Question

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

$\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{;}\mathbf{}\mathbf{y}\left(0\right)\mathbf{=}\mathbf{2}\mathbf{;}\mathbf{}\mathbf{}\mathbf{}\mathbf{y}\left(0.5\right)$

Short answer

The four decimal approximation of the indicated value $y\left(0.5\right)$using the RK4 method is 1.10307

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,

$\begin{array}{l}y\text{'}=F\left(x,y\right)\\ F\left(x,y\right)=4x-2y\end{array}$

The given boundary conditions can be written as,

$\begin{array}{l}{y}_0=f\left(x_0\right)\\ x_0=0\\ y_0=2\\ h=0.1\end{array}$

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,

role="math" localid="1668145911604" $\begin{array}{l}{k}_1=hF(x_0,y_0)\\ =0.1\times \left(4\times 0-2\times 2\right)\\ =0.1\times \left(-4\right)\\ =-0.4\end{array}$

The value of is calculated as,

role="math" localid="1668146080075" $$\begin{gathered} k_2=hF(x_0+\frac{h}{2},y_0+k_1) \\ \begin{array}{l}=0.1\times F\left(0+\frac{0.1}{2},2+\frac{-0.4}{2}\right)\\ =0.1\times F\left(1.05,1.8\right)\\ =-0.34\end{array} \end{gathered}$$

The value of $k_3$is calculated as,

$\begin{array}{l}{k}_3=hF\left(x_0+\frac{h}{2},y_0+\frac{k_2}{2}\right)\\ =0.1\times F\left(0+\frac{0.1}{2},2+\frac{-0.34}{2}\right)\\ =0.1\times F\left(1.05,1.83\right)\\ =-0.346\end{array}$

The value of$k_4$ is calculated as,

$\begin{array}{l}{k}_4=hF\left(x_0+h,y_0+k_3\right)\\ =0.1\times F\left(0+0.1,2+\left(-0.346\right)\right)\\ =0.1\times F\left(0.1,1.654\right)\\ =-0.2908\end{array}$

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

role="math" localid="1668147444894" $\begin{array}{rcl}{y}_2& =& y_1+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 1.6562+\frac{1}{6}\left(\left(-0.29124\right)+2\left(-0.242116\right)+2\left(-0.2470284\right)+\left(-2.01814\right)\right)\\ & =& 1.4110\end{array}$

Compute the second order Runga-Kutta method

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,

$\begin{array}{l}{k}_1=hF(x_0,y_0)\\ =0.1\times \left(4\times 0.1-2\times 1.6652\right)\\ =0.1\times \left(-2.0932\right)\\ =-0.29124\end{array}$

The value of $k_2$is calculated as,

$$\begin{gathered} k_2=hF(x_0+\frac{h}{2},y_0+k_1) \\ \begin{array}{l}=0.1\times F\left(0.1+\frac{0.1}{2},1.6562+\frac{-0.29124}{2}\right)\\ =0.1\times F\left(0.15,1.51058\right)\\ =-0.242116\end{array} \end{gathered}$$

The value of $k_3$is calculated as,

$\begin{array}{l}{k}_3=hF\left(x_0+\frac{h}{2},y_0+\frac{k_2}{2}\right)\\ =0.1\times F\left(0+\frac{0.1}{2},1.6562+\frac{-0.242116}{2}\right)\\ =0.1\times F\left(0.15,1.535142\right)\\ =-0.2470284\end{array}$

The value of is calculated as,

role="math" localid="1668147872178" $\begin{array}{l}{k}_4=hF\left(x_0+h,y_0+k_3\right)\\ =0.1\times F\left(0.1+0.1,1.6561+\left(-0.2470\right)\right)\\ =0.1\times F\left(0.2,1.4090\right)\\ =-0.2018\end{array}$

Now substitute the value $k_1,k_2,k_3$of $k_4$and in $\begin{array}{rcl}& & y_2\end{array}$.

role="math" localid="1668148063104" $\begin{array}{rcl}{y}_2& =& y_1+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 1.6562+\frac{1}{6}\left(\left(-0.29124\right)+2\left(-0.242116\right)+2\left(-0.2470284\right)+\left(-2.01814\right)\right)\\ & =& 1.4110\end{array}$

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

role="math" localid="1668150680236" $x_2=x_1+h=0.1+0.1=0.2$

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

$\begin{array}{c}{k}_1=hF(x_0,y_0)\\ =0.1\times \left(4\times 0.2-2\times 1.4110\right)\\ =0.1\times \left(-2.0932\right)\\ =-0.2022\end{array}$

The value of$k_2$ is calculated as,

$$\begin{gathered} k_2=hF(x_0+\frac{h}{2},y_0+k_1) \\ \begin{array}{l}=0.1\times F\left(0.2+\frac{0.1}{2},1.4110+\frac{-0.2022}{2}\right)\\ =0.1\times F\left(0.25,1.3099\right)\\ \end{array} \end{gathered}$$

