A calculator to solve differential equations of the second order using a numerical method based on the Fourth Order Runge-Kutta Method [1] and a calculator-based analytical method is presented for comparison and educational purposes.

The fourth-order Runge-Kutta method applied to a first-order differential equation is presented and then the same method is applied to a second order linear differential equation. Finally, an interactive tutorial to study the effects of the step size \( h \) used in the Runga-Kutta method is included.

Step 1: \[ k_1 = h \cdot f(y_n, t_n) \] Here, \( y_n \) and \( t_n \) are the current values of \( y \) and \( t \).

Step 2: \[ k_2 = h \cdot f\left(y_n + \dfrac{k_1}{2}, t_n + \dfrac{h}{2}\right) \] Step 3: \[ k_3 = h \cdot f\left(y_n + \dfrac{k_2}{2}, t_n + \dfrac{h}{2}\right) \] Step 4: \[ k_4 = h \cdot f(y_n + k_3, t_n + h) \] Step 5:

Update \( y \) and \( t \) The fourth-order Runge-Kutta method combines the above quantities (slopes) to approximate the value of \( y \) at the end of the interval \([t, t+h]\). The weighted average of these slopes is calculated using the formula: \[ \dfrac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4) \] This average slope is then multiplied by \( h \) and added to \( y_n \) to update the value of \( y \) at \( t_{n+1} \). \[ y_{n+1} = y_n + h (\dfrac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4)) \] \[ t_{n+1} = t_n + h \]

Let \[ \dfrac{dy}{dt} = v \] Thus, the second order differential equation may be written as system as follows: \[ \dfrac{dy}{dt} = v \] \[ \dfrac{dv}{dt} = \dfrac{d^2y}{dt^2} = f(y, v, t) \] We can then use the 4th Order Runge-Kutta method to solve this system of first-order differential equations as follows:

Step 1:

Compute \( k_{1y} \) and \( k_{1v} \) \[ k_{1y} = h \cdot v \] \[ k_{1v} = h \cdot f(y_n, v_n, t_n) \]

Step 2:

Compute \( k_{2y} \) and \( k_{2v} \) \[ k_{2y} = h \cdot (v_n + \dfrac{k_{1v}}{2}) \] \[ k_{2v} = h \cdot f\left(y_n + \dfrac{k_{1y}}{2}, v_n + \dfrac{k_{1v}}{2}, t_n + \dfrac{h}{2}\right) \]

Step 3:

Compute \( k_{3y} \) and \( k_{3v} \) \[ k_{3y} = h \cdot (v_n + \dfrac{k_{2v}}{2}) \] \[ k_{3v} = h \cdot f\left(y_n + \dfrac{k_{2y}}{2}, v_n + \dfrac{k_{2v}}{2}, t_n + \dfrac{h}{2}\right) \]

Step 4:

Compute \( k_{4y} \) and \( k_{4v} \) \[ k_{4y} = h \cdot (v_n + k_{3v}) \] \[ k_{4v} = h \cdot f(y_n + k_{3y}, v_n + k_{3v}, t_n + h) \]

Step 5:

Update \( y \) and \( v \) \[ y_{n+1} = y_n + \dfrac{1}{6}(k_{1y} + 2k_{2y} + 2k_{3y} + k_{4y}) \] \[ v_{n+1} = v_n + \dfrac{1}{6}(k_{1v} + 2k_{2v} + 2k_{3v} + k_{4v}) \] \[ t_{n+1} = t_n + h \]

We now present a calculator that approximates the solution to the second-order differential equations of the form
\[
a \dfrac{d^2y}{dt^2} + b \dfrac{dy}{dt} + c y = 0 \]
using the fourth-order Runge-Kutta method described above and the solution is also given analytically for comparison.

The above differential equation is also solved analytically to make comparisons.

Let \[ \dfrac{dy}{dt} = v \]
We therefore need to apply the 4th Order Runge-Kutta method to the following system.
\[ \dfrac{dy}{dt} = v \]
\[ \dfrac{dv}{dt} = \dfrac{d^2y}{dt^2} = \dfrac{1}{a} (- b \; v - c \; y ) \]
The calculator uses the fourth order Runge-Kutta method and also solves the same differential equation analytically and compare the results graphically.

Enter the coefficients \(a, b \) and \( c \) as well as the initial conditions \( y(0) \) and \( dy/dt(0) \) and the step size \( h \), then solve and compare the graphs of the numerical and analytical solutions obtained.

Note

1) You may use the mousse and hover above the graphs to obtain a reading of points on the graphs and therefore compare them.

2) Use the mousse to display one graph at a time or both by clicking on the graphs in the legend shown on the right side.

3) The errors in using the Runge-Kutta method depend on the step size \( h \). This calculator allows to change \( h \) and therefore compares the numerical and the analytical methods and studies the effect of \( h \) on the errors due to the approximation used in the 4th Order Runge-Kutta method.

4) You may also change the number \( n \) of steps used in the calculations. The graph extends from \( x = 0 \) to \( x = n \times h \).

Enter the following values for the constants \( b , c \) and \( d \) as follows:

\( a = 1 \) , \( b = 0.5 \) and \( c = 5 \) which should give an oscilating function with decreasing amplitude.

Now enter values of \( h \) starting from \( h = 0.01 \) and then increase \( h \) up to 1. Which values of \( h \) give a better approximation to the solution when compared to the analytical method?

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