Question

Consider the first order differential equation, $\frac{d{x}_1}{dt}=-x_2$with an initial value$x\left(0\right)=10$ . With an integration step size of seconds, determine the value of using (a) Euler’s method, (b) the modified Euler’s method.

Short answer

(a) The value of$x\left(0.5\right)$ by using the Euler’s method is $5.9049$.

(b) The value of $x\left(0.5\right)$by using the modified Euler’s method is $6.0715$.

Step-by-step solution

Write the given data from the question.

Write the differential equation,

$\frac{d{x}_1}{dt}=-x_2$

Initial value,$x\left(0\right)=10$

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

 Step 2: Determine the equation to calculate the value of 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.

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

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

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

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

Calculate the value of x0.5by using Euler’s method.

(a)

Solve the ordinary differential equation.

Substitute $0.1\text{sec}$for $\Delta t$and $-x\left(t\right)$for $\frac{dx\left(t\right)}{dt}$into equation (1).

$x\left(t+0.1\right)=x\left(t\right)-0.1x\left(t\right)$ …… (3)

Calculate the value of$x\left(0.1\right)$

Substitute 0 for into equation (3).

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

Substitute 10 for $x\left(0\right)$into above equation

$$\begin{gathered} x\left(0.1\right)=10-0.1\times 10 \\ x\left(0.1\right)=10-1 \\ x\left(0.1\right)=9 \end{gathered}$$

Calculate the value of $x\left(0.2\right)$.

Substitute $0.1$for into equation (3).

$\begin{array}{rcl}x\left(0.1+0.1\right)& =& x\left(0.1\right)-0.1x\left(0.1\right)\\ x\left(0.2\right)& =& x\left(0.1\right)-0.1x\left(0.1\right)\\ & & \end{array}$

Substitute 9 for into$x\left(0.1\right)$ above equation

$$\begin{gathered} x\left(0.2\right)=9-0.1\times 9 \\ x\left(0.2\right)=9-0.9 \\ x\left(0.2\right)=8.1 \end{gathered}$$

Calculate the value of $x\left(0.3\right)$.

Substitute 0.2 for into equation (3).

$\begin{array}{rcl}x\left(0.2+0.1\right)& =& x\left(0.2\right)-0.1x\left(0.2\right)\\ x\left(0.3\right)& =& x\left(0.2\right)-0.1x\left(0.2\right)\\ & & \end{array}$

Substitute $8.1$for$x\left(0.2\right)$ into above equation

role="math" localid="1656070654945" $$\begin{gathered} x\left(0.3\right)=8.1-0.1\times 8.1 \\ x\left(0.3\right)=8.1-0.81 \\ x\left(0.3\right)=7.29 \end{gathered}$$

Calculate the value of $x\left(0.4\right)$.

Substitute$0.3$ for into equation (3).

$\begin{array}{rcl}x\left(0.3+0.1\right)& =& x\left(0.3\right)-0.1x\left(0.3\right)\\ x\left(0.4\right)& =& x\left(0.3\right)-0.1x\left(0.3\right)\\ & & \end{array}$

Substitute $7.29$for$x\left(0.3\right)$ into above equation

$$\begin{gathered} x\left(0.4\right)=7.29-0.1\times 7.29 \\ x\left(0.4\right)=7.29-0.729 \\ x\left(0.4\right)=6.561 \end{gathered}$$

Calculate the value of $x\left(0.5\right)$.

Substitute for $x\left(0.4\right)$into equation (3).

$\begin{array}{rcl}x\left(0.4+0.1\right)& =& x\left(0.4\right)-0.1x\left(0.4\right)\\ x\left(0.5\right)& =& x\left(0.4\right)-0.1x\left(0.4\right)\\ & & \end{array}$

Substitute $6.561$for into above equation

$$\begin{gathered} x\left(0.5\right)=6.561-0.1\times 6.561 \\ x\left(0.5\right)=6.561-0.6561 \\ x\left(0.5\right)=5.9049 \end{gathered}$$

Hence the value of $x\left(0.5\right)$by using the Euler’s method is $5.9049$.

Calculate the value of by using modified Euler’s method.

