmit a logical “1.”)

If the resistance in the RL circuit of Figure 3.13(a) is zero, show that the current I (t) is directly proportional to the integral of the applied voltage E(t). Similarly, show that if the resistance in the RC circuit of Figure 3.13(b) is zero, the current is directly proportional to the derivative of the applied voltage.

The power generated or dissipated by a circuit element equals the voltage across the element times the current through the element. Show that the power dissipated by a resistor equal  l²R, the power associated with an inductor equals the derivative of $\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{LI}}^{\mathbf{2}}$  and the power associated with a capacitor equals the derivative of  $\frac{\mathbf{1}}{\mathbf{2}}\mathbf{C}{{\mathbf{E}}_{\mathbf{c}}}^{\mathbf{2}}$.

Derive a power balance equation for the RL and RC circuits. (See Problem 5.) Discuss the significance of the signs of the three power terms.

An industrial electromagnet can be modeled as an RL circuit, while it is being energized with a voltage source. If the inductance is 10 H and the wire windings contain   of resistance, how long does it take a constant applied voltage to energize the electromagnet to within 90% of its final value (that is, the current equals 90% of its asymptotic value)?

A 10^-8-F capacitor (10 nano-farads) is charged to 50V and then disconnected. One can model the charge leakage of the capacitor with a RC circuit with no voltage source and the resistance of the air between the capacitor plates. On a cold dry day, the resistance of the air gap is $\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{13}}\mathbf{\Omega }$; on a humid day, the resistance is $\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{\Omega }$. How long will it take the capacitor voltage to dissipate to half its original value on each day?

Show that when Euler’s method is used to approximate the solution of the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{5}\mathbf{y}$ y(0) = 1  , at x = 1, then the approximation with step size h is ${\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{5}\mathbf{)}}^{\frac{\mathbf{1}}{\mathbf{h}}}$.

Show that when Euler’s method is used to approximate the solution of the initial value problem ${\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{y}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{3}$,at x = 2, then the approximation with step size h is $\mathbf{3}\mathbf{(}\mathbf{1}\mathbf{-}\frac{\mathbf{h}}{\mathbf{2}}{\mathbf{)}}^{\frac{\mathbf{2}}{\mathbf{h}}}$ .

Show that when the trapezoid scheme given in formula (8) is used to approximate the solution $\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{x}}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{y}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$ , at x = 1, then we get ${\mathbf{y}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{=}\left(\frac{\mathbf{1}\mathbf{+}\frac{\mathbf{h}}{\mathbf{2}}}{\mathbf{1}\mathbf{-}\frac{\mathbf{h}}{\mathbf{2}}}\right){\mathbf{y}}_{\mathbf{n}}$,n = 0, 1, 2, . . . , which leads to the approximation ${\left(\frac{1+\frac{h}{2}}{1-\frac{h}{2}}\right)}^{\frac{\mathbf{1}}{\mathbf{h}}}$for the constant e. Compute this approximation for h = 1, ${\mathbf{10}}^{\mathbf{-}\mathbf{1}}\mathbf{,}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{,}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{,}\mathbf{and}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}$and compare your results with those in Tables 3.4 and 3.5.

Show that when the improved Euler’s method is used to approximate the solution of the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{4}\mathbf{y}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}$, at $\mathbf{x}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}$ , then the approximation with step size h is $\frac{\mathbf{1}}{\mathbf{3}}{\left(1+4h+8h^2\right)}^{\frac{\mathbf{1}}{\mathbf{2}\mathbf{h}}}$ .

Use the improved Euler’s method subroutine with step size h = 0.1 to approximate the solution to the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{x}\mathbf{=}{\mathbf{y}}^{\mathbf{2}}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}$ , at the points x = 1.1, 1.2, 1.3, 1.4, and 1.5. (Thus, input N = 5.) Compare these approximations with those obtained using Euler’s method (see Exercises 1.4, Problem 5, page 28).

Use the improved Euler’s method subroutine with step size h = 0.2 to approximate the solution to the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{(}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{y}\mathbf{)}\mathbf{,}\mathbf{y}\mathbf{(}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{1}$  at the points x = 1.2, 1.4, 1.6, and 1.8. (Thus, input N = 4.) Compare these approximations with those obtained using Euler’s method (see Exercises 1.4, Problem 6, page 28).

