Question

In Problems 7-12 use the Runge-Kutta method to approximate$\mathit{x}\mathbf{(}\mathbf{0}\mathbf{.2}\mathbf{)}$and$\mathit{y}\mathbf{(}\mathbf{0}\mathbf{.2}\mathbf{)}$.First use$\mathit{h}\mathbf{=}\mathbf{0}\mathbf{.2}$and then use$\mathit{h}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}$. Use a numerical solver and$\mathit{h}\mathbf{=}\mathbf{0}\mathbf{.1}$to graph the solution in the neighborhood of$\mathit{t}\mathbf{=}\mathbf{0}\mathbf{.}$

$\begin{array}{c}{\mathbf{x}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{y}\\ {\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{x}\\ \mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{6}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\end{array}$

Short answer

The solution of the problem calculated using Runge-Kutta method is

$y_2=3.4199,x_2=8.3055$for$h=0.1$and$y_2=3.4199,x_2=8.3055$for$h=0.2$

Step-by-step solution

Define Runge-Kutta Method.

To cancel out lower-order error terms, the Runge-Kutta technique of integrating differential equations uses a trial step at the middle of an interval. When paired with a clever adaptive step-size procedure, this method is relatively simple and resilient, making it an excellent general option for solving differential equations. The formula used in this method is,

$y_{n+1}=y_n+\frac{1}{6}{k}_1+\frac{1}{3}{k}_2+\frac{1}{3}{k}_3+\frac{1}{6}{k}_4+O\left(h^5\right)$

Find the value of y1,u1 for h=0.2

We have the first order differential equation with initial condition,

$\begin{array}{c}{x}^{\text{'}}=2x-y\\ y^{\text{'}}=x\\ x\left(0\right)=6,y\left(0\right)=2\end{array}$

Using Runge-Kutta method using the step find the approximate values using the formula,

$\begin{array}{c}{y}_{n+1}=y_n+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)\text{(1a)}\\ x_{n+1}=x_n+\frac{h}{6}\left(m_1+2m_2+2m_3+m_4\right)\text{(2a)}\end{array}$

We have to find the constants $\left(k_1,k_2,k_3\text{,}{k}_4\right)$and $\left(m_1,m_2,m_3\text{,}{m}_4\right)$at $n=0\text{,}h=0.2$.

Since we have $x_0=6\text{,}{y}_0=2$we get,

$\begin{array}{c}{k}_1=f\left(t_0,x_0,y_0\right)\\ =x_0\\ =6\\ m_1=g\left(t_0,t_0,y_0\right)\\ =2x_0-y_0\\ =2\times 6-2\\ =12-2\\ =10\end{array}$

$\begin{array}{c}{k}_2=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)\\ =x_0+\frac{1}{2}h{m}_1\\ =6+\frac{1}{2}\times 0.2\times 10\\ =6+1\\ =7\end{array}$

$\begin{array}{c}{m}_2=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)\\ =2\left(x_0+\frac{1}{2}h{m}_1\right)-\left(y_0+\frac{1}{2}h{k}_1\right)\\ =2\left(6+\frac{1}{2}\times 0.2\times 10\right)-\left(2+\frac{1}{2}\times 0.2\times 6\right)\\ =2\left(7\right)-\left(2.6\right)\\ =14-2.6\\ =11.4\end{array}$

$\begin{array}{c}{k}_3=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_2,y_0+\frac{1}{2}h{k}_2\right)\\ =x_0+\frac{1}{2}h{m}_2\\ =6+\frac{1}{2}\times 0.2\times 11.4\\ =6+1.14\\ =7.14\end{array}$

$\begin{array}{c}{m}_3=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_2,y_0+\frac{1}{2}h{k}_2\right)\\ =2\left(x_0+\frac{1}{2}h{m}_2\right)-\left(y_0+\frac{1}{2}h{k}_2\right)\\ =2\left(6+\frac{1}{2}\times 0.2\times 11.4\right)-\left(2+\frac{1}{2}\times 0.2\times 7\right)\\ =2\left(7.14\right)-\left(2.7\right)\\ =14.28-2.7\\ =11.58\end{array}$

$\begin{array}{c}{k}_4=f\left(t_0+h,x_0+h{m}_3,y_0+\frac{1}{2}h{k}_3\right)\\ =x_0+h{m}_3\\ =6+0.2\times 11.58\\ =6+2.316\\ =8.316\end{array}$

$\begin{array}{c}{m}_4=g\left(t_0+h,x_0+h{m}_3,y_0+h{k}_3\right)\\ =2\left(x_0+h{m}_3\right)-\left(y_0+h{k}_3\right)\\ =2(6+0.2\times 11.58)-(2+0.2\times 7.14)\\ =2\left(8.316\right)-\left(3.428\right)\\ =16.632-3.428\\ =13.204\end{array}$

After that when n=0, substitute the value of constants to get the value of y₁,x₁.

