Question

Question: In Problems 3-12 use the RK4 method with$\mathit{h}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}$to obtain a four decimal approximation of the indicated value.

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

Short answer

Therefore, the four decimal approximation of the indicated value $y\left(0.5\right)$using the RK4 method is 0.5493

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,

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

The given boundary conditions can be written as,

$\begin{array}{l}{y}_0=\left(x_0\right)\\ x_0=0\\ y_0=0.5\\ h=0.1\end{array}$

For $n=0$, 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,

$\begin{array}{l}{k}_1=hF(x_0,y_0)\\ =0.1\times {\left(0-0.5\right)}^2\\ =0.1\times \left(0.25\right)\\ =0.025\end{array}$

The value of $k_2$is calculated as,

$$\begin{gathered} k_2=hF(x_0+\frac{h}{2},y_0+k_1) \\ \begin{array}{l}=0.1\times F\left(0+\frac{0.1}{2},0.5+\frac{0.025}{2}\right)\\ =0.1\times F\left(0.05,0.5125\right)\\ =0.0212\end{array} \end{gathered}$$

The value of $k_3$is calculated as,

$$\begin{gathered} k_3=hF(x_0+\frac{h}{2},y_0+\frac{k_2}{2}) \\ \begin{array}{l}=0.1\times F\left(0+\frac{0.1}{2},0.5+\frac{0.0213}{2}\right)\\ =0.1\times F\left(0.05,0.5106\right)\\ =0.0212\end{array} \end{gathered}$$

The value of $k_4$is calculated as,

$\begin{array}{l}{k}_4=hF\left(x_0+h,y_0+k_3\right)\\ =0.1\times F\left(0+0.1,0.5+0.0212\right)\\ =0.1\times F\left(0.1,0.5212\right)\\ =0.0177\end{array}$

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

$\begin{array}{rcl}{y}_1& =& y_0+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 0.5+\frac{1}{6}\left(\left(0.025\right)+2\left(0.0213\right)+2\left(0.0212\right)+\left(0.0177\right)\right)\\ & =& 0.5210\end{array}$

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,

$\begin{array}{l}{k}_1=hF(x_1,y_1)\\ =0.1\times {\left(0.1-0.5210\right)}^2\\ =0.1\times \left(0.1775\right)\\ =0.0177\end{array}$

The value of $k_2$is calculated as,

role="math" localid="1668428323941" $$\begin{gathered} k_2=hF(x_1+\frac{h}{2},y_1+\frac{k_1}{2}) \\ \begin{array}{l}=0.1\times F\left(0.1+\frac{0.1}{2},0.5210+\frac{0.0177}{2}\right)\\ =0.1\times F\left(0.15,0.5299\right)\\ =0.0144\end{array} \end{gathered}$$

The value of $k_3$is calculated as,

$$\begin{gathered} k_3=hF(x_1+\frac{h}{2},y_1+\frac{k_2}{2}) \\ \begin{array}{l}=0.1\times F\left(0.1+\frac{0.1}{2},0.5210+\frac{0.0144}{2}\right)\\ =0.1\times F\left(0.15,0.5282\right)\\ =0.0143\end{array} \end{gathered}$$

The value of $k_4$is calculated as,

$\begin{array}{l}{k}_4=hF\left(x_1+h,y_1+k_3\right)\\ =0.1\times F\left(0.1+0.1,0.5210+0.0143\right)\\ =0.1\times F\left(0.2,0.5353\right)\\ =0.0112\end{array}$

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

$\begin{array}{rcl}{y}_2& =& y_1+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 0.5210+\frac{1}{6}\left(\left(0.0177\right)+2\left(0.0144\right)+2\left(0.0143\right)+\left(0.0200\right)\right)\\ & =& 0.5368\end{array}$

Compute the second order Runga-Kutta method

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

$x_2=x_1+h=0.1+0.1=0.2$

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 of $k_1$is calculated as,

$\begin{array}{l}{k}_1=hF(x_2,y_2)\\ =0.1\times {\left(0.2-0.5368\right)}^2\\ =0.1\times \left(0.1135\right)\\ =0.0113\end{array}$

The value of ${\mathrm{k}}_2$is calculated as

$$\begin{gathered} k_2=hF(x_2+\frac{h}{2},y_2+\frac{k_1}{2}) \\ \begin{array}{l}=0.1\times F\left(0.2+\frac{0.1}{2},0.5368+\frac{0.0113}{2}\right)\\ =0.1\times F\left(0.25,0.5415\right)\\ =0.0085\end{array} \end{gathered}$$