Use the improved Euler’s method subroutine with step size h = 0.2 to approximate the solution to  at the points x = 0, 0.2, 0.4, …., 2.0. Use your answers to make a rough sketch of the solution on [0, 2].

Use the improved Euler’s method with tolerance to approximate the solution to  $\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{t}\mathbf{ }\mathbf{sin}\mathbf{\left(}\mathbf{tx}\mathbf{\right)}\mathbf{,}\mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$, at t = 1. For a tolerance of $\mathbf{\epsilon }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{01}$ , use a stopping procedure based on the absolute error.

Use the improved Euler’s method with tolerance to approximate the solution to $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{sin} \mathbf{y}\mathbf{,} \mathbf{y} \mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$, at  $\mathbf{x}\mathbf{=}\mathbf{\pi }$. For a tolerance of  $\mathbf{\epsilon }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{01}$, use a stopping procedure based on the absolute error.

Use the improved Euler’s method with tolerance to approximate the solution to  $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{y}\mathbf{+}{\mathbf{y}}^{\mathbf{3}}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$, at x = 1. For a tolerance of $\mathbf{\epsilon }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{003}$ , use a stopping procedure based on the absolute error.

In Example 1, page 126, the improved Euler’s method approximation to \({\bf{e}}\) with step size \({\bf{h}}\) was shown to be \({\bf{\;}}{\left( {{\bf{1 + h + }}\frac{{{{\bf{h}}^{\bf{2}}}}}{{\bf{2}}}} \right)^{\frac{{\bf{1}}}{{\bf{h}}}}}\).First prove that the error \({\bf{e - \;}}{\left( {{\bf{1 + h + }}\frac{{{{\bf{h}}^{\bf{2}}}}}{{\bf{2}}}} \right)^{\frac{{\bf{1}}}{{\bf{h}}}}}\)approaches zero as \({\bf{h}} \to 0\). Then use L’Hopital’s rule to show the \(\mathop {{\bf{lim}}}\limits_{{\bf{h}} \to 0} \frac{{{\bf{error}}}}{{{{\bf{h}}^{\bf{2}}}}}{\bf{ = }}\frac{{\bf{e}}}{6} \approx 0.45305\).Compare this constant with the entries in the last column of Table 3.5.

By experimenting with the improved Euler’s method subroutine, find the maximum value over the interval [0,2] of the solution to the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{sin}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{)}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}$ Where does this maximum value occur? Give answers to two decimal places.

The solution to the initial value problem $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{+}\mathbf{2}{\mathbf{)}}^{\mathbf{2}}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}$crosses the x-axis at a point in the interval [0,14] .By experimenting with the improved Euler’s method subroutine, determine this point to two decimal places.

The solution to the initial value problem $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\frac{\mathbf{y}}{\mathbf{x}}\mathbf{=}{\mathbf{x}}^{\mathbf{3}}{\mathbf{y}}^{\mathbf{2}}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{3}$ has a vertical asymptote (“blows up”) at some point in the interval [1,2] By experimenting with the improved Euler’s method subroutine, determine this point to two decimal places.

Use Euler’s method (4) with h = 0.1 to approximate the solution to the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\mathbf{20}\mathbf{y}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$, on the interval $\mathbf{0}\mathbf{⩽}\mathbf{x}\mathbf{⩽}\mathbf{1}$  (that is, at x = 0, 0.1, . . . , 1.0).Compare your answers with the actual solution y = e^-20x. What went wrong? Next, try the step size h = 0.025 and also h = 0.2. What conclusions can you draw concerning the choice of step size?

