Question

The following set of differential equations can be used to represent that behaviour of a simple spring-mass system, with $x_1\left(t\right)$the mass’s position and$x_2\left(t\right)$ its velocity:

role="math" localid="1656152792032" $$\begin{gathered} \frac{d{x}_1}{dt}=x_2 \\ \frac{d{x}_2}{dt}=-x_1 \end{gathered}$$

For the initial condition of$x_1\left(0\right)=0,x_2\left(0\right)=0$ and a step size$0.1$ seconds, determine the values $x_1\left(0.3\right),$and $x_2\left(0.3\right),$using (a) Euler’s method, (b) the modified Euler’s method.

Short answer

(a)The value of $x_1\left(0.3\right)$is $0.97$ and value of$x_2\left(0.3\right)$ is $-0.299$by modified Euler’s method.

(b) The value of$x_1\left(0.3\right)$ is$0.9552$ and$x_2\left(0.3\right)$ is$-0.296$ by using modified Euler’s method.

Step-by-step solution

Write the given data from the question.

Write the differential equation,

$$\begin{gathered} \frac{d{x}_1}{dt}=-x_2 \\ \frac{d{x}_2}{dt}=-x_1 \end{gathered}$$

Initial conditions,

$$\begin{gathered} x_1\left(0\right)=1.0 \\ {x}_2\left(0\right)=0 \end{gathered}$$

Integration step size,$\Delta t=0.1\text{sec}$

Determine the equation to calculate the value of x10.3, and x20.3,by using Euler’s method and modified Euler’s method. 

The equation to calculate the value of by Euler’s method is given as follows.

$x\left(t+\Delta t\right)=x\left(t\right)+\left(\frac{dx\left(t\right)}{dt}\right)\Delta t$ ....(1)

The equation to calculate the value of$x\left(0.5\right)$by modified Euler’s method is given as follows

$x\left(t+\Delta t\right)=x\left(t\right)+\left(\frac{\frac{dx\left(t\right)}{dt}+\frac{d\tilde{x}\left(t\right)}{dt}}{2}\right)\Delta t$.....(2)

Here$d\tilde{x}\left(t\right)$is the variable.

Calculate the value of  x10.3,andx20.3, by using Euler’s method

For$x_1$

Solve the differential equation.

$x_1\left(t+\Delta t\right)=x_1\left(t\right)+\left(\frac{d{x}_1\left(t\right)}{dt}\right)\Delta t$

Substitute $0.1\text{sec}$for $\Delta t$and $x_2\left(t\right)$for $\frac{d{x}_1}{dt}$into above equation.

$x_1\left(t+0.1\right)=x_1\left(t\right)+0.1x_2\left(t\right)$ …… (3)

Calculate the value of $x_1\left(0.1\right)$.

Substitute 0 for t into equation (3).

$\begin{array}{rcl}{x}_1\left(0+0.1\right)& =& x_1\left(0\right)+0.1x_2\left(0\right)\\ x_1\left(0.1\right)& =& x_1\left(0\right)+0.1x_2\left(0\right)\\ & & \end{array}$

Substitute 1 for $x_1\left(0\right)$and 0 for $x_2\left(0\right)$into above equation

$$\begin{gathered} x\left(0.1\right)=1-0.1\times 0 \\ x\left(0.1\right)=1-0 \\ x\left(0.1\right)=1 \end{gathered}$$

For $x_2$

Solve the differential equation.

$x_2\left(t\right)\left(t+∆t\right)=x_2\left(t\right)+\left(\frac{d{x}_2\left(t\right)}{dt}\right)∆t$

Substitute 0.1 sec for $\Delta t$and $-x_1\left(t\right)$for $\frac{d{x}_2}{dt}$into above equation.

$x_2\left(t+0.1\right)=x_2\left(t\right)-0.1x_1\left(t\right)$ …… (4)

Calculate the value of $x_2\left(0.1\right)$.

