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 $\mathit{h}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}$to graph the solution in a neighborhood of $\mathit{t}\mathbf{=}\mathbf{0}$.

$\begin{array}{l}\mathbf{x}\mathbf{\text{'}}\mathbf{=}\mathbf{6}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{+}\mathbf{6}\mathbf{t}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{4}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{-}\mathbf{10}\mathbf{t}\mathbf{+}\mathbf{4}\\ \mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.5}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.2}\end{array}$

Short answer

$\begin{array}{ccc}h& y_1& x_1\\ 0.2& 2.3286& 2.1599\\ 0.1& 0.9582& 0.9582\end{array}$

Step-by-step solution

Consider the equations and the conditions

The equations and the initial conditions are,

$\begin{array}{l}x\text{'}=6x+y+6t\\ y\text{'}=4x+3y-10t+4\\ x\left(0\right)=0.5,y\left(0\right)=0.2\end{array}$

Compute the constants

Consider$h=0.2,t=0$

The values are,

$\begin{array}{l}{k}_1=f(t_0,x_0,y_0)=4x+3y-10t+4=2+0.6-0+4=6.6\\ m_1=g(t_0,x_0,y_0)=6x+y+6t=3+0.2+0=3.2\end{array}$

$\begin{array}{l}{k}_2=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=4\left(x+\frac{1}{2}h{m}_1\right)+3\left(y+\frac{1}{2}h{k}_1\right)-10\left(t+\frac{1}{2}h\right)+4\\ =3.28+2.58-1+4=8.86\\ m_2=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=6\left(x+\frac{1}{2}h{m}_1\right)+\left(y+\frac{1}{2}h{k}_1\right)+6\left(t+\frac{1}{2}h\right)\\ =4.92+0.86+0.6=6.38\end{array}$

$\begin{array}{l}{k}_3=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_2,y_0+\frac{1}{2}h{k}_2\right)=4\left(x+\frac{1}{2}h{m}_2\right)+3\left(y+\frac{1}{2}h{k}_2\right)-10\left(t+\frac{1}{2}h\right)+4\\ =4.552+3.258-1+4=10.81\\ m_3=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_2,y_0+\frac{1}{2}h{k}_2\right)=6\left(x+\frac{1}{2}h{m}_2\right)+\left(y+\frac{1}{2}h{k}_2\right)+6\left(t+\frac{1}{2}h\right)\\ =6.828+1.086+0.6=8.514\end{array}$

$\begin{array}{l}{k}_4=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_3,y_0+\frac{1}{2}h{k}_3\right)=4\left(x+\frac{1}{2}h{m}_3\right)+3\left(y+\frac{1}{2}hk\right)-10\left(t+\frac{1}{2}h\right)+4\\ =8.832+7.086-2+4=17.918\\ m_4=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_3,y_0+\frac{1}{2}h{k}_3\right)=6\left(x+\frac{1}{2}h{m}_3\right)+\left(y+\frac{1}{2}h{k}_3\right)+6\left(t+\frac{1}{2}h\right)\\ =13.248+2.362+1.2=16.81\end{array}$

Compute the equations

Substitute the values of variables in the equation.

$\begin{array}{l}{y}_1=y_0+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ \Rightarrow y_1=0.2+\frac{0.2}{6}(6.6+2(8.86)+2(10.81)+17.918)\\ \therefore y_1=2.3286\end{array}$

$\begin{array}{l}{x}_1=x_0+\frac{h}{6}\left(m_1+2m_2+2m_3+m_4\right)\\ \Rightarrow x_1=0.5+\frac{0.2}{6}(3.2+2(6.38)+2(8.514)+16.81)\\ \therefore x_1=2.1599\end{array}$

Compute the constants

Consider$h=0.1,t=0$

The values are,

$\begin{array}{l}{k}_1=f(t_0,x_0,y_0)=4x+3y-10t+4=2+0.6-0+4=6.6\\ m_1=g(t_0,x_0,y_0)=6x+y+6t=3+0.2+0=3.2\end{array}$

$\begin{array}{l}{k}_2=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=4\left(x+\frac{1}{2}h{m}_1\right)+3\left(y+\frac{1}{2}h{k}_1\right)-10\left(t+\frac{1}{2}h\right)+4\\ =2.64+0.99-0.5+4=7.13\\ m_2=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=6\left(x+\frac{1}{2}h{m}_1\right)+\left(y+\frac{1}{2}h{k}_1\right)+6\left(t+\frac{1}{2}h\right)\\ =3.96+0.53+0.3=4.79\end{array}$

$\begin{array}{l}{k}_3=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_2,y_0+\frac{1}{2}h{k}_2\right)=4\left(x+\frac{1}{2}h{m}_2\right)+3\left(y+\frac{1}{2}h{k}_2\right)-10\left(t+\frac{1}{2}h\right)+4\\ =2.958+1.6695-0.5+4=8.1275\\ m_3=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_2,y_0+\frac{1}{2}h{k}_2\right)=6\left(x+\frac{1}{2}h{m}_2\right)+\left(y+\frac{1}{2}h{k}_2\right)+6\left(t+\frac{1}{2}h\right)\\ =4.437+0.5565+0.3=5.2935\end{array}$

$\begin{array}{l}{k}_4=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_3,y_0+\frac{1}{2}h{k}_3\right)=4\left(x+\frac{1}{2}h{m}_3\right)+3\left(y+\frac{1}{2}h{k}_3\right)-10\left(t+\frac{1}{2}h\right)+4\\ =3.0587+1.819-0.5+4=8.3777\\ m_4=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_3,y_0+\frac{1}{2}h{k}_3\right)=6\left(x+\frac{1}{2}h{m}_3\right)+\left(y+\frac{1}{2}h{k}_3\right)+6\left(t+\frac{1}{2}h\right)\\ =4.588+0.606+0.3=5.494\end{array}$

Compute the equations

Substitute the values of variables in the equation.

$\begin{array}{l}{y}_1=y_0+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ \Rightarrow y_1=0.2+\frac{0.1}{6}(6.6+2(7.13)+2(8.1275)+8.3777)\\ \therefore y_1=0.9582\end{array}$

$\begin{array}{l}{x}_1=x_0+\frac{h}{6}\left(m_1+2m_2+2m_3+m_4\right)\\ \Rightarrow x_1=0.5+\frac{0.1}{6}(3.2+2(4.79)+2(5.2935)+5.494)\\ \therefore x_1=0.981\end{array}$

$\therefore y_1=0.9582,x_1=0.981$