Question

Question: Assume that${\mathbf{w}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{3}}{\mathbf{4}}$in (4). Use the resulting second-order Runge-Kutta method to approximate$\mathit{y}\left(0.5\right)$, where$\mathit{y}\mathbf{\left(}\mathit{x}\mathbf{\right)}$is the solution of the initial-value problem in Problem 1. Compare this approximate value with the approximate value obtained in Problem 11 in Exercises 9.1.

Short answer

Thus, the value of $y\left(0.5\right)$using improved Euler's method is 3.8254 and the value of $y\left(0.5\right)$using second order Runge-Kutta method is $3.8002$.

Step-by-step solution

Runga-Kutta method

The solution of a linear differential equation of the formis given as follows:

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

here

$$\begin{gathered} k_1=f\left(x_n,y_n\right) \\ {k}_2=f\left(\left(x_n+\frac{h}{2}\right),\left(y_n+\frac{h{k}_1}{2}\right)\right) \\ {k}_3=f\left(\left(x_n+\frac{h}{2}\right),\left(y_n+\frac{h{k}_2}{2}\right)\right) \\ {k}_4=f\left(\left(x_n+h\right),\left(y_n+h{k}_3\right)\right) \end{gathered}$$

Determine the value of w1,w2,α and β

The linear differential equation is$y\text{'}={\left(x+y-1\right)}^2$and$y\left(0\right)=2$. The value of$w_2$is$\frac{3}{4}$.

The formula for the second order Runge-kutta method is as follows:

$y_{n+1}=y_n+h\left(w_1{k}_1+w_2{k}_2\right)........\left(1\right)$

The value of$k_1$and$k_2$is calculated as follows:

$\begin{array}{l}{k}_1=f(x_n,y_n)\\ k_2=f(x_n+\alpha h,y_n+\beta h{k}_1)\end{array}$

The relation between the constants$w_1,w_2,\alpha$and$\beta$andis as follows:

$\begin{array}{l}{w}_1+w_2=1\\ w_2\alpha =\frac{1}{2}\\ w_2\beta =\frac{1}{2}\end{array}$

Substitute $\frac{3}{4}$for in expression $w_1+w_2=1$to obtain the value of $w_1$as follows:

$\begin{array}{l}{w}_1+\frac{3}{4}=1\\ w_1=1-\frac{3}{4}\\ w_1=\frac{1}{4}\end{array}$

Substitute $\frac{3}{4}$for $w_2$in expression $w_2\alpha =\frac{1}{2}$to obtain the value of $\alpha$as follows:

$\begin{array}{l}\left(\frac{3}{4}\right)\alpha =\frac{1}{2}\\ \alpha =\frac{4}{6}\\ \alpha =\frac{2}{3}\end{array}$

Substitute $\frac{3}{4}$for $w_2$in expression $w_2\beta =\frac{1}{2}$to obtain the value of $\beta$as follows:

$\begin{array}{l}\left(\frac{3}{4}\right)\beta =\frac{1}{2}\\ \beta =\frac{4}{6}\\ \beta =\frac{2}{3}\end{array}$

Calculate the value of k1k2k3 and k4

Calculate the value of$k_1{k}_2{k}_3$and$k_4$as follows:

Thus, the formula for the second order Runge-kutta method is written as follows:

$y_{n+1}=y_n+h\left(\frac{1}{4}+\frac{3}{4}{k}_2\right)............\left(2\right)$

The value of$k_1$and$k_2$is calculated as follows:

$\begin{array}{l}{k}_1=f\left(x_n,y_n\right)\\ k_2=f\left(x_n+\frac{2}{3}h,y_n+\frac{2}{3}h{k}_1\right)\end{array}$

The value of$k_1$and$k_2$is calculated as follows:

$\begin{array}{l}{k}_1=f(x_n,y_n)\\ k_2=f(x_n+\alpha h,y_n+\beta h{k}_1)\end{array}$

The differential equation$y\text{'}={\left(x+y-1\right)}^2$is of the form$y\text{'}=f\left(xy\right)$.

This implies that$f\left(xy\right)={\left(x+y-1\right)}^2$.

So, the value of$x_0$is 0 and the value of$y_0$is 2 .

To calculate the value of$y_1$substitute$n=0$in equation (2).

$y_1=y_0+h\left(\frac{1}{4}{k}_1+\frac{3}{4}{k}_2\right)\dots \dots \left(3\right)$

Calculate the value of$k_1$and$k_2$as follows:

$\begin{array}{l}{k}_1=f\left(x_0,y_0\right)\\ ={(0+2-1)}^2\\ =1^2\\ =1\end{array}$

Therefore, the value of$k_1$is 1.

role="math" localid="1668082561493" $\begin{array}{rcl}{k}_2& =& f\left(x_0+\frac{2}{3}h,y_0+\frac{2}{3}h{k}_1\right)\\ & =& f\left(\left(0.066\right),\left(2.066\right)\right)\\ & =& {\left(0.066+2.066-1\right)}^2\\ & =& 1.21484\end{array}$

Substitute the value$k_1$and$k_2$in equation (3).

$\begin{array}{rcl}{y}_1& =& y_0+h\left(\frac{1}{4}{k}_1+\frac{3}{4}{k}_2\right)\\ & =& 2+0.1\left(\frac{1}{4}+\frac{3}{4}1.28124\right)\\ & =& 2+0.1120\\ & =& 2.12110\end{array}$

Therefore, the value of$y_1$is 2.12110.

It is given that the initial value is $y\left(0\right)=2$.

Obtain the exact solution of the differential equation

The solution of differential equation of the formby the Improved Euler's method is given as follows:

$y_{n+1}=y_n+\mathrm{hf}\left(x_n{y}_n\right)-------\left(4\right)$

$y_{n+1}=y_n+\frac{h}{2}\left(f(x_n{y}_n\right)+f\left(x_{n+1}{y}_{n+1}\right))-------\left(5\right)$

Substitute 0 forin equation (4) to obtain the equation of$y_1^{*}$as shown below.

$y{*}_1=y_0+\mathrm{hf}(x_0,y_0)$

Substitute 0 for$x_0,2$for$y_0$and $0.1$ forin the above equation.

$\begin{array}{l}y{*}_1=2+(0.1)f(0,2)\\ =2+\left(0.1\right)·{\left(0+2-1\right)}^2\\ =2+0.1\\ =2.1\end{array}$

Substitute 0 for n in equation (5).

$y_1=y_0+\frac{h}{2}\left(f(x_0{y}_0\right)+f(x_1{y}_1*))$

Substitute 0 for$x_0,2$for$y_0,0.1$for$x_1,2.1$for$y{*}_1$and 0.1 forin the above equation.

$y_1=2+\frac{0.1}{2}\left(f\left(0,2\right)\right)+f\left(0.1,2.1\right)$

$\begin{array}{l}=2+\left(0.05\right){(0+2-1)}^2+{\left(0.1+2.1-1\right)}^2\\ =2+\left(0.05X2.44\right)\\ =2.122\end{array}$

Therefore, the approximated value of $y_1$or$y\left(0.1\right)$ is 2.1220.

Similarly, calculate the values of $y\left(0.2\right),y\left(0.3\right),y\left(0.4\right)$and $y\left(0.5\right)$as shown in

Thus, the value of$y\left(0.5\right)$using improved Euler's method is 3.8254 and the value of $y\left(0.5\right)$using second order Runge-Kutta method is 3.8002.