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.

${\mathit{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathit{x}\mathit{y}\mathbf{+}\sqrt{\mathbf{y}}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{;}\mathbf{}\mathit{y}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{)}$

Short answer

The four decimal approximation of the indicated value$y\left(0.5\right)$using the RK4 method is 1.7561

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,

$$\begin{gathered} y^{\text{'}}=F(x,y) \\ F(x,y)=xy+\sqrt{y} \end{gathered}$$

The given boundary conditions can be written as,

$$\begin{gathered} y_0=f\left(x_0\right) \\ {x}_0=0 \\ h=0.1 \\ {y}_0=1 \end{gathered}$$

For n=0,

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

get the values of$k_1,k_2,k_3,k_4$,

$$\begin{gathered} k_1=hF\left(x_0,y_0\right) \\ =0.1\times \left(\right(0\times 1)+(\sqrt{1}\left)\right) \\ =0.1\times \left(1\right)=0.1 \end{gathered}$$

The value of$k_2$is calculated as,

$$\begin{gathered} k_2=hF\left(x_0+\frac{h}{2},y_0+\frac{k_1}{2}\right) \\ =0.1\times F\left(0+\frac{0.1}{2},1+\frac{0.1}{2}\right) \\ =0.1\times F(0.05,1.05) \\ =0.108 \end{gathered}$$

The value of$k_3$is calculated as,

$$\begin{gathered} k_3=hF\left(x_0+\frac{h}{2},y_0+\frac{k_2}{2}\right) \\ =0.1\times F\left(0+\frac{0.1}{2},1+\frac{0.10772}{2}\right) \\ =0.1\times F(0.05,1.05386) \\ =0.10793 \end{gathered}$$

The value of$k_4$is calculated as,

$$\begin{gathered} k_4=hF\left(x_0+h,y_0+k_3\right) \\ =0.1\times F(0+0.1,1+0.10793) \\ =0.1\times F(0.1,1.108) \\ =0.1164 \end{gathered}$$

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

$$\begin{gathered} y_1=y_0+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right) \\ =1+\frac{1}{6}(0.1+2(0.108)+2(0.10793)+(0.1164\left)\right) \\ =1+\frac{1}{6}\times (0.1+0.116+0.221586+0.1164) \\ =1.1079 \end{gathered}$$

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$k_1,k_2,k_3,k_4$,

$$\begin{gathered} k_1=hF\left(x_1,y_1\right) \\ =0.1\times \left(\right(0.1\left)\right(1.1079)+(\sqrt{1.1079}\left)\right) \\ =0.1\times (0.1000) \\ =0.11634 \end{gathered}$$

The value of$k_2$is calculated as,

$$\begin{gathered} k_2=hF\left(x_1+\frac{h}{2},y_1+\frac{k_1}{2}\right) \\ =0.1\times F\left(0.1+\frac{0.1}{2},1.1079+\frac{(0.11634)}{2}\right) \\ =0.1\times F(0.15,1.6896) \\ =0.15533 \end{gathered}$$

The value of$k_3$is calculated as,

$$\begin{gathered} k_3=hF\left(x_1+\frac{h}{2},y_1+\frac{k_2}{2}\right) \\ =0.1\times F\left(0.1+\frac{0.1}{2},1.1079+\frac{(0.15533)}{2}\right) \\ =0.1\times F(0.15,1.88454) \\ =0.1655465 \end{gathered}$$

The value of$k_4$is calculated as,

$$\begin{gathered} k_4=hF\left(x_1+h,y_1+k_3\right) \\ =0.1\times F(0.1+0.1,1.1079+(0.1655465\left)\right) \\ =0.1\times F(0.2,1.27345) \\ =0.138316 \end{gathered}$$

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

$$\begin{gathered} y_2=y_1+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right) \\ =1.1079+\frac{1}{6}\left(\right(0.11634)+2(0.15533)+2(0.1655465)+(0.138316\left)\right) \\ =1.2337 \end{gathered}$$

Compute the second order Runga-Kutta method

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

$$\begin{gathered} x_2=x_1+h \\ =0.1+0.1 \\ =0.2 \end{gathered}$$

Now 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

$$\begin{gathered} k_1=hF\left(x_2,y_2\right) \\ =0.1\times \left(\right(0.2\left)\right(1.2337)+(\sqrt{1.2337}\left)\right) \\ =0.1\times (1.35746) \\ =0.135746 \end{gathered}$$

The value of$k_2$is calculated as,

$$\begin{gathered} k_2=hF\left(x_2+\frac{h}{2},y_2+\frac{k_1}{2}\right) \\ =0.1\times F\left(0.2+\frac{0.1}{2},1.2337+\frac{(0.135746)}{2}\right) \\ =0.1\times F(0.25,1.91243) \end{gathered}$$

