Chapter 6, Problem 1CR

To determine

The value of $y\left(1.1\right),y\left(1.2\right),y\left(1.3\right),y\left(1.4\right)\text{ and }y\left(1.5\right)$ for the differential equation $y^{\prime }=2\ln xy$ by using Euler’s method, Improved Euler’s method and Runga-Kutta method of fourth order.

Explanation of Solution

Given:

The linear differential equation is $y^{\prime }=2\ln xy$ such that $y\left(1\right)=2$ and the value of step size is $h=0.1$ and $h=0.05$.

Concept used:

Runga-Kutta method:

The solution of a linear differential equation of the form $y^{\prime }=f\left(x,y\right)$ is given as follows:

$y_{n+1}=y_n+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)$, where

$\begin{array}{l}{k}_1=f\left(x_n,y_n\right)\\ k_2=f\left(\left(x_n+\frac{h}{2}\right),\left(y_n+\frac{h{k}_1}{2}\right)\right)\\ k_3=f\left(\left(x_n+\frac{h}{2}\right),\left(y_n+\frac{h{k}_2}{2}\right)\right)\\ k_4=f\left(\left(x_n+h\right),\left(y_n+h{k}_3\right)\right)\end{array}$

Euler’s method

The approximated solution for the first order linear differential equation of the form $y^{\prime }=f\left(x,y\right)$ with initial value as $y\left(x_0\right)=y_0$ with a step size of $h$ is given as follows:

$y_{n+1}=y_n+hf\left(x_n,y_n\right)$, where $x_n=x_0+nh$.

Improved Euler’s method:

As per the Improved Euler’s method the solution of a linear differential equation of the form $y^{\prime }=f\left(x,y\right)$ is given as follows:

$\begin{array}{l}{y}_{n+1}^{*}=y_n+hf\left(x_n,y_n\right)\\ y_{n+1}=y_n+\frac{h}{2}\left(f\left(x_n,y_n\right)+f\left(x_{n+1},y_{n+1}^{*}\right)\right)\end{array}$

Calculation:

The linear differential equation is given as follows:

$y^{\prime }=2\ln \left(xy\right)$

The value of step size is given as $h=0.1$ and $h=0.05$.

The initial value is given as $y\left(1\right)=2$.

The given differential equation is of the form $y^{\prime }=f\left(x,y\right)$.

$f\left(x,y\right)=\ln \left(xy\right)$

This implies that $x_0=1$ and $y_0=2$.

Obtain the solution of the given differential equation by Euler’s method for $h=0.1$.

As per the Euler’s method the solution of a linear differential equation of the form $y^{\prime }=f\left(x,y\right)$ is given as follows:

$y_{n+1}=y_n+hf\left(x_n,y_n\right)\dots \left(1\right)$

Substitute $0$ for $n$ in equation $\left(1\right)$.

$y_1=y_0+hf\left(x_0,y_0\right)$

Substitute the value of $x_0,y_0\text{ and }h$ in the above equation.

$\begin{array}{c}{y}_1=2+\left(0.1\right)f\left(1,2\right)\\ =2+\left(0.1\right)\cdot 2\cdot \ln \left(2\right)\\ =2+0.1386\\ =2.1386\end{array}$

Therefore, the value of $y_1\text{ or }y\left(1.1\right)$ is $2.1386$.

Similarly, use the above procedure and the value of $y\left(1.2\right),y\left(1.3\right).y\left(1.4\right)\text{ and }y\left(1.5\right)$.

$x_n$	$y_n$(Euler’s Method) $h=0.1$	$y_n$(Euler’s Method) $h=0.05$
$1.00$	$2.0000$	$2.0000$
$1.05$		$2.0693$
$1.10$	$2.1386$	$2.1469$
$1.15$		$2.2328$
$1.20$	$2.3097$	$2.3272$
$1.25$		$2.4299$
$1.30$	$2.5136$	$2.5409$
$1.35$		$2.6604$
$1.40$	$2.7504$	$2.7883$
$1.45$		$2.9245$
$1.50$	$3.0201$	$3.0690$

Table $1$

Table $1$ represents the value of $y\left(1.2\right),y\left(1.3\right).y\left(1.4\right)\text{ and }y\left(1.5\right)$ calculated by Euler’s method.

Obtain the solution of the given differential equation by Improved Euler’s method for $h=0.1$.

