Quiz Chapter 08.04: Runge-Kutta 4th Order Method

MULTIPLE CHOICE TEST

RUNGE-KUTTA 4th ORDER METHOD

ORDINARY DIFFERENTIAL EQUATIONS

(More on Ordinary Differential Equations)

Pick the most appropriate answer

1. To solve the ordinary differential equation $3 \dfrac{dy}{dx} + xy^{2} = \sin x, \, y(0) = 5$ , by Runge-Kutta $4$ ^th order method, you need to rewrite the equation as

\dfrac{dy}{dx} = \sin x - xy^{2}, \, y(0) = 5

\dfrac{dy}{dx} = \dfrac{1}{3} \left( \sin x - xy^{2} \right), \, y(0) = 5

\dfrac{dy}{dx} = \dfrac{1}{3} \left( - \cos x - \dfrac{xy^{3}}{3} \right), \, y(0) = 5

\dfrac{dy}{dx} = \dfrac{1}{3} \sin x, \, y(0) = 5

2. Given

- - $3 \dfrac{dy}{dx} + 5y^{2} = \sin x, \, y(0.3) = 5$

and using a step size of $h=0.3$ , the value of $y(0.9)$ using Runge-Kutta $4$ ^th order method is most nearly

-0.25011 \times 10^{40}

-4297.4

-1261.5

0.88498

3. Given

- - $3 \dfrac{dy}{dx} + y^{2} = e^{x}, \, y(0.3) = 5$

and using a step size of $h=0.3$ , the best estimate of $\dfrac{dy}{dx} (0.9)$ using Runge-Kutta $4$ ^th order method is most nearly

-1.6604

-1.1785

-0.45831

2.7270

4. The velocity (m/s) of a parachutist is given as a function of time (seconds) by

- - $v(t) = 55.8 \tanh (0.17t), \, t \geq 0$

Using Runge-Kutta $4$ ^th order method with a step size of $5$ seconds, the distance traveled by the body from $t=2$ to $t=12$ seconds is estimated most nearly as

341.43

428.97

429.05

703.50

5. Runge-Kutta method can be derived from using the first three terms of Taylor series of writing the value of $y_{i+1}$ , that is the value of $y$ at $x_{i+1}$ , in terms of $y_{i}$ and all the derivatives of $y$ at $x_{i}$ . If $h=x_{i+1} - x_{i}$ , the explicit expression for $y_{i+1}$ if the first five terms of the Taylor series are chosen for the ordinary differential equation

- - $\dfrac{dy}{dx} + 5y = 3e^{-2x}, \, y(0) = 7$

would be

y_{i+1} = y_{i} + \left( 3e^{-2x_{i}} - 5y_{i} \right) h + \dfrac{5h^{2}}{2}

y_{i+1} = y_{i} + \left( 3e^{-2x_{i}} - 5y_{i} \right) h + \left( -21e^{-2x_{i}} + 25y_{i} \right) \dfrac{h^{2}}{2} + \left( -483e^{-2x_{i}} + 625y_{i} \right) \dfrac{h^{3}}{6} + \left(-300909e^{-2x_{i}} + 390625 y_{i} \right) \dfrac{h^{4}}{24}

y_{i+1} = y_{i} + \left( 3e^{-2x_{i}} - 5y_{i} \right) h + \left( -6e^{-2x_{i}} \right) \dfrac{h^{2}}{2} + \left( 12e^{-2x_{i}} \right) \left( 12e^{-2x_{i}} \right) \dfrac{h^{3}}{6} + \left( -24e^{-2x_{i}} \right) \dfrac{h^{4}}{24}

y_{i+1} = y_{i} + \left( 3e^{-2x_{i}} - 5y_{i} \right) h + \left( -6e^{-2x_{i}} + 5 \right) \dfrac{h^{2}}{2} + \left( 12e^{-2x_{i}} \right) \dfrac{h^{3}}{6} + \left( -24e^{-2x_{i}} \right) \dfrac{h^{4}}{24}

6. A hot, solid cylinder is immersed in a cool oil bath as part of a quenching process. This process makes the temperature of the cylinder, $\theta_{c}$ , and the bath, $\theta_{b}$ , change with time.

If the initial temperature of the bar and the oil bath is given as $600^{\circ}$ C and $27^{\circ}$ C, respectively, and:

- - radius of cylinder $=3$ cm
  - density of cylinder $=2700$ kg/m³
  - specific heat of cylinder $=895$ J/(kg-K)
  - convection heat transfer coefficient $=100$ W/(m²-K)
  - specific heat of oil $=1910$ J/(kg-K)
  - mass of oil $=2$ kg

The coupled ordinary differential equations governing the heat transfer are given by

\begin{matrix}362.4 \dfrac{d \theta_{c}}{dt} + \theta_{c} = \theta{b} \\ \\ 675.5 \dfrac{d \theta_{b}}{dt} + \theta_{b} = \theta_{c}\\ \end{matrix}

\begin{matrix}362.4 \dfrac{d \theta_{c}}{dt} - \theta_{c} = \theta{b} \\ \\ 675.5 \dfrac{d \theta_{b}}{dt} - \theta_{b} = \theta_{c} \\ \end{matrix}

\begin{matrix} 675.5 \dfrac{d \theta_{c}}{dt} + \theta_{c} = \theta{b} \\ \\ 362.4 \dfrac{d \theta_{b}}{dt} + \theta_{b} = \theta_{c} \\ \end{matrix}

\begin{matrix} 675.5 \dfrac{d \theta_{c}}{dt} - \theta_{c} = \theta{b} \\ \\ 362.4 \dfrac{d \theta_{b}}{dt} - \theta_{b} = \theta_{c} \\ \end{matrix}