Chapter 2.6, Problem 1P

Program Plan Intro

Program Description: Purpose of the problem is to construct a table for the approximation solution and the actual solution of $y^{\prime }=-y$ up to five decimal places by the use of Runge-Kutta’s method.

Summary Introduction:

Purpose will use Runge-Kutta’s method to construct the table of the approximation solution and the actual solution $y=y\left(x\right)$ is the solution of a differential equation $\frac{dy}{dx}=f\left(x,y\right)$ with $y\left(x_0\right)=y_0$ where approximate value can be calculated with the formula $y_{i+1}=y_i+k\cdot h$ and the value of $k$ can be calculated as $k=\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)$ where the value of $k_1=f\left(x_i,y_i\right)$ , $k_2=f\left(x_i+\frac{1}{2}h,y_i+\frac{1}{2}h{k}_1\right)$ , $k_3=f\left(x_i+\frac{1}{2}h,y_i+\frac{1}{2}h{k}_2\right)$ and $k_4=f\left(x_{i+1},y_i+h{k}_3\right)$ .

Explanation of Solution

Given information:

In the interval of $\left[0,0.5\right]$ the function have 2 subintervals as $h=0.25$ .

The differential equation is $\frac{dy}{dx}=-y$ and $y\left(x\right)=2e^{-x}$ with $y(0)=2$ .

Calculation:

The initial value problem $\frac{dy}{dx}=f\left(x,y\right)$ , $y\left(x_0\right)=y_0$ can be calculated with RungeKutta’s method with the size of $h$ .

The differential equation is $\frac{dy}{dx}=-y$ and $y\left(x\right)=2e^{-x}$ with $y(0)=2$ .

The value of $k_1$ is $f\left(x_i,y_i\right)$ .

Therefore, the value of $k_1$ can be expressed as,

  $k_1=-y_i$

Substitute 0 for $i$ in the predicted formula to obtain the formula of $u_1$ as,

  $k_1=-y_0$ ........ (1)

Substitute $2$ for $y_0$ in equation (1) to obtain the value of $k_1$ as,

  $k_1=-2$

The value of $k_2$ is $f\left(x_i+\frac{1}{2}h,y_i+\frac{1}{2}h{k}_1\right)$ .

Therefore, the value of $k_2$ can be expressed as,

  $k_2=-\left(y_i+\frac{1}{2}h{k}_1\right)$

Substitute 0 for $i$ in the predicted formula to obtain the formula of $k_2$ as,

  $k_2=-\left(y_0+\frac{1}{2}h{k}_1\right)$ ........ (2)

Substitute $2$ for $y_0$ , $0.25$ for $h$ and $-2$ for $k_1$ in equation (2) to obtain the value of $k_2$ as,

  $\begin{array}{l}{k}_2=-\left(2+\frac{1}{2}\left(0.25\right)\left(-2\right)\right)\\ k_2=-1.75\end{array}$

The value of $k_3$ is $f\left(x_i+\frac{1}{2}h,y_i+\frac{1}{2}h{k}_2\right)$ .

Therefore, the value of $k_3$ can be expressed as,

  $k_3=-\left(y_i+\frac{1}{2}h{k}_2\right)$

Substitute 0 for $i$ in the predicted formula to obtain the formula of $k_3$ as,

  $k_3=-\left(y_0+\frac{1}{2}h{k}_2\right)$ ........ (3)

Substitute $2$ for $y_0$ , $0.25$ for $h$ and $-1.75$ for $k_2$ in equation (3) to obtain the value of $k_3$ as,

  $\begin{array}{l}{k}_3=-\left(2+\frac{1}{2}\left(0.25\right)\left(-1.75\right)\right)\\ k_3=-1.78125\end{array}$

The value of $k_4$ is $f\left(x_{i+1},y_i+h{k}_3\right)$ .

Therefore, the value of $k_4$ can be expressed as,

  $k_4=-\left(y_i+h{k}_3\right)$

Substitute 0 for $i$ in the predicted formula to obtain the formula of $k_4$ as,

  $k_4=-\left(y_0+h{k}_3\right)$ ........ (4)

Substitute $2$ for $y_0$ , $0.25$ for $h$ and $-1.78125$ for $k_3$ in equation (4) to obtain the value of $k_4$ as,

  $\begin{array}{l}{k}_4=-\left(2+\left(0.25\right)\left(-1.78125\right)\right)\\ k_4=-1.55469\end{array}$

The value of $k$ can be calculated as,

  $\begin{array}{l}k=\frac{1}{6}\left(-2+2\left(-1.75\right)+2\left(-1.78125\right)+\left(-1.55469\right)\right)\\ k=-1.76953\end{array}$

The approximate value can be calculated as,

  $\begin{array}{l}{y}_{i+1}=y_i+k\cdot h\\ y_1=2+\left(-1.76953\right)\cdot \left(0.25\right)\\ y_1=1.55762\end{array}$

Now actual value can be calculated by substituting the given values of $x_i$ in the $y\left(x\right)=2e^{-x}$ .

Substitute $0.25$ for $x$ in the equation $y\left(x\right)=2e^{-x}$ to obtain the actual solution of $y$ for $x=0.25$ .

  $\begin{array}{l}y\left(0.25\right)=2e^{-\left(0.25\right)}\\ y\left(0.25\right)=1.55760\end{array}$

Similarly, further values can also be calculated using the above steps.

$k_1$	$k_2$	$k_3$	$k_4$	$x_i$	$\text{approx value }{y}_i$	$\begin{array}{l}\text{Actual value}\\ y\left(x_i\right)\end{array}$
$-2$	$-1.75$	$-1.78125$	$-1.55469$	$0.25$	$1.55762$	$1.55760$
$-1.55762$	$-1.36292$	$-1.38725$	$1.2108$	$0.5$	$1.21309$	$1.21306$

Therefore, the above table shows all the values of approximation value $y_i$ and actual value $y_i$ .