As per the Improved Euler’s method the solution of a linear differential equation of the form $y^{\prime }=f\left(x,y\right)$ is given as follows:

$\begin{array}{l}{y}_{n+1}^{*}=y_n+hf\left(x_n,y_n\right)\dots \left(2\right)\\ y_{n+1}=y_n+\frac{h}{2}\left(f\left(x_n,y_n\right)+f\left(x_{n+1},y_{n+1}^{*}\right)\right)\dots \left(3\right)\end{array}$

Substitute $0$ for $n$ in equation $\left(2\right)$.

$y_1^{*}=y_0+hf\left(x_0,y_0\right)$

Substitute the value of $x_0,y_0\text{ and }h$ in the above equation.

$\begin{array}{c}{y}_1^{*}=2+\left(0.1\right)f\left(1,2\right)\\ =2+\left(0.1\right)\cdot 2\cdot \ln \left(2\right)\\ =2+0.1386\\ =2.1386\end{array}$

Substitute $0$ for $n$ in equation $\left(3\right)$.

$y_1=y_0+\frac{h}{2}\left(f\left(x_0,y_0\right)+f\left(x_1,y_1^{*}\right)\right)$

Substitute the value of $x_0,y_0\text{,}{x}_1\text{,}{y}_1^{*}\text{ and }h$ in the above equation.

$\begin{array}{c}{y}_1=2+\frac{0.1}{2}\left(f\left(1,2\right)+f\left(1.1,2.1386\right)\right)\\ =2+\left(0.05\right)\left(2\ln \left(2\right)+2\ln \left(1.1\cdot 2.1386\right)\right)\\ =2+\left(0.05\right)\left(1.3863+1.7109\right)\\ =2.1549\end{array}$

Therefore, the value of $y_1\text{ or }y\left(1.1\right)$ is $2.1549$.

Similarly, with above procedure used the value of $y\left(1.2\right),y\left(1.3\right).y\left(1.4\right)\text{ and }y\left(1.5\right)$.

$x_n$	$y_n$ (Improved Euler’s Method) $h=0.1$	$y_n$ (Improved Euler’s Method) $h=0.05$
$1.00$	$2.0000$	$2.0000$
$1.05$		$2.0735$
$1.10$	$2.1549$	$2.1554$
$1.15$		$2.2459$
$1.20$	$2.3439$	$2.3450$
$1.25$		$2.4527$
$1.30$	$2.5672$	$2.5689$
$1.35$		$2.6937$
$1.40$	$2.8246$	$2.8269$
$1.45$		$2.9686$
$1.50$	$3.1157$	$3.1187$

Table $2$

Table $2$ represents the value of $y\left(1.2\right),y\left(1.3\right).y\left(1.4\right)\text{ and }y\left(1.5\right)$ calculated by Improved Euler’s method.

Obtain the solution of the given differential equation by Euler’s modified method for $h=0.1$.

As per the Runga-Kutta method of fourth order the value of $y_{n+1}$ is given as follows:

$y_{n+1}=y_n+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\dots \left(4\right)$

Here, the value of $k_1,k_2,k_3\text{ and }{k}_4$ is given as follows:

$\begin{array}{l}{k}_1=f\left(x_n,y_n\right)\dots \left(5\right)\\ k_2=f\left(\left(x_n+\frac{h}{2}\right),\left(y_n+\frac{h{k}_1}{2}\right)\right)\dots \left(6\right)\\ k_3=f\left(\left(x_n+\frac{h}{2}\right),\left(y_n+\frac{h{k}_2}{2}\right)\right)\dots \left(7\right)\\ k_4=f\left(\left(x_n+h\right),\left(y_n+h{k}_3\right)\right)\dots \left(8\right)\end{array}$

Substitute $0$ for $n$ in equation $\left(4\right)$.

$y_1=y_0+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)\dots \left(9\right)$

Calculate the value of $k_1$ as follows:

$\begin{array}{c}{k}_1=f\left(x_0,y_0\right)\\ =f\left(1,2\right)\\ =2\ln \left(2\right)\\ =1.3863\end{array}$

Calculate the value of $k_2$ as follows:

$\begin{array}{c}{k}_2=f\left(\left(x_0+\frac{h}{2}\right),\left(y_0+\frac{h{k}_1}{2}\right)\right)\\ =f\left(\left(1+0.05\right),\left(2+\left(\left(0.05\right)\cdot \left(1.3863\right)\right)\right)\right)\\ =f\left(1.05,2.069315\right)\end{array}$

Further solve the above equation.