The value of $k_3$is calculated as,

$$\begin{gathered} k_3=hF(x_2+\frac{h}{2},y_2+\frac{k_2}{2}) \\ \begin{array}{l}=0.1\times F\left(0.2+\frac{0.1}{2},0.5368+\frac{0.0085}{2}\right)\\ =0.1\times F\left(0.25,0.5410\right)\\ =0.0085\end{array} \end{gathered}$$

The value of $k_4$is calculated as,

$\begin{array}{l}{k}_4=hF\left(x_2+h,y_2+k_3\right)\\ =0.1\times F\left(0.2+0.1,0.5368+0.0085\right)\\ =0.1\times F\left(0.3,0.5453\right)\\ =0.0060\end{array}$

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

$\begin{array}{rcl}{y}_3& =& y_2+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 0.5368+\frac{1}{6}\left(\left(0.0113\right)+2\left(0.0085\right)+2\left(0.0085\right)+\left(0.0060\right)\right)\\ & =& 0.5443\end{array}$

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 of $k_1$is calculated as,

$\begin{array}{l}{k}_1=hF(x_3,y_3)\\ =0.1\times {\left(0.3-0.5443\right)}^2\\ =0.1\times \left(0.0597\right)\\ =0.0059\end{array}$

The value of $k_2$is calculated as,

$$\begin{gathered} k_2=hF(x_3+\frac{h}{2},y_3+\frac{k_1}{2}) \\ \begin{array}{l}=0.1\times F\left(0.3+\frac{0.1}{2},0.5443+\frac{0.0059}{2}\right)\\ =0.1\times F\left(0.35,0.5473\right)\\ =0.0039\end{array} \end{gathered}$$

The value of $k_3$is calculated as,

role="math" localid="1668430349021" $$\begin{gathered} k_3=hF(x_3+\frac{h}{2},y_3+\frac{k_2}{2}) \\ \begin{array}{l}=0.1\times F\left(0.3+\frac{0.1}{2},0.5443+\frac{0.0039}{2}\right)\\ =0.1\times F\left(0.35,0.5473\right)\\ =0.0039\end{array} \end{gathered}$$

The value of $k_4$is calculated as,

$\begin{array}{l}{k}_4=hF\left(x_3+h,y_3+k_3\right)\\ =0.1\times F\left(0.3+0.1,0.5368+0.0039\right)\\ =0.1\times F\left(0.4,0.5481\right)\\ =0.0022\end{array}$

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

$\begin{array}{rcl}{y}_4& =& y_3+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 0.5443+\frac{1}{6}\left(\left(0.0059\right)+2\left(0.0039\right)+2\left(0.0039\right)+\left(0.0022\right)\right)\\ & =& 0.5482\end{array}$

Compute the fourth order of Runga-Kutta method

For $n=4$, the value of $x_4$s 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,

$\begin{array}{l}{k}_1=hF(x_4,y_4)\\ =0.1\times {\left(0.4-0.5482\right)}^2\\ =0.1\times \left(0.0219\right)\\ =0.0022\end{array}$

The value of $k_2$is calculated as,

$$\begin{gathered} k_2=hF(x_4+\frac{h}{2},y_4+\frac{k_1}{2}) \\ \begin{array}{l}=0.1\times F\left(0.4+\frac{0.1}{2},0.5482+\frac{0.0022}{2}\right)\\ =0.1\times F\left(0.45,0.5493\right)\\ =0.00099\end{array} \end{gathered}$$

The value of $k_3$is calculated as,

$$\begin{gathered} k_3=hF(x_4+\frac{h}{2},y_4+\frac{k_2}{2}) \\ \begin{array}{l}=0.1\times F\left(0.4+\frac{0.1}{2},0.5482+\frac{0.00099}{2}\right)\\ =0.1\times F\left(0.45,0.5487\right)\\ =0.00099\end{array} \end{gathered}$$

The value of $k_4$is calculated as,

$\begin{array}{l}{k}_4=hF\left(x_3+h,y_3+k_3\right)\\ =0.1\times F\left(0.4+0.1,0.5482+0.00099\right)\\ =0.1\times F\left(0.5,0.5492\right)\\ =0.00024\end{array}$

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

$\begin{array}{rcl}{y}_5& =& y_4+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 0.5482+\frac{1}{6}\left(\left(0.3037\right)+2\left(0.00099\right)+2\left(0.00099\right)+\left(0.00024\right)\right)\\ & =& 0.5493\end{array}$

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 using the RK4 method is 0.5493