Question

Question: In Problems 3-12 use the RK4 method withto obtain a four decimal approximation of the indicated value.

$\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{x}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{y}\left(0.5\right)$

Short answer

Therefore, the four decimal approximation of the indicated value $y\left(0.5\right)$using the RK4 method is 0.1266

Step-by-step solution

Runga-Kutta method

The general formulas used in the given problem are,

$$\begin{gathered} k_1=f\left(x_n,y_n\right) \\ {k}_2=f\left(\left(x_n+\frac{h}{2}\right),\left(y_n+\frac{h{k}_1}{2}\right)\right) \\ {k}_3=f\left(\left(x_n+\frac{h}{2}\right),\left(y_n+\frac{h{k}_2}{2}\right)\right) \\ {k}_4=f\left(\left(x_n+h\right),\left(y_n+h{k}_3\right)\right) \end{gathered}$$

And,

$y_{n+1}=y_n+\frac{h}{6}\left(k_1+k_2+k_3+k_4\right)$

Determine the First order Runga-Kutta method

The given differential equation can be written as,

$\begin{array}{l}y\text{'}=F\left(x,y\right)\\ F\left(x,y\right)=x+y^2\end{array}$

The given boundary conditions can be written as,

$\begin{array}{l}{y}_0=f\left(x_0\right)\\ x_0=0\\ y_0=0\\ h=0.1\end{array}$

For,

Substitute the value of $x_0,y_0$and $h$in the following equations to get the values of $k_1,k_2,k_3,k_4$,

The value of $k_1$is calculated as,

$\begin{array}{l}{k}_1=hF\left(x_0,y_0\right)\\ =0.1\left(\left(0\right)+\left(0^2\right)\right)\\ =0.1\times \left(0\right)=0\end{array}$

The value of $k_2$is calculated as,

$$\begin{gathered} k_2=hF(x_0+\frac{h}{2},y_0+k_1) \\ \begin{array}{l}=0.1\times F\left(0+\frac{0.1}{2},0+\frac{0}{2}\right)\\ =0.1\times F\left(0.05,0\right)\\ =0.005\end{array} \end{gathered}$$

The value of $k_3$is calculated as,

$$\begin{gathered} k_3=hF(x_0+\frac{h}{2},y_0+\frac{k_2}{2}) \\ \begin{array}{l}=0.1\times F\left(0+\frac{0.1}{2},0+\frac{0.005}{2}\right)\\ =0.1\times F\left(0.05,0.0025\right)\\ =0.0050\end{array} \end{gathered}$$

The value of $k_4$is calculated as,

$\begin{array}{l}{k}_4=hF\left(x_0+h,y_0+k_3\right)\\ =0.1\times F\left(0+0.1,0+0.0050\right)\\ =0.1\times F\left(0.1,0.0050\right)\\ =0.0100\end{array}$

Now substitute the value of $k_1,k_2,k_3,k_4$ in $y_1$.

$\begin{array}{rcl}{y}_1& =& y_0+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 0+\frac{1}{6}\left(\left(0\right)+2\left(0.005\right)+2\left(0.0050\right)+\left(0.0100\right)\right)\\ & =& 0.0050\end{array}$

For $n=1$, the value of $x_1$is calculated as,

$x_1=x_0+h=0+0.1=0.1$

Substitute the value of $x_1{y}_1$and $h$in the following equations to get the values of $k_1,k_2,k_3,k_4$

The value of $k_1$is calculated as,

localid="1668412846573" $\begin{array}{l}{k}_1=hF\left(x_1,y_1\right)\\ =0.1\left(\left(0.1\right)+{\left(0.005\right)}^2\right)\\ =0.1\times \left(0.1000\right)\\ =0.0100\end{array}$

The value of $k_2$is calculated as,

localid="1668412877343" $$\begin{gathered} k_2=hF(x_1+\frac{h}{2},y_1+\frac{k_1}{2}) \\ \begin{array}{l}=0.1\times F\left(0.1+\frac{0.1}{2},0.005+\frac{0.0100}{2}\right)\\ =0.1\times F\left(0.15,0.0100\right)\\ =0.0150\end{array} \end{gathered}$$

