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

$\begin{array}{l}\mathbf{x}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{7}\mathbf{t}\\ \mathbf{x}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{3}\mathbf{t}\\ \mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}\end{array}$

Short answer

$\begin{array}{ccc}h& y_1& x_1\\ 0.2& -2.078& 0.3248\\ 0.1& 0.1585& 0.154\end{array}$

Step-by-step solution

Consider the equations and the conditions

The equations and the initial conditions are,

$\begin{array}{l}x\text{'}+4x-y\text{'}=7t\\ x\text{'}+y\text{'}-2y=3t\\ x\left(0\right)=1,y\left(0\right)=-2\end{array}$

Add the equations.

$x\text{'}=5t-2x+y$

Subtract the equations.

$y\text{'}=2x-2t+y$

Compute the constants

Consider $h=0.2,t=0$

The values are,

$\begin{array}{l}{k}_1=f(t_0,x_0,y_0)=2x-2t+y=2-0-2=0\\ m_1=g(t_0,x_0,y_0)=5t-2x+y=0-2-2=-4\end{array}$

$\begin{array}{l}{k}_2=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=2\left(x+\frac{1}{2}h{m}_1\right)-2\left(t+\frac{1}{2}h\right)+\left(y+\frac{1}{2}h{k}_1\right)\\ =2\left(1.2\right)-2\left(0.1\right)-2=0.2\\ m_2=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=5\left(t+\frac{1}{2}h\right)-2\left(x+\frac{1}{2}h{m}_1\right)+\left(y+\frac{1}{2}h{k}_1\right)\\ =5\left(0.1\right)-2\left(1.2\right)-2=-3.9\end{array}$

$\begin{array}{l}{k}_3=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=2\left(x+\frac{1}{2}h{m}_2\right)-2\left(t+\frac{1}{2}h\right)+\left(y+\frac{1}{2}h{k}_2\right)\\ =2\left(0.61\right)-2\left(0.1\right)-1.98=-0.96\\ m_3=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=5\left(t+\frac{1}{2}h\right)-2\left(x+\frac{1}{2}h{m}_2\right)+\left(y+\frac{1}{2}h{k}_2\right)\\ =5\left(0.1\right)-2\left(0.61\right)-1.98=-2.7\end{array}$

$\begin{array}{l}{k}_4=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=2\left(x+\frac{1}{2}h{m}_3\right)-2\left(t+\frac{1}{2}h\right)+\left(y+\frac{1}{2}h{k}_3\right)\\ =2\left(0.73\right)-2\left(0.1\right)-2.096=-0.836\\ m_4=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=5\left(t+\frac{1}{2}h\right)-2\left(x+\frac{1}{2}h{m}_3\right)+\left(y+\frac{1}{2}h{k}_3\right)\\ =5\left(0.1\right)-2\left(0.73\right)-2.096=-3.056\end{array}$

Compute the equations

Substitute the values of variables in the equation.

$\begin{array}{l}{y}_1=y_0+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ \Rightarrow y_1=-2+\frac{0.2}{6}(0+2(0.2)+2(-0.96)-0.836)\\ \therefore y_1=-2.078\end{array}$

$\begin{array}{l}{x}_1=x_0+\frac{h}{6}\left(m_1+2m_2+2m_3+m_4\right)\\ \Rightarrow x_1=1+\frac{0.2}{6}(-4+2(-3.9)+2(-2.7)-3.056)\\ \therefore x_1=0.3248\end{array}$

Compute the constants

Consider $h=0.1,t=0$

The values are,

$\begin{array}{l}{k}_1=f(t_0,x_0,y_0)=2x-2t+y=2-0-2=0\\ m_1=g(t_0,x_0,y_0)=5t-2x+y=0-2-2=-4\end{array}$

$\begin{array}{l}{k}_2=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=2\left(x+\frac{1}{2}h{m}_1\right)-2\left(t+\frac{1}{2}h\right)+\left(y+\frac{1}{2}h{k}_1\right)\\ =2\left(0.8\right)-2\left(0.05\right)-2=-0.5\\ m_2=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=5\left(t+\frac{1}{2}h\right)-2\left(x+\frac{1}{2}h{m}_1\right)+\left(y+\frac{1}{2}h{k}_1\right)\\ =5\left(0.05\right)-2\left(0.8\right)-2=-3.35\end{array}$

$\begin{array}{l}{k}_3=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=2\left(x+\frac{1}{2}h{m}_2\right)-2\left(t+\frac{1}{2}h\right)+\left(y+\frac{1}{2}h{k}_2\right)\\ =2\left(0.8325\right)-2\left(0.05\right)-2.025=-0.46\\ m_3=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=5\left(t+\frac{1}{2}h\right)-2\left(x+\frac{1}{2}h{m}_2\right)+\left(y+\frac{1}{2}h{k}_2\right)\\ =5\left(0.05\right)-2\left(0.8325\right)-2.025=-3.44\end{array}$

$\begin{array}{l}{k}_4=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=2\left(x+\frac{1}{2}h{m}_3\right)-2\left(t+\frac{1}{2}h\right)+\left(y+\frac{1}{2}h{k}_3\right)\\ =2\left(0.828\right)-2\left(0.1\right)-2.023=-0.567\\ m_4=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=5\left(t+\frac{1}{2}h\right)-2\left(x+\frac{1}{2}h{m}_3\right)+\left(y+\frac{1}{2}h{k}_3\right)\\ =5\left(0.1\right)-2\left(0.828\right)-2.023=-3.179\end{array}$

Compute the equations

Substitute the values of variables in the equation.

$\begin{array}{l}{y}_1=y_0+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ \Rightarrow y_1=0.2+\frac{0.1}{6}(0+2(-0.5)+2(-0.46)-0.567)\\ \therefore y_1=0.1585\end{array}$

$\begin{array}{l}{x}_1=x_0+\frac{h}{6}\left(m_1+2m_2+2m_3+m_4\right)\\ \Rightarrow x_1=0.5+\frac{0.1}{6}(-4+2(-3.35)+2(-3.44)-3.179)\\ \therefore x_1=0.154\end{array}$

$\therefore y_1=0.1585,x_1=0.154$