$\begin{array}{c}{k}_2=2\ln \left(1.05\cdot 2.069315\right)\\ =1.5520\end{array}$

Calculate the value of $k_1$ as follows:

$\begin{array}{c}{k}_3=f\left(\left(x_{}+\frac{h}{2}\right),\left(y_0+\frac{h{k}_2}{2}\right)\right)\\ =f\left(\left(1.05\right),\left(2.0776\right)\right)\\ =2\ln \left(1.05\cdot 2.0776\right)\\ =1.5600\end{array}$

Calculate the value of $k_4$ as follows:

$\begin{array}{c}{k}_4=f\left(\left(x_0+h\right),\left(y_0+h{k}_3\right)\right)\\ =f\left(\left(1.1\right),\left(2.1560\right)\right)\\ =2\ln \left(1.1\cdot 2.1560\right)\\ =1.7271\end{array}$

Substitute the value of $k_1,k_2,k_3,k_4\text{ and }{y}_0$ in equation $\left(9\right)$.

$\begin{array}{c}{y}_1=y_0+\frac{0.1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ =2+\frac{0.1}{6}\left(1.3863+2\cdot 1.5520+2\cdot 1.5600+1.7271\right)\\ =2+0.1556\\ =2.1556\end{array}$

Therefore, the value of $y_1\text{ or }y\left(1.1\right)$ is $2.1556$.

Similarly, with above procedure used the value of $y\left(1.2\right),y\left(1.3\right).y\left(1.4\right)\text{ and }y\left(1.5\right)$.

$x_n$	$y_n$ (Runga-Kutta $4th$ order) $h=0.1$	$y_n$ (Runga-Kutta $4th$ order) $h=0.05$
$1.00$	$2.0000$	$2.0000$
$1.05$		$2.0736$
$1.10$	$2.1556$	$2.1556$
$1.15$		$2.2462$
$1.20$	$2.3454$	$2.3454$
$1.25$		$2.4532$
$1.30$	$2.5695$	$2.5695$
$1.35$		$2.6944$
$1.40$	$2.8278$	$2.8278$
$1.45$		$2.9696$
$1.50$	$3.1197$	$3.1197$

Table $3$

Table $3$ represents the value of $y\left(1.2\right),y\left(1.3\right).y\left(1.4\right)\text{ and }y\left(1.5\right)$ calculated by Runga-Kutta method.

The table which shows a comparison between the values of $y\left(1.2\right),y\left(1.3\right)$, $y\left(1.4\right)$ and $y\left(1.5\right)$ obtained by Euler’s method, improved Euler’s method and Runga-Kutta method of fourth order is as below.

$x_n$	$y_n$ (Euler’s Method) $h=0.1$	$y_n$ (Euler’s Method) $h=0.05$	$y_n$ (Improved Euler’s Method) $h=0.1$	$y_n$ (Improved Euler’s Method) $h=0.05$	$y_n$ (Runga-Kutta $4th$ order) $h=0.1$	$y_n$ (Runga-Kutta $4th$ order) $h=0.05$
$1.00$	$2.0000$	$2.0000$	$2.0000$	$2.0000$	$2.0000$	$2.0000$
$1.05$		$2.0693$		$2.0735$		$2.0736$
$1.10$	$2.1386$	$2.1569$	$2.1549$	$2.1554$	$2.1556$	$2.1556$
$1.15$		$2.2328$		$2.2459$		$2.2462$
$1.20$	$2.3097$	$2.3272$	$2.3439$	$2.3450$	$2.3454$	$2.3454$
$1.25$		$2.4299$		$2.4527$		$2.4532$
$1.30$	$2.5136$	$2.5409$	$2.5672$	$2.5689$	$2.5695$	$2.5695$
$1.35$		$2.6604$		$2.6937$		$2.6944$
$1.40$	$2.7504$	$2.7883$	$2.8246$	$2.8269$	$2.8278$	$2.8278$
$1.45$		$2.9245$		$2.9686$		$2.9696$
$1.50$	$3.0201$	$3.0690$	$3.1157$	$3.1187$	$3.1197$	$3.1197$

Table $4$

Thus, table $4$ shows the comparison of the values of $y\left(1.2\right),y\left(1.3\right)$, $y\left(1.4\right)$ and $y\left(1.5\right)$ obtained by Euler’s method, improved Euler’s method and Runga-Kutta method of fourth order.