$x\left(t+0.1\right)=x\left(t\right)+\left(\frac{-x\left(t\right)-\tilde{x}\left(t\right)}{2}\right)\times 0.1$b)

The variable $\tilde{x}\left(t\right)$is defined as,

localid="1656084442483" $\tilde{x}\left(t\right)=x\left(t\right)+\frac{dx\left(t\right)}{dt}\Delta t$

Substitute localid="1656084471034" $0.1$forlocalid="1656084461231" $\Delta t$andlocalid="1656084449532" $-x\left(t\right)$for localid="1656084454921" $\frac{dx\left(t\right)}{dt}$into above equation.

localid="1656084483967" $\tilde{x}\left(t\right)=x\left(t\right)-x\left(t\right)\times 0.1$ …… (4)

localid="1656084489993" $\frac{d\tilde{x}\left(t\right)}{dt}=-\tilde{x}\left(t\right)$

Solve the modified form of Euler’s method.

Substitute andlocalid="1656084495667" $-\tilde{x}\left(t\right)$forlocalid="1656084510241" $\frac{dx\left(t\right)}{dt}$into equation (2).

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

Substitute $0.1$for $\Delta t$into above equation.

$x\left(t+0.1\right)=x\left(t\right)+\left(\frac{-x\left(t\right)-\tilde{x}\left(t\right)}{2}\right)\times 0.1$ $x\left(t+0.1\right)=x\left(t\right)+\left(\frac{-x\left(t\right)-\tilde{x}\left(t\right)}{2}\right)\times 0.1$ …… (5)

Calculate the value of variable localid="1656084545440" $\tilde{x}\left(0\right)$

Substitute 0 for into equation (4).

$\tilde{x}\left(0\right)=x\left(0\right)-x\left(0\right)\times 0.1$

Substitute for into above equation.

$$\begin{gathered} \tilde{x}\left(0\right)=10-10\times 0.1 \\ \tilde{x}\left(0\right)=10-1 \\ \tilde{x}\left(0\right)=9 \end{gathered}$$$$\begin{gathered} \tilde{x}\left(0\right)=10-10\times 0.1 \\ \tilde{x}\left(0\right)=10-1 \\ \tilde{x}\left(0\right)=9 \end{gathered}$$

Calculate the value of$x\left(0,1\right)$

Substitute 0 for into equation (5).

Substitute 10 for $x\left(0\right)$, 9 for $\tilde{x}\left(0\right)$into above equation.

Calculate the value of variable $\tilde{x}\left(0.1\right)$.

Substitute$0.1$for into equation (4).

$\tilde{x}\left(0.1\right)=x\left(0.1\right)-x\left(0.1\right)\times 0.1$

Substitute 9.05 for $x\left(0.1\right)$into above equation.

localid="1656092132555" $$\begin{gathered} \tilde{x}\left(0.1\right)=9.05-9.05\times 0.1 \\ \tilde{x}\left(0.1\right)=9.05-0.905 \\ \tilde{x}\left(0.1\right)=8.145 \end{gathered}$$

Calculate the value of $x\left(0.2\right)$.

Substitute $0.1$for t into equation (5).

$\begin{array}{rcl}x\left(0.1+0.1\right)& =& x\left(0.1\right)+\left(\frac{-x\left(0.1\right)-\tilde{x}\left(0.1\right)}{2}\right)\times 0.1\\ x\left(0.2\right)& =& x\left(0.1\right)+\left(\frac{-x\left(0.1\right)-\tilde{x}\left(0.1\right)}{2}\right)\times 0.1\\ & & \end{array}$

Substitute $9.05$for $x\left(0.1\right)$,$8.145$for $\tilde{x}\left(0.1\right)$into above equation.

$$\begin{gathered} x\left(0.2\right)=9.05+\left(\frac{-9.05-8.145}{2}\right)\times 0.1 \\ x\left(0.2\right)=9.05-0.859 \\ x\left(0.2\right)=8.191 \end{gathered}$$

Calculate the value of variable$\tilde{x}\left(0.2\right)$.

Substitute 0.2 for into equation (4).

