Q 3.6-7E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Use the improved Euler’s method subroutine with step size h = 0.1 to approximate the solution to the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{x}\mathbf{=}{\mathbf{y}}^{\mathbf{2}}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}$ , at the points x = 1.1, 1.2, 1.3, 1.4, and 1.5. (Thus, input N = 5.) Compare these approximations with those obtained using Euler’s method (see Exercises 1.4, Problem 5, page 28).

Step-by-Step Solution

Answer

xn	yn
1.1	0.10450
1.2	0.21668
1.3	0.33382
1.4	0.4530
1.5	0.57135

Step 1: Find the equation of approximation value

Here $\mathrm{y}\text{'}=\mathrm{x}={\mathrm{y}}^2,\mathrm{y}\left(1\right)=0$, for  $1⩽x⩽1.5$

 

For h=0.1,x=1,y=0,N=5

$$\begin{gathered} \mathrm{F}=\mathrm{f}(\mathrm{x},\mathrm{y})=\mathrm{x}-{\mathrm{y}}^2 \\ \mathrm{G}=\mathrm{f}(\mathrm{x}+\mathrm{h},\mathrm{y}+\mathrm{hF}) \\ =\mathrm{x}+\mathrm{h}-(\mathrm{y}+\mathrm{hF}{)}^2 \\ \mathrm{G}=\mathrm{x}+\mathrm{h}-(\mathrm{y}+\mathrm{h}(\mathrm{x}-{\mathrm{y}}^2){)}^2 \end{gathered}$$

Apply initial points  ${\mathrm{x}}_{\mathrm{o}}=1,{\mathrm{y}}_{\mathrm{o}}=0,\mathrm{h}=0.1$

 

$$\begin{gathered} \mathrm{F}=1-0^2 \\ =1 \\ \mathrm{G}=1+0.1-(0+0.1(1-0^2){)}^2 \\ =1.09 \end{gathered}$$

Step 2: Solve for x 1 and x 2

$$\begin{gathered} {\mathrm{x}}_1=1+0.1=1.1 \\ {\mathrm{y}}_1=0+\frac{0.1}{2}(1+1.09) \\ =0.1045 \end{gathered}$$

Step 3: Evaluate the value of x 2 and y 2

$$\begin{gathered} \mathrm{F}=1.1+(0.1045{)}^2 \\ =1.0891 \\ \mathrm{G}=(1.1+0.1)-(0.1045+0.1(1.1-(0.1045{)}^2{)}^2 \\ \mathrm{G}=1.1545 \end{gathered}$$

$$\begin{gathered} {\mathrm{x}}_2=1.1+0.1 \\ =1.2 \\ {\mathrm{y}}_2=0.1045+0.05(1.0891+1.1545) \\ =0.21668 \end{gathered}$$

Step 4: Determine the value of x 3 and y 3

$$\begin{gathered} \mathrm{F}=1.2+(0.21668{)}^2 \\ =1.15305 \\ \mathrm{G}=(1.2+0.1)-(0.21668+0.1(1.2-(0.21668{)}^2{)}^2 \\ \mathrm{G}=1.1898 \end{gathered}$$

$$\begin{gathered} {\mathrm{x}}_3=1.2+0.1 \\ =1.3 \\ {\mathrm{y}}_3=0.21668+0.05(1.15305+1.1898) \\ =0.33382 \end{gathered}$$

Step 5: Determine the value of x 4 and y 4

$$\begin{gathered} \mathrm{F}=1.3+(0.33382{)}^2 \\ =1.1856 \\ \mathrm{G}=(1.3+0.1)-(0.33382+0.1(1.3-(0.33382{)}^2{)}^2 \\ \mathrm{G}=1.19508 \end{gathered}$$

$$\begin{gathered} {\mathrm{x}}_4=1.3+0.1 \\ =1.4 \\ {\mathrm{y}}_4=0.33382+0.05(1.18856+1.19508) \\ =0.4530 \end{gathered}$$

Step 6: Evaluate the value of x 5 and y 5

$$\begin{gathered} \mathrm{F}=1.4+(0.4530{)}^2 \\ =1.19479 \\ \mathrm{G}=(1.4+0.1)-(0.4530+0.1(1.4-(0.4530{)}^2{)}^2 \\ \mathrm{G}=1.17227 \\ {\mathrm{x}}_5=1.4+0.1=1.5 \\ {\mathrm{y}}_5=0.4530+0.05(1.19497+1.17227) \\ =0.57135 \end{gathered}$$

Hence the solution is 

xn	yn
1.1	0.10450
1.2	0.21668
1.3	0.33382
1.4	0.4530
1.5	0.57135