Question

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

${\mathit{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathit{y}\mathbf{-}{\mathit{y}}^{\mathbf{2}}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{;}\mathbf{}\mathit{y}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{)}$

Short answer

The four decimal approximation of the indicated value $y(0.5)$using the RK4 method is 1.333.

Step-by-step solution

Runga-Kutta method

The general formulas used in the given problem are,

$$\begin{gathered} k_1=hF\left(x_n,y_n\right) \\ {k}_2=hF\left(x_n+\frac{h}{2},y_n+\frac{k_1}{2}\right) \\ {k}_3=hF\left(x_n+\frac{h}{2},y_n+\frac{k_2}{2}\right) \\ {k}_4=hF\left(x_n+h,y_n+k_3\right) \end{gathered}$$

And,

$y_{n+1}=y_n+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)$

Determine the First order Runga-Kutta method

The given differential equation can be written as,

$$\begin{gathered} y^{\text{'}}=F(x,y) \\ F(x,y)=x{y}^2-\frac{y}{x} \end{gathered}$$

The given boundary conditions can be written as,

$$\begin{gathered} y_0=f\left(x_0\right) \\ {x}_0=1 \\ {y}_0=1 \\ h=0.1 \end{gathered}$$

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 ${\mathrm{k}}_1$is calculated as,

$$\begin{gathered} k_1=h \\ F\left(x_0,y_0\right)=0.1\times \left(1\times {\left(1\right)}^2-\frac{1}{1}\right) \\ =0.1\times \left(0\right) \\ =0 \end{gathered}$$

The value of $k_2$is calculated as,

$$\begin{gathered} k_2=hF\left(x_0+\frac{h}{2},y_0+\frac{k_1}{2}\right) \\ =0.1\times F\left(1+\frac{0.1}{2},1+\frac{0}{2}\right) \\ =0.1\times F(1.05,1) \\ =0.0098 \end{gathered}$$

The value of $k_3$is calculated as,

$$\begin{gathered} k_3=hF\left(x_0+\frac{h}{2},y_0+\frac{k_2}{2}\right) \\ =0.1\times F\left(1+\frac{0.1}{2},1+\frac{(0.0098)}{2}\right) \\ =0.1\times F(1.05.1.005) \\ =0.0103 \end{gathered}$$

The value of $k_4$is calculated as,

$$\begin{gathered} k_4=hF\left(x_0+h,y_0+k_3\right) \\ =0.1\times F(1+0.1,1+0.0103) \\ =0.1\times F(1.1,1.0103) \\ =0.0204 \end{gathered}$$

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

$$\begin{gathered} y_1=y_0+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right) \\ =1+\frac{1}{6}(0+(2\times 0.0098)+(2\times 0.0103)+0.0204) \\ =1+\frac{1}{6}\times (0+0.0196+0.0206+0.0204) \\ =1.0101 \end{gathered}$$

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

$$\begin{gathered} x_1=x_0+h \\ =1+0.1 \\ =1.1 \end{gathered}$$

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,

$$\begin{gathered} k_1=hF\left(x_1,y_1\right) \\ =0.1\times \left(1.1\times {(1.0101)}^2-\frac{1.0101}{1.1}\right) \\ =0.1\times (0.2041) \\ =0.0204 \end{gathered}$$

The value of$k_2$is calculated as,

$k_2=hF\left(x_1+\frac{h}{2},y_1+\frac{k_1}{2}\right)$

$$\begin{gathered} =0.1\times F\left(1.1+\frac{0.1}{2},1.0101+\frac{0.0204}{2}\right) \\ =0.1\times F(1.15,1.0203) \\ =0.0310 \end{gathered}$$

The value of $k_3$is calculated as,

$$\begin{gathered} k_3=hF\left(x_1+\frac{h}{2},y_1+\frac{k_2}{2}\right) \\ =0.1\times F\left(1.1+\frac{0.1}{2},1.0101+\frac{0.0310}{2}\right) \\ =0.1\times F(1.15,1.0256) \\ =0.0318 \end{gathered}$$

The value of $k_4$is calculated as,

$$\begin{gathered} k_4=hF\left(x_1+h,y_1+k_3\right) \\ =0.1\times F(1.1+0.1,1.0101+0.0318) \\ =0.1\times F(1.2,1.0419) \\ =0.0434 \end{gathered}$$

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

$$\begin{gathered} y_2=y_1+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right) \\ =1.0101+\frac{1}{6}(0.0204+(2\times 0.0310)+(2\times 0.0318)+0.0434) \\ =1.0101+\frac{1}{6}\times (0.0204+0.062+0.0636+0.0434) \\ =1.0417 \end{gathered}$$

Compute the second order Runga-Kutta method

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

$$\begin{gathered} x_2=x_1+h \\ =1.1+0.1 \\ =1.2 \end{gathered}$$

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,

$$\begin{gathered} k_1=hF\left(x_2,y_2\right) \\ =0.1\times \left(1.2\times {(1.0417)}^2-\frac{1.0417}{1.2}\right) \\ =0.1\times (0.4429) \\ =0.0443 \end{gathered}$$

The value of$k_2$is calculated as,

$$\begin{gathered} k_2=hF\left(x_2+\frac{h}{2},y_2+\frac{k_1}{2}\right) \\ =0.1\times F\left(1.2+\frac{0.1}{2},1.0417+\frac{0.0443}{2}\right) \\ =0.1\times F(1.25,1.069) \\ =0.0574 \end{gathered}$$