Building Temperature. In Section 3.3 we modeled the temperature inside a building by the initial value problem (13)\(\frac{{{\bf{dT}}}}{{{\bf{dt}}}}{\bf{ = K}}\,\,\left[ {{\bf{M}}\,{\bf{(t) - T}}\,{\bf{(t)}}} \right]{\bf{ + H}}\,{\bf{(t) + U}}\,{\bf{(t),}}\,\,{\bf{T}}\,{\bf{(}}{{\bf{t}}_{\bf{o}}}{\bf{) = }}{{\bf{T}}_{\bf{o}}}\) , where M is the temperature outside the building, T is the temperature inside the building, H is the additional heating rate, U is the furnace heating or air conditioner cooling rate, K is a positive constant, and \({{\bf{T}}_{\bf{o}}}\) is the initial temperature at time \({{\bf{t}}_{\bf{o}}}\) . In a typical model, \({{\bf{t}}_{\bf{o}}}{\bf{ = 0}}\) (midnight),\({{\bf{T}}_{\bf{o}}}{\bf{ = 6}}{{\bf{5}}^{\bf{o}}}\), \({\bf{H}}\left( {\bf{t}} \right){\bf{ = 0}}{\bf{.1}}\), \({\bf{U(t) = 1}}{\bf{.5}}\left[ {{\bf{70 - T(t)}}} \right]\) and \({\bf{M(t) = 75 - 20cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}\) . The constant K is usually between\(\frac{{\bf{1}}}{{\bf{4}}}{\bf{and}}\frac{{\bf{1}}}{{\bf{2}}}\), depending on such things as insulation. To study the effect of insulating this building, consider the typical building described above and use the improved Euler’s method subroutine with\({\bf{h = }}\frac{{\bf{2}}}{{\bf{3}}}\) to approximate the solution to (13) on the interval \(0 \le {\bf{t}} \le 24\) (1 day) for \({\bf{k = 0}}{\bf{.2,}}\,{\bf{0}}{\bf{.4}}\), and 0.6.

Falling Body. In Example 1 of Section 3.4, page 110, we modeled the velocity of a falling body by the initial value problem \({\bf{m}}\frac{{{\bf{dv}}}}{{{\bf{dt}}}}{\bf{ = mg - bv,v(0) = }}{{\bf{v}}_{\bf{o}}}{\bf{ = 0}}\)under the assumption that the force due to air resistance is –bv. However, in certain cases the force due to air resistance behaves more like\({\bf{ - b}}{{\bf{v}}^{\bf{r}}}\), where \({\bf{(r > 1)}}\) is some constant. This leads to the model \({\bf{m}}\frac{{{\bf{dv}}}}{{{\bf{dt}}}}{\bf{ = mg - b}}{{\bf{v}}^{\bf{r}}}{\bf{,v(0) = }}{{\bf{v}}_{\bf{o}}}\) (14).To study the effect of changing the parameter r in (14), take \({\bf{m =  1,}}\,\,{\bf{g =  9}}{\bf{.81,}}\,\,{\bf{b =  2}}\) and \({{\bf{v}}_{\bf{o}}}{\bf{ = 0}}\).Then use the improved Euler’s method subroutine with \({\bf{h = 0}}{\bf{.2}}\) to approximate the solution to (14) on the interval \(0 \le {\bf{t}} \le 5\) for \({\bf{r = 1}}{\bf{.0,}}\,\,{\bf{1}}{\bf{.5}}\) and 2.0. What is the relationship between these solutions and the constant solution\({\bf{v(t) = }}{\left( {\frac{{{\bf{9}}{\bf{.81}}}}{{\bf{2}}}} \right)^{\frac{{\bf{1}}}{{\bf{r}}}}}\)?

Determine the recursive formulas for the Taylor method of order 2 for the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{cos}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{)}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{\pi }$.

Determine the recursive formulas for the Taylor method of order 2 for the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{xy}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}$.

Determine the recursive formulas for the Taylor method of order 4 for the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$.

Determine the recursive formulas for the Taylor method of order 4 for the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{y}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$ .

Use the Taylor methods of orders 2 and 4 with h = 0.25 to approximate the solution to the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{-}\mathbf{y}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$, at x = 1. Compare these approximations to the actual solution $\mathbf{y}\mathbf{=}\mathbf{x}\mathbf{+}{\mathbf{e}}^{\mathbf{-}\mathbf{x}}$ evaluated at x = 1.

Use the Taylor methods of orders 2 and 4 with h = 0.25 to approximate the solution to the initial value problem  $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{y}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$, at x = 1. Compare these approximations to the actual solution $\mathbf{y}\mathbf{=}\mathbf{1}\mathbf{+}{\mathbf{e}}^{\mathbf{-}\mathbf{x}}$ evaluated at x = 1.

Use the fourth-order Runge–Kutta subroutine with h = 0.25 to approximate the solution to the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{y}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$ , at x = 1. Compare this approximation with the one obtained in Problem 6 using the Taylor method of order 4.