The value of $k_3$is calculated as,

localid="1668412859255" $$\begin{gathered} k_3=hF(x_1+\frac{h}{2},y_1+\frac{k_2}{2}) \\ \begin{array}{l}=0.1\times F\left(0.1+\frac{0.1}{2},0.005+\frac{0.0150}{2}\right)\\ =0.1\times F\left(0.15,0.01252\right)\\ =0.01502\end{array} \end{gathered}$$

The value of $k_4$is calculated as,

$\begin{array}{l}{k}_4=hF\left(x_1+h,y_1+k_3\right)\\ =0.1\times F\left(0.1+0.1,0.005+0.01502\right)\\ =0.1\times F\left(0.2,0.0201\right)\\ =0.0201\end{array}$

Now substitute the value of $k_1,k_2,k_3,k_4$in $y_2$.

$\begin{array}{rcl}{y}_2& =& y_1+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 0.005+\frac{1}{6}\left(\left(0.01502\right)+2\left(0.01502\right)+2\left(0.001502\right)+\left(0.0201\right)\right)\\ & =& 0.020\end{array}$

Compute the second order Runga-Kutta method

For$n=2$, the value of$x_2$is calculated as,

$\begin{array}{l}{x}_2=x_1+h\\ =0.1+0.1\\ =0.2\end{array}$

Substitute the value of$x_2,y_2$and$h$in the following equations to get the values of$k_1,k_2,k_3,k_4$,

The value ofis calculated as

$\begin{array}{l}{k}_1=hF\left(x_2,y_2\right)\\ =0.1\left(\left(0.2\right)+{\left(0.020\right)}^2\right)\\ =0.1\times \left(0.1001\right)\\ =0.0201\end{array}$

The value ofis calculated as,

$$\begin{gathered} k_2=hF(x_2+\frac{h}{2},y_2+\frac{k_1}{2}) \\ \begin{array}{l}=0.1\times F\left(0.2+\frac{0.1}{2},0.020+\frac{0.0201}{2}\right)\\ =0.1\times F\left(0.25,0.03002\right)\\ =0.0251\end{array} \end{gathered}$$

The value ofis calculated as,

$$\begin{gathered} k_3=hF(x_2+\frac{h}{2},y_2+\frac{k_2}{2}) \\ \begin{array}{l}=0.1\times F\left(0.2+\frac{0.1}{2},0.020+\frac{0.0251}{2}\right)\\ =0.1\times F\left(0.25,0.1455\right)\\ =0.02712\end{array} \end{gathered}$$

The value ofis calculated as,

$\begin{array}{l}{k}_4=hF\left(x_2+h,y_2+k_3\right)\\ =0.1\times F\left(0.2+0.1,0.020+0.02712\right)\\ =0.1\times F\left(0.3,0.04712\right)\\ =0.03023\end{array}$

Now substitute the value of$k_1,k_2,k_3,k_4$andin$y_3$.

$\begin{array}{rcl}{y}_2& =& y_1+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 0.020+\frac{1}{6}\left(\left(0.0201\right)+2\left(0.0251\right)+2\left(0.02712\right)+\left(0.03023\right)\right)\\ & =& 0.045\end{array}$

Compute the third order of Runga-Kutta method

For$n=3$, the value of$X_3$is calculated as,

$\begin{array}{l}{x}_3=x_2+h\\ =0.2+0.1\\ =0.3\end{array}$

Substitute the value of$x_3,y_3$and$h$in the following equations to get the values of$k_1,k_2,k_3,k_4$

The value ofis calculated as,

role="math" localid="1668420211863" $\begin{array}{l}{k}_1=hF\left(x_3,y_3\right)\\ =0.1\left(\left(0.3\right)+{\left(0.0451\right)}^2\right)\\ =0.1\times \left(0.1001\right)\\ =0.0302\end{array}$

