Question

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

$\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{1}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{;}\mathbf{}\mathbf{}\mathbf{y}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{y}\left(0.5\right)$

Short answer

The four decimal approximation of the indicated value $y\left(0.5\right)$using the RK4 method is 0.5463

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,

$\begin{array}{l}y\text{'}=F\left(x,y\right)\\ F\left(x,y\right)=1+y^2\end{array}$

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 $n=0$, 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,

$\begin{array}{l}{k}_1=hF(x_0,y_0)\\ =0.1\times \left(1+{\left(0\right)}^2\right)\\ =0.1\times 1\\ =0.1\end{array}$

The value of$k_2$is calculated as,

$$\begin{gathered} k_2=hF(x_0+\frac{h}{2},y_0+\frac{k_1}{2}) \\ \begin{array}{l}=0.1\times F\left(0+\frac{0.1}{2},0+\frac{0.1}{2}\right)\\ =0.1\times F\left(0.05,0.05\right)\\ =0.10025\end{array} \end{gathered}$$

The value of$k_3$is calculated as,

$\begin{array}{l}{k}_3=hF\left(x_0+\frac{h}{2},y_0+\frac{k_2}{2}\right)\\ =0.1\times F\left(0+\frac{0.1}{2},0+\frac{1.0025}{2}\right)\\ =0.1\times F\left(0.05,0.050125\right)\\ =0.100251\end{array}$

The value of $k_4$is calculated as,

$\begin{array}{l}{k}_4=hF\left(x_0+h,y_0+k_3\right)\\ =0.1\times F\left(0+0.1,0+0.100251\right)\\ =0.1\times F\left(0.1,0.100251\right)\\ =0.101101\end{array}$

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

$\begin{array}{rcl}{y}_1& =& y_0+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 0+\frac{1}{6}\left(0.1+2\left(1.0025\right)+2\left(0.100251\right)+\left(0.10101\right)\right)\\ & =& 0.1003\end{array}$

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

$\begin{array}{l}{x}_1=x_0+h\\ =0+0.1\\ =0.1\end{array}$

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,

$\begin{array}{l}{k}_1=hF(x_1,y_1)\\ =0.1\times \left(1+{\left(0.10034\right)}^2\right)\\ =0.1\times \left(1.0101\right)\\ =0.0101\end{array}$

The value of $k_2$is calculated as,

$$\begin{gathered} k_2=hF(x_1+\frac{h}{2},y_1+\frac{k_1}{2}) \\ \begin{array}{l}=0.1\times F\left(0.1+\frac{0.1}{2},0.10034+\frac{0.1010}{2}\right)\\ =0.1\times F\left(0.15,1.5084\right)\\ =0.1023\end{array} \end{gathered}$$

The value of $k_3$is calculated as,

$\begin{array}{l}{k}_3=hF\left(x_1+\frac{h}{2},y_1+\frac{k_2}{2}\right)\\ =0.1\times F\left(0.1+\frac{0.1}{2},0.10034+\frac{1.0023}{2}\right)\\ =0.1\times F\left(0.15,0.1515\right)\\ =0.1023\end{array}$

The value of$k_4$ is calculated as,

$\begin{array}{l}{k}_4=hF\left(x_1+h,y_1+k_3\right)\\ =0.1\times F\left(0.1+0.1,0.10034+0.1023\right)\\ =0.1\times F\left(0.2,0.2026\right)\\ =0.1041\end{array}$

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

role="math" localid="1668153972843" $\begin{array}{rcl}{y}_2& =& y_1+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 0.10034+\frac{1}{6}\left(0.1010+2\left(1.023\right)+2\left(0.1023\right)+\left(0.1041\right)\right)\\ & =& 0.2027\end{array}$

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,

$\begin{array}{l}{k}_1=hF(x_2,y_2)\\ =0.1\times \left(1+{\left(0.2027\right)}^2\right)\\ =0.1\times \left(1.041\right)\\ =0.1041\end{array}$

The value of$k_2$is calculated as,

$$\begin{gathered} k_2=hF(x_2+\frac{h}{2},y_2+\frac{k_1}{2}) \\ \begin{array}{l}=0.1\times F\left(0.2+\frac{0.1}{2},0.2027+\frac{0.1041}{2}\right)\\ =0.1\times F\left(0.25,0.2559\right)\\ =0.1066\end{array} \end{gathered}$$