$\begin{array}{c}{y}_1=y_0+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ =2+\frac{0.2}{6}[6+2(7)+2(7.14)+8.316]\\ =2+\frac{0.2}{6}\times \left(42.596\right)\\ =2+1.41987\\ =3.4199\end{array}$

and

$\begin{array}{c}{x}_1=x_0+\frac{h}{6}\left(m_1+2m_2+2m_3+m_4\right)\\ =6+\frac{0.2}{6}[10+2(11.4)+2(11.58)+13.204]\\ =6+\frac{0.2}{6}\times \left(69.164\right)\\ =6+2.3055\\ =8.3055\end{array}$

Find the value of y2 ffor h=0.1

Using method using the step h=0 along with the conditions use the below formula to find the values.

$\begin{array}{c}{y}_{n+1}=y_n+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)\text{(1a)}\\ x_{n+1}=x_n+\frac{h}{6}\left(m_1+2m_2+2m_3+m_4\right)\text{(2a)}\end{array}$

$\begin{array}{c}{k}_1=f\left(t_0,x_0,y_0\right)\\ =x_0\\ =6\\ m_1=g\left(t_0,t_0,y_0\right)\\ =2x_0-y_0\\ =2\times 6-2\\ =12-2\\ =10\end{array}$

$\begin{array}{c}{k}_2=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)\\ =x_0+\frac{1}{2}h{m}_1\\ =6+\frac{1}{2}\times 0.1\times 10\\ =6+0.5\\ =6.5\end{array}$

$\begin{array}{c}{m}_2=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)\\ =2\left(x_0+\frac{1}{2}h{m}_1\right)-\left(y_0+\frac{1}{2}h{k}_1\right)\\ =2\left(6+\frac{1}{2}\times 0.1\times 10\right)-\left(2+\frac{1}{2}\times 0.1\times 6\right)\\ =2\left(6.5\right)-\left(2.3\right)\\ =13-2.3\\ =10.7\end{array}$

$\begin{array}{c}{k}_3=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_2,y_0+\frac{1}{2}h{k}_2\right)\\ =x_0+\frac{1}{2}h{m}_2\\ =6+\frac{1}{2}\times 0.1\times 10.7\\ =6+0.535\\ =6.535\end{array}$

$\begin{array}{c}{m}_3=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_2,y_0+\frac{1}{2}h{k}_2\right)\\ =2\left(x_0+\frac{1}{2}h{m}_2\right)-\left(y_0+\frac{1}{2}h{k}_2\right)\\ =2\left(6+\frac{1}{2}\times 0.1\times 10.7\right)-\left(2+\frac{1}{2}\times 0.1\times 6.5\right)\\ =2\left(6.535\right)-\left(2.325\right)\\ =13.07-2.325\\ =10.745\end{array}$

$\begin{array}{c}{k}_4=f\left(t_0+h,x_0+h{m}_3,y_0+\frac{1}{2}h{k}_3\right)\\ =x_0+h{m}_3\\ =6+0.1\times 10.745\\ =6+1.0745\\ =7.0745\end{array}$

$\begin{array}{c}{m}_4=g\left(t_0+h,x_0+h{m}_3,y_0+h{k}_3\right)\\ =2\left(x_0+h{m}_3\right)-\left(y_0+h{k}_3\right)\\ =2(6+0.1\times 10.745)-(2+0.1\times 6.535)\\ =2\left(7.0745\right)-\left(2.6535\right)\\ =14.149-2.6535\\ =11.4955\end{array}$

After that when n=0 substitute the vlue of constants to get the value of y_1,x₁

$\begin{array}{c}{y}_1=y_0+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ =2+\frac{0.1}{6}[6+2(6.5)+2(6.535)+7.0745]\\ =2+\frac{0.1}{6}\times \left(39.1445\right)\\ =2+0.6524\\ =2.6524\end{array}$

$\begin{array}{c}{x}_1=x_0+\frac{h}{6}\left(m_1+2m_2+2m_3+m_4\right)\\ =6+\frac{0.1}{6}[10+2(10.7)+2(10.745)+11.4955]\\ =6+\frac{0.1}{6}\times \left(64.3855\right)\\ =6+1.0731\\ =7.0731\end{array}$

After that we can find the value of x and y using other constants as shown in the table below. Then we get the value of $y_2=3.4199$

Hence the value is .$y_2=3.4199$