Substitute 0 for t into equation (4).

$\begin{array}{rcl}{x}_2\left(0+0.1\right)& =& x_2\left(0\right)-0.1x_1\left(0\right)\\ x_2\left(0.1\right)& =& x_2\left(0\right)-0.1x_1\left(0\right)\\ & & \end{array}$

Substitute 1 for $x_1\left(0\right)$and 0 for $x_2\left(0\right)$into above equation

$$\begin{gathered} x\left(0.1\right)=0-0.1\times 1 \\ x\left(0.1\right)=0-0.1 \\ x\left(0.1\right)=-0.1 \end{gathered}$$

Repeat the above process and obtain the value of $x_1\left(0.3\right)$and$x_2\left(0.3\right)$

	0	0.1	0.2	0.3
x1	1	1	0.99	0.97
x2	0	-0.1	-0.2	-0.299

Hence the value of $x_1\left(0.3\right)$is $0.97$and value of $x_2\left(0.3\right)$is -0.299.

Calculate the value of  x10.3,and by using modified Euler’s method.

(b)

For$x_1$

The variable $\widetilde{x_2}\left(t\right)$is defined as,

$\widetilde{x_2}\left(t\right)=x_2\left(t\right)+\frac{d{x}_2\left(t\right)}{dt}\Delta t$

Substitute $0.1$for $\Delta t$and $-x_1\left(t\right)$for $\frac{d{x}_2\left(t\right)}{dt}$ into above equation.

$\widetilde{x_2}\left(t\right)=x_2\left(t\right)-x_1\left(t\right)\times 0.1$...(5)

$\frac{d\widetilde{x_2}\left(t\right)}{dt}=\widetilde{x_2}\left(t\right)$

Solve the modified form of Euler’s method.

Substitute,$\widetilde{x_2}\left(t\right)$for $\frac{d\widetilde{x_2}\left(t\right)}{dt}$and $x_2\left(t\right)$for$\frac{d{x}_1\left(t\right)}{dt}$into above equation.

$x_1\left(t+\Delta t\right)=x_1\left(t\right)+\left(\frac{x_2\left(t\right)+\tilde{x}\left(t\right)}{2}\right)\Delta t$

Substitute 0.1 for $∆t$into above equation.

$x_1\left(t+0.1\right)=x_1\left(t\right)+\left(\frac{x_2\left(t\right)+\widetilde{x_2}\left(t\right)}{2}\right)\times 0.1$ …… (6)

For $x_2\left(t\right)$,

The variable $\widetilde{x_1}\left(t\right)$is defined as,

$\widetilde{x_1}\left(t\right)=x_2\left(t\right)+\frac{d{x}_1\left(t\right)}{dt}\Delta t$

Substitute $0.1$for $\Delta t$and $-x_1\left(t\right)$for $\frac{d{x}_2\left(t\right)}{dt}$into above equation.

$\widetilde{x_1}\left(t\right)=x_1\left(t\right)+x_2\left(t\right)\times 0.1$ …… (7)

$\frac{d\widetilde{x_1}\left(t\right)}{dt}=-\widetilde{x_1}\left(t\right)$

Solve the modified form of Euler’s method.

$x_2\left(t+\Delta t\right)=x_2\left(t\right)+\left(\frac{\frac{d{x}_2\left(t\right)}{dt}+\frac{d\widetilde{x_1}\left(t\right)}{dt}}{2}\right)\Delta t$

Substitute$-\widetilde{x_1}\left(t\right)$, for$\frac{d\widetilde{x_1}\left(t\right)}{dt}$and$-x_1\left(t\right)$for $\frac{d{x}_2\left(t\right)}{dt}$into above equation.

$x_2\left(t+\Delta t\right)=x_2\left(t\right)+\left(\frac{-x_1\left(t\right)-\widetilde{x_1}\left(t\right)}{2}\right)\Delta t$

Substitute for into above equation.

