Question

Use the fourth-order Runge–Kutta algorithm to approximate the solution to the initial value problem$\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{xy}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{1}$at x = 2. For a tolerance of, use a stopping procedure based on the absolute error.

Short answer

$\mathit{\varphi }\mathbf{(}\mathbf{2}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{7014}$

Step-by-step solution

Find the values of ki, i = 1, 2, 3, 4

Since $\mathrm{y}\text{'}=1-\mathrm{xy},\mathrm{y}\left(1\right)=1$and $\mathrm{x}={\mathrm{x}}_0=1,\mathrm{y}={\mathrm{y}}_{\mathrm{o}}=1$and h = 1, M = 10

$$\begin{gathered} {\mathrm{k}}_1=\mathrm{h}\text{f}(\mathrm{x},\mathrm{y})=1(1-0)=0 \\ {\mathrm{k}}_2=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_1}{2}\right)=-\frac{1}{2} \\ {\mathrm{k}}_3=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_2}{2}\right)=-\frac{1}{8} \\ {\mathrm{k}}_4=\mathrm{hf}\left(\mathrm{x}+\mathrm{h},\mathrm{y}+{\mathrm{k}}_3\right)=-\frac{3}{4} \end{gathered}$$

Find the values of x and y

$$\begin{gathered} \mathrm{x}=1+1=2 \\ \mathrm{y}=1+\frac{1}{6}\left({\mathrm{k}}_1+2{\mathrm{k}}_2+2{\mathrm{k}}_3+{\mathrm{k}}_4\right) \\ =1+\frac{1}{6}(0-2\left(\frac{1}{2}\right)-2\left(\frac{1}{8}\right)-\left(\frac{3}{4}\right)) \\ =0.6667 \end{gathered}$$

Therefore $\varphi \left(2\right)=0.6667$.

Since$\left|1-0.6667\right|=0.3333>0.001$

Find the other values

Apply the same procedure for h=0.5, and h=0.25 respectively.

$\varphi \left(2\right)=0.700799$

Thus, $\left|0.6667-0.7008\right|=0.0341>0.001$

$\varphi \left(2\right)=0.701385$

Hence, $\left|0.6667-0.7014\right|=0.0006<0.001$

So,$\varphi \left(2\right)=0.7014$.

Hence the solution is $\mathit{\varphi }\left(2\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{7014}$