Quiz Chapter 08.04: Runge-Kutta 4th Order Method

 MULTIPLE CHOICE TEST RUNGE-KUTTA 4th ORDER METHOD ORDINARY DIFFERENTIAL EQUATIONS

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

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

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

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

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

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