The value ofis calculated as,

role="math" localid="1668420199805" $$\begin{gathered} k_2=hF(x_3+\frac{h}{2},y_3+\frac{k_1}{2}) \\ \begin{array}{l}=0.1\times F\left(0.3+\frac{0.1}{2},0.0451+\frac{0.0302}{2}\right)\\ =0.1\times F\left(0.35,0.0602\right)\\ =0.0354\end{array} \end{gathered}$$

The value ofis calculated as,

role="math" localid="1668420229196" $$\begin{gathered} k_3=hF(x_3+\frac{h}{2},y_3+\frac{k_2}{2}) \\ \begin{array}{l}=0.1\times F\left(0.3+\frac{0.1}{2},0.0451+\frac{0.0354}{2}\right)\\ =0.1\times F\left(0.35,0.0628\right)\\ =0.0354\end{array} \end{gathered}$$

The value ofis calculated as,

role="math" localid="1668420243644" $\begin{array}{l}{k}_4=hF\left(x_3+h,y_3+k_3\right)\\ =0.1\times F\left(0.3+0.1,0.0451+0.0354\right)\\ =0.1\times F\left(0.4,0.0805\right)\\ =0.04065\end{array}$

Now substitute the value of $k_1,k_2,k_3,k_4$in $y_4$.

role="math" localid="1668420352363" $\begin{array}{rcl}{y}_4& =& y_3+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 0.0451+\frac{1}{6}\left(\left(0.0302\right)+2\left(0.0354\right)+2\left(0.0354\right)+\left(0.04065\right)\right)\\ & =& 0.08050\end{array}$

Compute the fourth order of Runga-Kutta method

For $n=4$,the value of $x_4$is calculated as,

$\begin{array}{l}{x}_4=x_3+h\\ =0.3+0.1\\ =0.4\end{array}$

Substitute the value of $x_4,y_4$and $h$in the following equations to get the values of $k_1,k_2,k_3,k_4$,

The value of $k_1$is calculated as,

$\begin{array}{l}{k}_1=hF\left(x_3,y_3\right)\\ =0.1\left(\left(0.4\right)+{\left(0.08050\right)}^2\right)\\ =0.1\times \left(0.4064\right)\\ =0.04065\end{array}$

The value of$k_2$is calculated as,

$$\begin{gathered} k_2=hF(x_4+\frac{h}{2},y_4+\frac{k_1}{2}) \\ \begin{array}{l}=0.1\times F\left(0.4+\frac{0.1}{2},0.0805+\frac{0.04065}{2}\right)\\ =0.1\times F\left(0.35,0.10083\right)\\ =0.04602\end{array} \end{gathered}$$

The value of$k_3$is calculated as,

$$\begin{gathered} k_3=hF(x_4+\frac{h}{2},y_4+\frac{k_2}{2}) \\ \begin{array}{l}=0.1\times F\left(0.4+\frac{0.1}{2},0.0805+\frac{0.04602}{2}\right)\\ =0.1\times F\left(0.35,0.10351\right)\\ =0.0461\end{array} \end{gathered}$$

The value of$k_4$is calculated as,

$\begin{array}{l}{k}_4=hF\left(x_3+h,y_3+k_3\right)\\ =0.1\times F\left(0.4+0.1,0.0805+0.0461\right)\\ =0.1\times F\left(0.5,0.1266\right)\\ =0.052\end{array}$

Now substitute the value of $k_1,k_2,k_3,k_4$in $y_5$.

$\begin{array}{rcl}{y}_5& =& y_4+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 0.0805+\frac{1}{6}\left(\left(0.04605\right)+2\left(0.04602\right)+2\left(0.0461\right)+\left(0.052\right)\right)\\ & =& 0.1266\end{array}$

For$n=5$the value of$x_5$is calculated as,

$\begin{array}{l}{x}_5=x_4+h\\ =0.4+0.1\\ =0.5\end{array}$

Therefore, the four decimal approximation of the indicated value $y\left(0.5\right)$using the RK4 method is 0.1266.