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

localid="1668176373600" $\begin{array}{c}\mathbf{x}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{4}\mathbf{t}\\ \mathbf{-}\mathbf{x}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{6}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{10}\\ \mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{3}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}\end{array}$

Short answer

$\begin{array}{ccc}h& y_1& x_1\\ 0.2& 0.092705& 1.97396\\ 0.1& -0.4527& 2.4727\end{array}$

Step-by-step solution

Consider the equations and the conditions

The equations and the initial conditions are,

$\begin{array}{l}x\text{'}+y\text{'}=4t\\ -x\text{'}+y\text{'}+y=6t^2+10\\ x\left(0\right)=3,y\left(0\right)=-1\end{array}$

Add the equations.

$y\text{'}=2t+3t^2+5-\frac{y}{2}$

Subtract the equations.

$x\text{'}=2t-3t^2-5+\frac{y}{2}$

Compute the constants

Consider $h=0.2,t=0$

The values are,

$\begin{array}{l}{k}_1=f(t_0,x_0,y_0)=2t+3t^2+5-\frac{y}{2}=0+0+5-\frac{-1}{2}=5.5\\ m_1=g(t_0,x_0,y_0)=2t-3t^2-5+\frac{y}{2}=0-0-5+\frac{-1}{2}=-5.5\end{array}$

localid="1668176399555" $\begin{array}{c}{k}_2=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=2\left(t+\frac{1}{2}h\right)+3{\left(t+\frac{1}{2}h\right)}^2+5-\frac{1}{2}\left(y+\frac{1}{2}h{k}_1\right)\\ =2\left(0.1\right)+3\left(0.01\right)+5+0.225=5.455\\ m_2=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=2\left(t+\frac{1}{2}h\right)t-3{\left(t+\frac{1}{2}h\right)}^2-5+\frac{1}{2}\left(y+\frac{1}{2}h{k}_1\right)\\ =2\left(0.1\right)-3\left(0.01\right)-5-0.225=-5.055\end{array}$

localid="1668176389048" $\begin{array}{c}{k}_3=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=2\left(t+\frac{1}{2}h\right)+3{\left(t+\frac{1}{2}h\right)}^2+5-\frac{1}{2}\left(y+\frac{1}{2}h{k}_2\right)\\ =2\left(0.1\right)+3\left(0.01\right)+5+0.22725=5.457\\ m_3=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=2\left(t+\frac{1}{2}h\right)t-3{\left(t+\frac{1}{2}h\right)}^2-5+\frac{1}{2}\left(y+\frac{1}{2}h{k}_2\right)\\ =2\left(0.1\right)-3\left(0.01\right)-5-0.22725=-5.057\end{array}$

localid="1668176408424" $\begin{array}{c}{k}_4=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=2\left(t+\frac{1}{2}h\right)+3{\left(t+\frac{1}{2}h\right)}^2+5-\frac{1}{2}\left(y+\frac{1}{2}h{k}_3\right)\\ =2\left(0.1\right)+3\left(0.01\right)+5+0.22715=5.45715\\ m_4=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=2\left(t+\frac{1}{2}h\right)t-3{\left(t+\frac{1}{2}h\right)}^2-5+\frac{1}{2}\left(y+\frac{1}{2}h{k}_3\right)\\ =2\left(0.1\right)-3\left(0.01\right)-5-0.22715=-5.05715\end{array}$

Compute the equations

Substitute the values of variables in the equation.

$\begin{array}{l}{y}_1=y_0+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ \Rightarrow y_1=-1+\frac{0.2}{6}(5.5+2(5.455)+2(5.457)+5.45715)\\ \therefore y_1=0.0927\end{array}$

$\begin{array}{l}{x}_1=x_0+\frac{h}{6}\left(m_1+2m_2+2m_3+m_4\right)\\ \Rightarrow x_1=1+\frac{0.2}{6}(-5.5+2(-5.055)+2(-5.057)-5.05715)\\ \therefore x_1=1.97396\end{array}$

Compute the constants

Consider $h=0.1,t=0$

The values are,

$\begin{array}{l}{k}_1=f(t_0,x_0,y_0)=2t+3t^2+5-\frac{y}{2}=0+0+5-\frac{-1}{2}=5.5\\ m_1=g(t_0,x_0,y_0)=2t-3t^2-5+\frac{y}{2}=0-0-5+\frac{-1}{2}=-5.5\end{array}$

localid="1668176436143" $\begin{array}{c}{k}_2=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=2\left(t+\frac{1}{2}h\right)+3{\left(t+\frac{1}{2}h\right)}^2+5-\frac{1}{2}\left(y+\frac{1}{2}h{k}_1\right)\\ =2\left(0.05\right)+3\left(0.0025\right)+5+0.3625=5.47\\ m_2=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=2\left(t+\frac{1}{2}h\right)t-3{\left(t+\frac{1}{2}h\right)}^2-5+\frac{1}{2}\left(y+\frac{1}{2}h{k}_1\right)\\ =2\left(0.05\right)-3\left(0.0025\right)-5-0.3625=-5.27\end{array}$

localid="1668176446761" $\begin{array}{c}{k}_3=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=2\left(t+\frac{1}{2}h\right)+3{\left(t+\frac{1}{2}h\right)}^2+5-\frac{1}{2}\left(y+\frac{1}{2}h{k}_2\right)\\ =2\left(0.05\right)+3\left(0.0025\right)+5+0.36325=5.470\\ m_3=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=2\left(t+\frac{1}{2}h\right)t-3{\left(t+\frac{1}{2}h\right)}^2-5+\frac{1}{2}\left(y+\frac{1}{2}h{k}_2\right)\\ =2\left(0.05\right)-3\left(0.0025\right)-5-0.36325=-5.27\end{array}$

localid="1668176454482" $\begin{array}{c}{k}_4=f\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=2\left(t+\frac{1}{2}h\right)+3{\left(t+\frac{1}{2}h\right)}^2+5-\frac{1}{2}\left(y+\frac{1}{2}h{k}_3\right)\\ =2\left(0.05\right)+3\left(0.0025\right)+5+0.36325=5.457\\ m_4=g\left(t_0+\frac{1}{2}h,x_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)=2\left(t+\frac{1}{2}h\right)t-3{\left(t+\frac{1}{2}h\right)}^2-5+\frac{1}{2}\left(y+\frac{1}{2}h{k}_3\right)\\ =2\left(0.05\right)-3\left(0.0025\right)-5-0.22715=-5.057\end{array}$

Compute the equations

Substitute the values of variables in the equation.

$\begin{array}{l}{y}_1=y_0+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ \Rightarrow y_1=-1+\frac{0.1}{6}(5.5+2(5.47)+2(5.470)+5.457)\\ \therefore y_1=-0.4527\end{array}$

$\begin{array}{l}{x}_1=x_0+\frac{h}{6}\left(m_1+2m_2+2m_3+m_4\right)\\ \Rightarrow x_1=3+\frac{0.1}{6}(-5.5+2(-5.27)+2(-5.27)-5.057)\\ \therefore x_1=2.4727\end{array}$

$\therefore y_1=-0.4527,x_1=2.4727$