The value of$k_3$is calculated as,

$$\begin{gathered} k_3=hF\left(x_2+\frac{h}{2},y_2+\frac{k_2}{2}\right) \\ =0.1\times F\left(1.2+\frac{0.1}{2},1.0417+\frac{0.0574}{2}\right) \\ =0.1\times F(1.25,1.076) \\ =0.0586 \end{gathered}$$

The value of$k_4$is calculated as,

$$\begin{gathered} k_4=hF\left(x_2+h,y_2+k_3\right) \\ =0.1\times F(1.2+0.1,1.0417+0.0586) \\ =0.1\times F(1.3,1.106)=0.0738 \end{gathered}$$

Now substitute the value of ${\mathrm{k}}_1,{\mathrm{k}}_2,{\mathrm{k}}_3$and $k_4$in $y_3$.

$$\begin{gathered} y_3=y_2+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right) \\ =1.0417+\frac{1}{6}(0.0443+(2\times 0.0574)+(2\times 0.0586)+0.0738) \\ =1.0417+\frac{1}{6}\times (0.0443+0.1148+0.1172+0.0738) \\ =1.0989 \end{gathered}$$

Compute the third order of the Runga-Kutta method

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

$$\begin{gathered} x_3=x_2+h \\ =1.2+0.1 \\ =1.3 \end{gathered}$$

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,

$$\begin{gathered} k_1=hF\left(x_2,y_2\right) \\ =0.1\times \left(1.3\times {(1.0989)}^2-\frac{1.0989}{1.3}\right) \\ =0.1\times (0.7245) \\ =0.0724 \end{gathered}$$

The value of$k_2$is calculated as,

$$\begin{gathered} k_2=hF\left(x_3+\frac{h}{2},y_3+\frac{k_1}{2}\right) \\ =0.1\times F\left(1.3+\frac{0.1}{2},1.0989+\frac{0.0724}{2}\right) \\ =0.1\times F(1.35,1.1351) \\ =0.0899 \end{gathered}$$

The value of$k_3$is calculated as,

$$\begin{gathered} k_3=hF\left(x_3+\frac{h}{2},y_3+\frac{k_2}{2}\right) \\ =0.1\times F\left(1.3+\frac{0.1}{2},1.0989+\frac{0.0899}{2}\right) \\ =0.1\times F(1.35,1.1438) \\ =0.0919 \end{gathered}$$

The value of$k_4$is calculated as,

$$\begin{gathered} k_4=hF\left(x_3+h,y_3+k_3\right) \\ =0.1\times F(1.3+0.1,1.0989+0.0919) \\ =0.1\times F(1.4,1.1908) \\ =0.1135 \end{gathered}$$

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

$$\begin{gathered} y_4=y_3+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right) \\ =1.0989+\frac{1}{6}(0.0724+(2\times 0.0899)+(2\times 0.0919)+0.1135) \\ =1.0989+\frac{1}{6}\times (0.0724+0.1798+0.1838+0.1135) \\ =1.1905 \end{gathered}$$

Compute the fourth order of Runga-Kutta method

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

$$\begin{gathered} x_4=x_3+h \\ =1.3+0.1 \\ =1.4 \end{gathered}$$

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,

$$\begin{gathered} k_1=hF\left(x_4,y_4\right) \\ =0.1\times \left(1.4\times {(1.1905)}^2-\frac{1.1905}{1.4}\right) \\ =0.1\times (1.1338) \\ =0.1134 \end{gathered}$$

The value of$k_2$is calculated as,

$$\begin{gathered} k_2=hF\left(x_4+\frac{h}{2},y_4+\frac{k_1}{2}\right) \\ =0.1\times F\left(1.4+\frac{0.1}{2},1.1905+\frac{0.1134}{2}\right) \\ =0.1\times F(1.45,1.2472) \\ =0.1395 \end{gathered}$$

The value of$k_3$is calculated as,

$$\begin{gathered} k_3=hF\left(x_4+\frac{h}{2},y_4+\frac{k_2}{2}\right) \\ =0.1\times F\left(1.4+\frac{0.1}{2},1.1905+\frac{0.1395}{2}\right) \\ =0.1\times F(1.45,1.2602) \\ =0.1434 \end{gathered}$$

The value of$k_4$is calculated as,

$$\begin{gathered} k_4=hF\left(x_4+h,y_4+k_3\right) \\ =0.1\times F(1.4+0.1,1.1905+0.1434) \\ =0.1\times F(1.5,1.334) \\ =0.1780 \end{gathered}$$

$$\begin{gathered} =0.1\times F(1.4+0.1,1.1905+0.1434) \\ =0.1\times F(1.5,1.334) \\ =0.1780 \end{gathered}$$

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

$$\begin{gathered} y_5=y_4+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right) \\ =1.1905+\frac{1}{6}(0.1134+(2\times 0.1395)+(2\times 0.1434)+0.1780) \\ =1.1905+\frac{1}{6}\times (0.1134+0.279+0.2868+0.1780) \\ =1.3333 \end{gathered}$$

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

$$\begin{gathered} x_5=x_4+h \\ =1.4+0.1 \\ =1.5 \end{gathered}$$

Therefore, the four decimal approximation of the indicated value $y(1.5)$ using the RK4 method is 1.333.