The value of $k_3$is calculated as,

$\begin{array}{l}{k}_3=hF\left(x_2+\frac{h}{2},y_2+\frac{k_2}{2}\right)\\ =0.1\times F\left(0.2+\frac{0.1}{2},0.2027+\frac{0.1065}{2}\right)\\ =0.1\times F\left(0.25,0.2599\right)\\ =0.1066\end{array}$

The value of $k_4$is calculated as,

$\begin{array}{l}{k}_4=hF\left(x_2+h,y_2+k_3\right)\\ =0.1\times F\left(0.2+0.1,0.2027+0.1066\right)\\ =0.1\times F\left(0.3,0.3092\right)\\ =0.1096\end{array}$

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

$\begin{array}{rcl}{y}_3& =& y_2+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 0.2027+\frac{1}{6}\left(0.1041+2\left(1.065\right)+2\left(0.1066\right)+\left(0.1096\right)\right)\\ & =& 0.3093\end{array}$

Compute the third order of Runga-Kutta method

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

$x_3=x_2+h=0.2+0.1=0.3$

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,

$\begin{array}{l}{k}_1=hF(x_3,y_3)\\ =0.1\times \left(1+{\left(0.3093\right)}^2\right)\\ =0.1\times \left(1.096\right)\\ =0.1096\end{array}$

The value of $k_2$is calculated as,

$$\begin{gathered} k_2=hF(x_3+\frac{h}{2},y_3+\frac{k_1}{2}) \\ \begin{array}{l}=0.1\times F\left(0.3+\frac{0.1}{2},0.3093+\frac{0.1096}{2}\right)\\ =0.1\times F\left(0.35,0.3641\right)\\ =0.1133\end{array} \end{gathered}$$

The value of is calculated as,

$$\begin{gathered} k_3=hF(x_3+\frac{h}{2},y_3+\frac{k_2}{2}) \\ \begin{array}{l}=0.1\times F\left(0.3+\frac{0.1}{2},0.3093+\frac{0.1133}{2}\right)\\ =0.1\times F\left(0.35,0.3659\right)\\ =0.1134\end{array} \end{gathered}$$

The value of$k_4$ is calculated as,

$\begin{array}{l}{k}_4=hF\left(x_3+h,y_3+k_3\right)\\ =0.1\times F\left(0.3+0.1,0.3091+0.1134\right)\\ =0.1\times F\left(0.4,0.4277\right)\\ =0.1179\end{array}$

Now substitute the value of role="math" localid="1668154842352" $k_1,k_2,k_3,k_4$in $y_4$.

$\begin{array}{rcl}{y}_4& =& y_3+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 0.3093+\frac{1}{6}\left(0.1096+2\left(0.1133\right)+2\left(0.1134\right)+\left(0.1179\right)\right)\\ & =& 0.4288\end{array}$

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,

$\begin{array}{l}{k}_1=hF(x_4,y_4)\\ =0.1\times \left(1+{\left(0.4288\right)}^2\right)\\ =0.1\times \left(1.179\right)\\ =0.1179\end{array}$

The value of $k_2$is calculated as,

$$\begin{gathered} k_2=hF(x_4+\frac{h}{2},y_4+\frac{k_1}{2}) \\ \begin{array}{l}=0.1\times F\left(0.4+\frac{0.1}{2},0.4288+\frac{0.1179}{2}\right)\\ =0.1\times F\left(0.45,0.4817\right)\\ =0.1232\end{array} \end{gathered}$$

The value of $k_3$is calculated as,

role="math" localid="1668155251734" $$\begin{gathered} k_3=hF(x_4+\frac{h}{2},y_4+\frac{k_2}{2}) \\ \begin{array}{l}=0.1\times F\left(0.4+\frac{0.1}{2},0.4288+\frac{0.1232}{2}\right)\\ =0.1\times F\left(0.45,0.4844\right)\\ =0.1235\end{array} \end{gathered}$$

The value of $k_4$is calculated as,

$\begin{array}{l}{k}_4=hF\left(x_4+h,y_4+k_3\right)\\ =0.1\times F\left(0.4+0.1,0.3091+0.1235\right)\\ =0.1\times F\left(0.5,0.5463\right)\\ =0.1298\end{array}$

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

$\begin{array}{rcl}{y}_5& =& y_4+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 0.4288+\frac{1}{6}\left(0.1179+2\left(0.1232\right)+2\left(0.1235\right)+\left(0.1298\right)\right)\\ & =& 0.5463\end{array}$

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

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

Therefore, the four decimal approximation of the indicated value$y\left(0.5\right)$ using the RK4 method is 0.5463