Use the fourth-order Runge–Kutta subroutine with h= 0.25 to approximate the solution to the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{2}\mathbf{y}\mathbf{-}\mathbf{6}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$, at x= 1. (Thus, input N= 4.) Compare this approximation to the actual solution $\mathbf{y}\mathbf{=}\mathbf{3}\mathbf{-}\mathbf{2}{\mathbf{e}}^{\mathbf{2}\mathbf{x}}$evaluated at x= 1.

Since $\mathrm{f}(\mathrm{x},\mathrm{y})=2\mathrm{y}-6$and x = 0, y = 1 and h = 0.25

$$\begin{gathered} {\mathrm{k}}_1=\mathrm{h}\text{f}(\mathrm{x},\mathrm{y})=0.25\left(2\right(1)-6)=-1 \\ {\mathrm{k}}_2=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_1}{2}\right)=-1.25 \\ {\mathrm{k}}_3=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_2}{2}\right)=-1.3125 \\ {\mathrm{k}}_4=\mathrm{hf}\left(\mathrm{x}+\mathrm{h},\mathrm{y}+{\mathrm{k}}_3\right)=-1.65625 \end{gathered}$$

$$\begin{gathered} \mathrm{x}=0+0.25=0.25 \\ \mathrm{y}=1+\frac{1}{6}\left({\mathrm{k}}_1+2{\mathrm{k}}_2+2{\mathrm{k}}_3+{\mathrm{k}}_4\right) \\ =1+\frac{1}{6}(-1-2(1.25)-2(1.3125)-1.65625) \\ =-0.29688 \end{gathered}$$

$$\begin{gathered} {\mathrm{k}}_1=\mathrm{h}\text{f}(\mathrm{x},\mathrm{y})=0.25\left(2\right(-0.29688)-6)=-1.62844 \\ {\mathrm{k}}_2=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_1}{2}\right)=-2.06055 \\ {\mathrm{k}}_3=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_2}{2}\right)=-2.16358 \\ {\mathrm{k}}_4=\mathrm{hf}\left(\mathrm{x}+\mathrm{h},\mathrm{y}+{\mathrm{k}}_3\right)=-2.73022 \end{gathered}$$

Now

$$\begin{gathered} \mathrm{x}=0.25+0.25=0.5 \\ \mathrm{y}=1+\frac{1}{6}\left({\mathrm{k}}_1+2{\mathrm{k}}_2+2{\mathrm{k}}_3+{\mathrm{k}}_4\right) \\ =1+\frac{1}{6}(-1.64844-2(2.06055-2(2.16358-2.73022) \\ =-2.43470 \end{gathered}$$

$$\begin{gathered} {\mathrm{k}}_1=\mathrm{h}\text{f}(\mathrm{x},\mathrm{y})=0.25\left(2\right(-2.43470)-6)=-2.71735 \\ {\mathrm{k}}_2=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_1}{2}\right)=-3.39670 \\ {\mathrm{k}}_3=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_2}{2}\right)=-3.56652 \\ {\mathrm{k}}_4=\mathrm{hf}\left(\mathrm{x}+\mathrm{h},\mathrm{y}+{\mathrm{k}}_3\right)=-4.50060 \\ \mathrm{x}=0.5+0.25=0.75 \\ \mathrm{y}=1+\frac{1}{6}\left({\mathrm{k}}_1+2{\mathrm{k}}_2+2{\mathrm{k}}_3+{\mathrm{k}}_4\right) \\ =1+\frac{1}{6}(-2.71735-2(3.39670)-2(3.56652)-4.50060) \\ =-5.95876 \end{gathered}$$

And

$$\begin{gathered} {\mathrm{k}}_1=\mathrm{h}\text{f}(\mathrm{x},\mathrm{y})=0.25\left(2\right(-5.95876)-6)=-4.47938 \\ {\mathrm{k}}_2=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_1}{2}\right)=-5.59922 \\ {\mathrm{k}}_3=\mathrm{hf}\left(\mathrm{x}+\frac{\mathrm{h}}{2},\mathrm{y}+\frac{{\mathrm{k}}_2}{2}\right)=-5.87918 \\ {\mathrm{k}}_4=\mathrm{hf}\left(\mathrm{x}+\mathrm{h},\mathrm{y}+{\mathrm{k}}_3\right)=-7.41895 \\ \mathrm{x}=0.75+0.25=1 \\ \mathrm{y}=1+\frac{1}{6}\left({\mathrm{k}}_1+2{\mathrm{k}}_2+2{\mathrm{k}}_3+{\mathrm{k}}_4\right) \\ =-5.95876+\frac{1}{6}(-4.47938-2(5.59922)-2(5.87918)-7.41895) \\ =-11.7679 \end{gathered}$$

Since at x = 1

$$\begin{gathered} \varphi \left(x\right)=3-2e^{2x} \\ \varphi \left(1\right)=-11.7781 \end{gathered}$$

Then$\left|\mathrm{y}\left(1\right)-\varphi \left(1\right)\right|=0.0120$

Hence the solution is$\varphi \left(1\right)=-11.7679$