Use the fourth-order Runge–Kutta subroutine with h = 0.25 to approximate the solution to the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{-}\mathbf{y}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$, at x = 1. Compare this approximation with the one obtained in Problem 5 using the Taylor method of order 4.

Use the fourth-order Runge–Kutta algorithm to approximate the solution to the initial value problem  $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{xy}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{1}$at x = 2. For a tolerance of , use a stopping procedure based on the absolute error.

The solution to the initial value problem\({\bf{y' = }}\frac{{\bf{2}}}{{{{\bf{x}}^{\bf{4}}}}}{\bf{ - }}{{\bf{y}}^{\bf{2}}}{\bf{,y(1) =  - 0}}{\bf{.414}}\), crosses the x-axis at a point in the interval \(\left[ {{\bf{1,2}}} \right]\).By experimenting with the fourth-order Runge–Kutta subroutine, determine this point to two decimal places

By experimenting with the fourth-order Runge-Kutta subroutine, find the maximum value over the interval \(\left[ {{\bf{1,2}}} \right]\) of the solution to the initial value problem\({\bf{y' = }}\frac{{{\bf{1}}{\bf{.8}}}}{{{{\bf{x}}^{\bf{4}}}}}{\bf{ - }}{{\bf{y}}^{\bf{2}}}{\bf{,y(1) =  - 1}}\) . Where does this maximum occur? Give your answers to two decimal places.

The solution to the initial value problem \(\frac{{{\bf{dy}}}}{{{\bf{dx}}}}{\bf{ = }}{{\bf{y}}^{\bf{2}}}{\bf{ - 2}}{{\bf{e}}^{\bf{x}}}{\bf{y + }}{{\bf{e}}^{{\bf{2x}}}}{\bf{ + }}{{\bf{e}}^{\bf{x}}}{\bf{,y(0) = 3}}\)has a vertical asymptote (“blows up”) at some point in the interval\(\left[ {{\bf{0,2}}} \right]\). By experimenting with the fourth-order Runge–Kutta subroutine, determine this point to two decimal places.

Use the fourth-order Runge–Kutta algorithm to approximate the solution to the initial value problem\({\bf{y' = ycosx,y(0) = 1}}\) , at \({\bf{x = \pi }}\). For a tolerance of \({\bf{\varepsilon  = 0}}{\bf{.01}}\) use a stopping procedure based on the absolute error.

Use the fourth-order Runge–Kutta subroutine with h = 0.1 to approximate the solution to\({\bf{y' = cos}}\;{\bf{5y - x,y(0) = 0}}\),at the points x = 0, 0.1, 0.2, . . ., 3.0. Use your answers to make a rough sketch of the solution on\(\left[ {{\bf{0,3}}} \right]\).

Use the fourth-order Runge–Kutta subroutine with h = 0.1 to approximate the solution to\({\bf{y' = 3cos(y - 5x),y(0) = 0}}\) , at the points x = 0, 0.1, 0.2, . . ., 4.0. Use your answers to make a rough sketch of the solution on [0, 4].

The Taylor method of order 2 can be used to approximate the solution to the initial value problem\({\bf{y' = y,y(0) = 1}}\) , at x = 1. Show that the approximation \({{\bf{y}}_{\bf{n}}}\) obtained by using the Taylor method of order 2 with the step size \(\frac{{\bf{1}}}{{\bf{n}}}\) is given by the formula\({{\bf{y}}_{\bf{n}}}{\bf{ = }}{\left( {{\bf{1 + }}\frac{{\bf{1}}}{{\bf{n}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{2}}{{\bf{n}}^{\bf{2}}}}}} \right)^{\bf{n}}}\). The solution to the initial value problem is\({\bf{y = }}{{\bf{e}}^{\bf{x}}}\), so \({{\bf{y}}_{\bf{n}}}\)is an approximation to the constant e.

If the Taylor method of order p is used in Problem 17, show that \({{\bf{y}}_{\bf{n}}}{\bf{ = }}{\left( {{\bf{1 + }}\frac{{\bf{1}}}{{\bf{n}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{2}}{{\bf{n}}^{\bf{2}}}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{6}}{{\bf{n}}^{\bf{3}}}}}{\bf{ + }}.....{\bf{ + }}\frac{{\bf{1}}}{{{\bf{p!}}{{\bf{n}}^{\bf{p}}}}}} \right)^{\bf{n}}}\), n = 1, 2, …