The value of $k_3$is calculated as,

$\begin{array}{l}{k}_3=hF\left(x_0+\frac{h}{2},y_0+\frac{k_2}{2}\right)\\ =0.1\times F\left(0.2+\frac{0.1}{2},1.4110+\frac{-0.16198}{2}\right)\\ =0.1\times F\left(0.25,1.33001\right)\\ =-1.66002\end{array}$

The value of $k_4$is calculated as,

$\begin{array}{l}{k}_4=hF\left(x_0+h,y_0+k_3\right)\\ =0.1\times F\left(0.2+0.1,1.4110+\left(-0.166002\right)\right)\\ =0.1\times F\left(0.3,1.244998\right)\\ =-0.1289996\end{array}$

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

$\begin{array}{rcl}{y}_2& =& y_1+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 1.4110+\frac{1}{6}\left(\left(-0.2022\right)+2\left(-0.16198\right)+2\left(-0.166002\right)+\left(-0.1289996\right)\right)\\ & =& 1.2465\end{array}$

Compute the third order of Runga-Kutta method

For $n=3$, the value of $x_3$is calculated as,

role="math" localid="1668150638001" $x_3=x_2+h=0.2+0.1=0.3$

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,

$\begin{array}{c}{k}_1=hF(x_3,y_3)\\ =0.1\times \left(4\times 0.3-2\times 0.12465\right)\\ =0.1\times \left(-1.293\right)\\ =-1.293\end{array}$

The value of $k_2$is calculated as,

role="math" localid="1668149422886" $$\begin{gathered} k_2=hF(x_3+\frac{h}{2},y_3+\frac{k_1}{2}) \\ \begin{array}{l}=0.1\times F\left(0.3+\frac{0.1}{2},1.2465+\frac{-0.1293}{2}\right)\\ =0.1\times F\left(0.35,1.1815\right)\\ =-0.00627\end{array} \end{gathered}$$

The value of $k_3$is calculated as,

$\begin{array}{l}{k}_3=hF\left(x_3+\frac{h}{2},y_3+\frac{k_2}{2}\right)\\ =0.1\times F\left(0.3+\frac{0.1}{2},1.2465+\frac{-0.09637}{2}\right)\\ =0.1\times F\left(0.35,1.198315\right)\\ =-0.0996633\end{array}$

The value of $k_4$is calculated as,

$\begin{array}{l}{k}_4=hF\left(x_3+h,y_3+k_3\right)\\ =0.1\times F\left(0.3+0.1,1.2465+\left(-0.09637\right)\right)\\ =0.1\times F\left(0.4,1.15013\right)\\ =-0.070026\end{array}$

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

role="math" localid="1668150482240" $\begin{array}{rcl}{y}_4& =& y_3+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 1.2465+\frac{1}{6}\left(\left(-0.1293\right)+2\left(-0.09637\right)+2\left(-0.0996633\right)+\left(-0.070026\right)\right)\\ & =& 1.1480\end{array}$

Compute the fourth order of Runga-Kutta method

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

$x_4=x_3+h=0.3+0.1=0.4$

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,

$\begin{array}{c}{k}_1=hF(x_4,y_4)\\ =0.1\times \left(4\times 0.4-2\times 1.1480\right)\\ =0.1\times \left(-0.696\right)\\ =-0.0696\end{array}$

The value of$k_2$ is calculated as,

$$\begin{gathered} k_2=hF(x_4+\frac{h}{2},y_4+\frac{k_1}{2}) \\ \begin{array}{l}=0.1\times F\left(0.4+\frac{0.1}{2},1.1480+\frac{-0.0696}{2}\right)\\ =0.1\times F\left(0.45,1.1132\right)\\ =-0.04264\end{array} \end{gathered}$$

The value of $k_3$is calculated as,

$\begin{array}{l}{k}_3=hF\left(x_4+\frac{h}{2},y_4+\frac{k_2}{2}\right)\\ =0.1\times F\left(0.4+\frac{0.1}{2},1.1480+\frac{-0.04264}{2}\right)\\ =0.1\times F\left(0.45,1.12668\right)\\ =-0.045336\end{array}$

The value of $k_4$is calculated as,

$\begin{array}{l}{k}_4=hF\left(x_4+h,y_4+k_3\right)\\ =0.1\times F\left(0.4+0.1,1.1480+\left(-0.045336\right)\right)\\ =0.1\times F\left(0.5,1.102664\right)\\ =-0.0205238\end{array}$

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

$\begin{array}{rcl}{y}_5& =& y_4+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 1.11480+\frac{1}{6}\left(\left(-0.0696\right)+2\left(-0.04624\right)+2\left(-0.045336\right)+\left(-0.0205238\right)\right)\\ & =& 1.1037\end{array}$

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

$\begin{array}{c}{x}_5=x_4+h\\ =1.4+0.1\\ =1.5\end{array}$

Therefore, the four decimal approximation of the indicated value$y\left(0.5\right)$ using the RK4 method is 1.1037