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 $\mathit{t}\mathbf{=}\mathbf{0}$.

$\begin{array}{l}\mathbf{x}\mathbf{\text{'}}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{4}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{y}\\ \mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\end{array}$

Short answer

$\begin{array}{ccc}h& y_1& x_1\\ 0.2& 3.3382& 2.0785\\ 0.1& 1.8963& 1.4006\end{array}$

Step-by-step solution

Consider the equations and the conditions

The equations and the initial conditions are,

$\begin{array}{l}x\text{'}=x+2y\\ y\text{'}=4x+3y\\ x\left(0\right)=1,y\left(0\right)=1\end{array}$

Compute the constants

Consider$h=0.2,t=0$

The values are,

$\begin{array}{l}{k}_1=f(t_0,x_0,y_0)=4x+3y=4+3=7\\ m_1=g(t_0,x_0,y_0)=x+2y=1+2=3\end{array}$

$\begin{array}{l}{k}_3=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_2,y_0+\frac{1}{2}h{k}_2\right)=4\left(x+\frac{1}{2}h{m}_2\right)+3\left(y+\frac{1}{2}h{k}_2\right)=5.88+6.09=11.97\\ m_3=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_2,y_0+\frac{1}{2}h{k}_2\right)=\left(x+\frac{1}{2}h{m}_2\right)+2\left(y+\frac{1}{2}h{k}_2\right)=1.47+4.06=5.53\end{array}$

$\begin{array}{l}{k}_4=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_3,y_0+\frac{1}{2}h{k}_3\right)=4\left(x+\frac{1}{2}h{m}_3\right)+3\left(y+\frac{1}{2}h{k}_3\right)=8.424+10.182=18.606\\ m_4=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_3,y_0+\frac{1}{2}h{k}_3\right)=\left(x+\frac{1}{2}h{m}_3\right)+2\left(y+\frac{1}{2}h{k}_3\right)=2.106+6.788=8.894\end{array}$

Compute the equations

Substitute the values of variables in the equation.

$\begin{array}{l}{y}_1=y_0+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ \Rightarrow y_1=5+\frac{0.2}{6}(7+2(10.3)+2(11.97)+18.606)\\ \therefore y_1=3.3382\end{array}$

$\begin{array}{l}{x}_1=x_0+\frac{h}{6}\left(m_1+2m_2+2m_3+m_4\right)\\ \Rightarrow x_1=-3+\frac{0.2}{6}(3+2(4.7)+2(5.53)+(8.894\left)\right)\\ \therefore x_1=2.0785\end{array}$

$\therefore y_1=4.2873,x_1=-3.9097$

Compute the constants

Consider$h=0.1,t=0$

The values are,

$\begin{array}{l}{k}_1=f(t_0,x_0,y_0)=4x+3y=4+3=7\\ m_1=g(t_0,x_0,y_0)=x+2y=1+2=3\end{array}$

$\begin{array}{l}{k}_2=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=4\left(x+\frac{1}{2}h{m}_1\right)+3\left(y+\frac{1}{2}h{k}_1\right)=4.6+3.05=8.65\\ m_2=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=\left(x+\frac{1}{2}h{m}_1\right)+2\left(y+\frac{1}{2}h{k}_1\right)=1.15+2.7=3.85\end{array}$

$\begin{array}{l}{k}_3=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_2,y_0+\frac{1}{2}h{k}_2\right)=4\left(x+\frac{1}{2}h{m}_2\right)+3\left(y+\frac{1}{2}h{k}_2\right)=4.77+4.2975=9.0675\\ m_3=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_2,y_0+\frac{1}{2}h{k}_2\right)=\left(x+\frac{1}{2}h{m}_2\right)+2\left(y+\frac{1}{2}h{k}_2\right)=1.1925+2.865=4.0575\end{array}$

$\begin{array}{l}{k}_4=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_3,y_0+\frac{1}{2}h{k}_3\right)=4\left(x+\frac{1}{2}h{m}_3\right)+3\left(y+\frac{1}{2}h{k}_3\right)=5.623+5.72025=11.3433\\ m_4=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_3,y_0+\frac{1}{2}h{k}_3\right)=\left(x+\frac{1}{2}h{m}_3\right)+2\left(y+\frac{1}{2}h{k}_3\right)=1.40575+3.8135=5.2193\end{array}$

Compute the equations

Substitute the values of variables in the equation.

$\begin{array}{l}{y}_1=y_0+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ \Rightarrow y_1=5+\frac{0.1}{6}(7+2(8.65)+2(9.0675)+11.3433)\\ \therefore y_1=1.8963\end{array}$

$\begin{array}{l}{x}_1=x_0+\frac{h}{6}\left(m_1+2m_2+2m_3+m_4\right)\\ \Rightarrow x_1=-3+\frac{0.1}{6}(3+2(3.85)+2(4.0575)+5.2193)\\ \therefore x_1=1.4006\end{array}$

$\therefore y_1=1.8963,x_1=1.4006$