$=0.1861$

The value of$k_3$is calculated as,

$$\begin{gathered} k_3=hF\left(x_2+\frac{h}{2},y_2+\frac{k_2}{2}\right) \\ =0.1\times F\left(0.2+\frac{0.1}{2},1.2337+\frac{(0.1861)}{2}\right) \\ =0.1\times F(0.25,1.32675) \\ =0.148354 \end{gathered}$$

The value of$k_4$is calculated as,

$$\begin{gathered} k_4=hF\left(x_2+h,y_2+k_3\right) \\ =0.1\times F(0.2+0.1,1.2337+(0.148354\left)\right) \\ =0.1\times F(0.3,1.3821) \\ =0.1591 \end{gathered}$$

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

$$\begin{gathered} y_3=y_2+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right) \\ =1.2337+\frac{1}{6}\left(\right(0.135746)+2(0.1861)+2(0.148354)+(0.1591\left)\right) \\ =1.2337+\frac{1}{6}\left(\right(0.135746)+2(0.1861)+2(0.148354)+(0.1591\left)\right) \\ =1.3807 \end{gathered}$$

Compute the third order of Runga-Kutta method

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

$$\begin{gathered} x_3=x_2+h \\ =0.2+0.1 \\ =0.3 \end{gathered}$$

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,

$$\begin{gathered} k_1=hF\left(x_3,y_3\right) \\ =0.1\times \left(\right(0.3\left)\right(1.3807)+(\sqrt{1.3807}\left)\right) \\ =0.1\times (1.595026) \\ =0.1595026 \end{gathered}$$

The value of $k_2$is calculated as,

$$\begin{gathered} k_2=hF\left(x_3+\frac{h}{2},y_3+\frac{k_1}{2}\right) \\ =0.1\times F\left(0.3+\frac{0.1}{2},1.3807+\frac{(0.1595026)}{2}\right) \\ =0.1\times F(0.35,1.4605) \\ =0.171965 \end{gathered}$$

The value of $k_3$is calculated as,

$$\begin{gathered} k_3=hF\left(x_3+\frac{h}{2},y_3+\frac{k_2}{2}\right) \\ =0.1\times F\left(0.3+\frac{0.1}{2},1.3807+\frac{(0.171965)}{2}\right) \\ =0.1\times F(0.35,1.4669) \\ =0.17245 \end{gathered}$$

The value of $k_4$is calculated as,

$$\begin{gathered} k_4=hF\left(x_3+h,y_3+k_3\right) \\ =0.1\times F(0.3+0.1,1.3807+(0.17245\left)\right) \\ =0.1\times F(0.4,1.55315) \\ =0.18675 \end{gathered}$$

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

$$\begin{gathered} y_4=y_3+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right) \\ =1.3807+\frac{1}{6}\left(\right(0.1595026)+2(0.171965)+2(0.17245)+(0.18675\left)\right) \\ =1.3807+\frac{1}{6}\times (0.0302+0.07073+0.0708+0.04065) \\ =1.5531 \end{gathered}$$

Compute the fourth order of the 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${\mathrm{k}}_1$is calculated as,

$$\begin{gathered} k_1=hF\left(x_4,y_4\right) \\ =0.1\times \left(\right(0.4\left)\right(1.5531)+(\sqrt{1.5531}\left)\right) \\ =0.1\times (1.867474) \\ =0.1867474 \end{gathered}$$

The value of$k_2$is calculated as,

$$\begin{gathered} k_2=hF\left(x_4+\frac{h}{2},y_4+\frac{k_1}{2}\right) \\ =0.1\times F\left(0.4+\frac{0.1}{2},1.5531+\frac{(0.1867474)}{2}\right) \\ =0.1\times F(0.45,1.6464737) \\ =0.20241 \end{gathered}$$

The value of$k_3$is calculated as,

$$\begin{gathered} k_3=hF\left(x_4+\frac{h}{2},y_4+\frac{k_2}{2}\right) \\ =0.1\times F\left(0.4+\frac{0.1}{2},1.5531+\frac{(0.20241)}{2}\right) \\ =0.1\times F(0.45,1.654305) \\ =0.203064 \end{gathered}$$

The value of$k_4$is calculated as,

$$\begin{gathered} k_4=hF\left(x_4+h,y_4+k_3\right) \\ =0.1\times F(0.4+0.1,1.5531+(0.203064\left)\right) \\ =0.1\times F(0.5,1.756164) \\ =0.22033 \end{gathered}$$

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

$$\begin{gathered} y_5=y_4+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right) \\ =1.5531+\frac{1}{6}\left(\right(0.1867474)+2(0.20241)+2(0.203064)+(0.22033\left)\right) \\ =1.5531+\frac{1}{6}\times (0.1867474+0.40482+0.406128+0.22033) \\ =1.7561 \end{gathered}$$

For$n=5$, the value of$x_5$is calculated as,$x_5=x_4+h=0.4+0.1=0.5$

Therefore, the four decimal approximation of the indicated value $y\left(0.5\right)$using the RK4 method is 1.7561