$x_2\left(t+0.1\right)=x_2\left(t\right)+\left(\frac{-x_1\left(t\right)-\widetilde{x_1}\left(t\right)}{2}\right)\times 0.1$ ....(8)

Calculate the value of variable $\widetilde{x_1}\left(0\right)$.

Substitute 0 for t into equation (7).

$\widetilde{x_1}\left(0\right)=x_1\left(0\right)+x_2\left(0\right)\times 0.1$

Substitute 1 for $x_1\left(0\right)$and 0 for $x_2\left(0\right)$into above equation.

$$\begin{gathered} \widetilde{x_1}\left(0\right)=1+0\times 0.1 \\ \widetilde{x_1}\left(0\right)=1+0 \\ \widetilde{x_1}\left(0\right)=1 \end{gathered}$$

Calculate the value of variable $\widetilde{x_2}\left(0\right)$

Substitute 0 for t into equation (5).

$\widetilde{x_2}\left(0\right)=x_2\left(0\right)+x_1\left(0\right)\times 0.1$

Substitute 1 for$x_1\left(0\right)$and 0 for $x_2\left(0\right)$into above equation.

$$\begin{gathered} \widetilde{x_2}\left(0\right)=0-0\times 0.1 \\ \widetilde{x_2}\left(0\right)=1-0.1 \\ \widetilde{x_2}\left(0\right)=-0.1 \end{gathered}$$

Calculate the value of $x_1\left(0.1\right)$.

Substitute 0 for t into equation (6).

$\begin{array}{rcl}{x}_1\left(0+0.1\right)& =& x_1\left(0\right)+\left(\frac{+x_2\left(0\right)+\widetilde{x_2}\left(0\right)}{2}\right)\times 0.1\\ x_1\left(0.1\right)& =& x_1\left(0\right)+\left(\frac{+x_2\left(0\right)+\widetilde{x_2}\left(0\right)}{2}\right)\times 0.1\\ & & \end{array}$

Substitute 1 for $x_1\left(0\right)$, $-0.1$for $\widetilde{x_2}\left(0\right)$into above equation.

$$\begin{gathered} x_1\left(0.1\right)=1+\left(\frac{0-0.1}{2}\right)\times 0.1 \\ {x}_1\left(0.1\right)=1-0.005 \\ {x}_1\left(0.1\right)=0.995 \end{gathered}$$

Calculate the value of $x_2\left(0.1\right)$.

Substitute 0 for t into equation (6).

$\begin{array}{rcl}{x}_2\left(0+0.1\right)& =& x_2\left(0\right)+\left(\frac{-x_1\left(0\right)-\widetilde{x_1}\left(0\right)}{2}\right)\times 0.1\\ x_2\left(0.1\right)& =& x_2\left(0\right)+\left(\frac{-x_1\left(0\right)-\widetilde{x_1}\left(0\right)}{2}\right)\times 0.1\\ & & \end{array}$

Substitute 0 for $x_2\left(0\right)$, 1 for $\widetilde{x_1}\left(0\right)$into above equation

$$\begin{gathered} x_1\left(0.1\right)=0+\left(\frac{-1-1}{2}\right)\times 0.1 \\ {x}_1\left(0.1\right)=0-0.1 \\ {x}_1\left(0.1\right)=-0.1 \end{gathered}$$

Repeat the above process and calculate the value of$x_1\left(0.3\right)$and$x_2\left(0.3\right)$

	0	0.1	0.2	0.3
$\widetilde{x_1}$	1	0.9845	0.9601	------
$\widetilde{x_2}$	-0.1	-0.1995	-0.297	------
$x_1$	1	0.995	0.98	0.9552
$x_2$	0	-0.1	-0.199	-0.296

Hence, the value of $x_1\left(0.3\right)$is $0.9552$and$x_2\left(0.3\right)$is $-0.296$by using modified Euler’s method.