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 to graph the solution in a neighborhood of $\mathit{t}\mathbf{=}\mathbf{0}$.

$\begin{array}{l}\mathbf{x}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\mathbf{y}\mathbf{+}\mathbf{t}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{t}\\ \mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{5}\end{array}$

Short answer

$\begin{array}{ccc}h& y_1& x_1\\ 0.2& 4.2873& -3.9097\\ 0.1& 4.63& -3.482\end{array}$

Step-by-step solution

Consider the equations and the conditions

The equations and the initial conditions are,

$\begin{array}{l}x\text{'}=-y+t\\ y\text{'}=x-t\\ x\left(0\right)=-3,y\left(0\right)=5\end{array}$

Compute the constants

Consider$h=0.2,t=0$

The values are,

$\begin{array}{l}{k}_1=f(t_0,x_0,y_0)=x-t=-3-0=-3\\ m_1=g(t_0,x_0,y_0)=-y+t=-5+0=-5\end{array}$

$\begin{array}{l}{k}_2=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=\left(x+\frac{1}{2}h{m}_1\right)-\left(t+\frac{1}{2}h\right)=-3-0.5-0.1=-3.6\\ m_2=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=-\left(y+\frac{1}{2}h{k}_1\right)+\left(t+\frac{1}{2}h\right)=-5+0.3+0.1=-4.6\end{array}$

$\begin{array}{l}{k}_3=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_2,y_0+\frac{1}{2}h{k}_2\right)=\left(x+\frac{1}{2}h{m}_2\right)-\left(t+\frac{1}{2}h\right)=-3.46-0.1=-3.54\\ m_3=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_2,y_0+\frac{1}{2}h{k}_2\right)=-\left(y+\frac{1}{2}h{k}_2\right)+\left(t+\frac{1}{2}h\right)=-5+0.36+0.1=-4.5\end{array}$

$\begin{array}{l}{k}_4=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_3,y_0+\frac{1}{2}h{k}_3\right)=\left(x+\frac{1}{2}h{m}_3\right)-\left(t+\frac{1}{2}h\right)=-3.9-0.2=-4.1\\ m_4=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_3,y_0+\frac{1}{2}h{k}_3\right)=-\left(y+\frac{1}{2}h{k}_3\right)+\left(t+\frac{1}{2}h\right)=-5+0.708+0.2=-4.092\end{array}$

Compute the equations

Substitute the values of variables in the equation.

$\begin{array}{l}{y}_1=y_0+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ \Rightarrow y_1=5+\frac{0.2}{6}(-3+2(-3.6)+2(-3.54)-4.1)\\ \therefore y_1=4.2873\end{array}$

$\begin{array}{l}{x}_1=x_0+\frac{h}{6}\left(m_1+2m_2+2m_3+m_4\right)\\ \Rightarrow x_1=-3+\frac{0.2}{6}(-5+2(-4.6)+2(-4.5)+(-4.092\left)\right)\\ \therefore x_1=-3.9097\end{array}$

Compute the constants

Consider$h=0.1,t=0$

The values are,

$\begin{array}{l}{k}_1=f(t_0,x_0,y_0)=x-t=-3-0=-3\\ m_1=g(t_0,x_0,y_0)=-y+t=-5+0=-5\end{array}$

$\begin{array}{l}{k}_2=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=\left(x+\frac{1}{2}h{m}_1\right)-\left(t+\frac{1}{2}h\right)=-3.25-0.05=-3.3\\ m_2=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=-\left(y+\frac{1}{2}h{k}_1\right)+\left(t+\frac{1}{2}h\right)=-4.85+0.05=-4.8\end{array}$

$\begin{array}{l}{k}_3=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_2,y_0+\frac{1}{2}h{k}_2\right)=\left(x+\frac{1}{2}h{m}_2\right)-\left(t+\frac{1}{2}h\right)=-3.24-0.05=-3.29\\ m_3=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_2,y_0+\frac{1}{2}h{k}_2\right)=-\left(y+\frac{1}{2}h{k}_2\right)+\left(t+\frac{1}{2}h\right)=-4.835+0.05=-4.785\end{array}$

$\begin{array}{l}{k}_4=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_3,y_0+\frac{1}{2}h{k}_3\right)=\left(x+\frac{1}{2}h{m}_3\right)-\left(t+\frac{1}{2}h\right)=-3.2393-0.05=-3.2893\\ m_4=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_3,y_0+\frac{1}{2}h{k}_3\right)=-\left(y+\frac{1}{2}h{k}_3\right)+\left(t+\frac{1}{2}h\right)=-4.8355+0.05=-4.7855\end{array}$

Compute the equations

Substitute the values of variables in the equation.

$\begin{array}{l}{y}_1=y_0+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ \Rightarrow y_1=5+\frac{0.1}{6}(-3+2(-3.3)+2(-3.29)+(-4.785\left)\right)\\ \therefore y_1=4.63\end{array}$

$\begin{array}{l}{x}_1=x_0+\frac{h}{6}\left(m_1+2m_2+2m_3+m_4\right)\\ \Rightarrow x_1=-3+\frac{0.1}{6}(-5+2(-4.8)+2(-4.785)+(-4.7855\left)\right)\\ \therefore x_1=-3.482\end{array}$

$\therefore y_1=4.63,x_1=-3.482$