$\tilde{x}\left(0.2\right)=x\left(0.2\right)-x\left(0.2\right)\times 0.1$

Substitute $8.191$for $x\left(0.2\right)$into above equation.

$$\begin{gathered} \tilde{x}\left(0.1\right)=8.191-8.191\times 0.1 \\ \tilde{x}\left(0.1\right)=9.05-0.8191 \\ \tilde{x}\left(0.1\right)=7.3719 \end{gathered}$$

Calculate the value of $x\left(0.3\right)$.

Substitute 0.2 for into equation (5).

$\begin{array}{rcl}x\left(0.2+0.1\right)& =& x\left(0.2\right)+\left(\frac{-x\left(0.2\right)-\tilde{x}\left(0.2\right)}{2}\right)\times 0.1\\ x\left(0.3\right)& =& x\left(0.2\right)+\left(\frac{-x\left(0.2\right)-\tilde{x}\left(0.2\right)}{2}\right)\times 0.1\\ & & \end{array}$

Substitute$8.191$for$x\left(0.2\right)$,$7.3719$for$\tilde{x}\left(0.2\right)$into above equation.

$$\begin{gathered} x\left(0.3\right)=8.191+\left(\frac{-8.191-7.3719}{2}\right)\times 0.1 \\ x\left(0.3\right)=9.05-0.778 \\ x\left(0.3\right)=7.413 \end{gathered}$$

Calculate the value of variable $\tilde{x}\left(0.3\right)$.

Substitute $0.3$for into equation (4).

$\tilde{x}\left(0.3\right)=x\left(0.3\right)-x\left(0.3\right)\times 0.1$

Substitute$7.413$for$x\left(0.3\right)$into above equation.

$$\begin{gathered} \tilde{x}\left(0.3\right)=7.413-7.413\times 0.1 \\ \tilde{x}\left(0.3\right)=7.413-0.7413 \\ \tilde{x}\left(0.3\right)=6.6717 \end{gathered}$$

Calculate the value of $x\left(0.4\right)$.

Substitute $0.3$for into equation (5).

$$\begin{gathered} x\left(0.3+0.1\right)=x\left(0.3\right)+\left(\frac{-x\left(0.3\right)-\tilde{x}\left(0.3\right)}{2}\right)\times 0.1 \\ x\left(0.4\right)=x\left(0.3\right)+\left(\frac{-x\left(0.3\right)-\tilde{x}\left(0.3\right)}{2}\right)\times 0.1 \end{gathered}$$

Substitute$7.413$for $x\left(0.3\right)$,$6.6717$for$\tilde{x}\left(0.3\right)$into above equation.

$$\begin{gathered} x\left(0.4\right)=7.413+\left(\frac{-7.413-6.6717}{2}\right)\times 0.1 \\ x\left(0.4\right)=7.413-0.7042 \\ x\left(0.4\right)=6.7088 \end{gathered}$$

Calculate the value of variable$\tilde{x}\left(0.4\right)$.

Substitute$0.4$for t into equation (4).

$\tilde{x}\left(0.4\right)=x\left(0.4\right)-x\left(0.4\right)\times 0.1$

Substitute for into above equation.

$$\begin{gathered} \tilde{x}\left(0.4\right)=6.7088-6.7088\times 0.1 \\ \tilde{x}\left(0.4\right)=6.7088-0.67088 \\ \tilde{x}\left(0.4\right)=6.0379 \end{gathered}$$

Calculate the value of $x\left(0.5\right)$.

Substitute $0.4$for into equation (5).

$\begin{array}{rcl}x\left(0.4+0.1\right)& =& x\left(0.4\right)+\left(\frac{-x\left(0.4\right)-\tilde{x}\left(0.4\right)}{2}\right)\times 0.1\\ x\left(0.5\right)& =& x\left(0.4\right)+\left(\frac{-x\left(0.4\right)-\tilde{x}\left(0.4\right)}{2}\right)\times 0.1\\ & & \end{array}$

Substitute $6.7088$for$x\left(0.4\right)$,$6.0379$for $\tilde{x}\left(0.4\right)$into above equation.

$$\begin{gathered} x\left(0.5\right)=6.7088+\left(\frac{-6.7088-6.0379}{2}\right)\times 0.1 \\ x\left(0.5\right)=6.7088-0.6373 \\ x\left(0.5\right)=6.0715 \end{gathered}$$

Hence, the value of $x\left(0.5\right)$by using the modified Euler’s method is$6.0715$.