Question

Use the improved Euler’s method subroutine with step size h= 0.2 to approximate the solution to the initial value problem$\mathbf{y}\mathbf{\text{'}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{(}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{y}\mathbf{)}\mathbf{,}\mathbf{y}\mathbf{(}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{1}$ at the points x= 1.2, 1.4, 1.6, and 1.8. (Thus, input N= 4.) Compare these approximations with those obtained using Euler’s method (see Exercises 1.4, Problem 6, page 28).

Short answer

xn	yn
1.2	1.48
1.4	2.24788
1.6	3.6518
1.8	6.88733

Step-by-step solution

Find the equation of approximation value

Here, $\mathrm{y}\text{'}=\frac{1}{\mathrm{x}}({\mathrm{y}}^2+\mathrm{y}),\mathrm{y}(1)=0$ for $1⩽x⩽1.8$

For h=0.2, x=1, y=1, N=4

$$\begin{gathered} \mathrm{F}=\mathrm{f}(\mathrm{x},\mathrm{y})=\frac{1}{\mathrm{x}}({\mathrm{y}}^2+\mathrm{y}) \\ \mathrm{G}=\mathrm{f}(\mathrm{x}+\mathrm{h},\mathrm{y}+\mathrm{hF}) \\ =\frac{1}{\mathrm{x}+0.2}\left({\left(\mathrm{y}+\frac{0.2}{\mathrm{x}}({\mathrm{y}}^2+\mathrm{y})\right)}^2+\mathrm{y}+\frac{0.2}{\mathrm{x}}({\mathrm{y}}^2+\mathrm{y})\right) \end{gathered}$$

Solve for x1 and y1

Apply initial points ${\mathrm{x}}_{\mathrm{o}}=1,{\mathrm{y}}_{\mathrm{o}}=1,\mathrm{h}=0.2$

$$\begin{gathered} \mathrm{F}(1,1)=2 \\ \mathrm{G}(1,1)=2.8 \end{gathered}$$

$$\begin{gathered} {\mathrm{x}}_1=1+0.2 \\ =1.2 \\ {\mathrm{y}}_1=1+\frac{0.2}{2}(2+2.8) \\ =1.48 \end{gathered}$$

Evaluate the value of  x2 and y2

$$\begin{gathered} \mathrm{F}(1.2,1.48)=3.05867 \\ \mathrm{G}(1.2,1.48)=4.61934 \end{gathered}$$

$$\begin{gathered} {\mathrm{x}}_2=1.2+0.2 \\ =1.4 \\ {\mathrm{y}}_2=1.48+0.1(3.05867+4.61934) \\ =2.24788 \end{gathered}$$

Determine the value of  x3 and y3

$$\begin{gathered} \mathrm{F}(1.4,2.24788)=5.21489 \\ \mathrm{G}(1.4,2.2488)=8.82538 \end{gathered}$$

$$\begin{gathered} {\mathrm{x}}_3=1.4+0.2 \\ =1.6 \\ {\mathrm{y}}_3=2.24788+0.1(5.21489+8.82538) \\ =3.6518 \end{gathered}$$

Determine the value of  x4 and y4

$$\begin{gathered} \mathrm{F}(1.6,3.6518)=10.6172 \\ \mathrm{G}(1.6,3.6518)=21.7381 \end{gathered}$$

$$\begin{gathered} {\mathrm{x}}_4=1.6+0.2 \\ =1.8 \\ {\mathrm{y}}_4=3.6518+0.1(10.6172+21.7381) \\ =6.88733 \end{gathered}$$

Hence the solution is

xn	yn
1.2	1.48
1.4	2.24788
1.6	3.6518
1.8	6.88733