Q 3.6-8E - Fundamentals Of Differential Equations And Boundary Value Problems

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).

Step-by-Step Solution

Answer

xn	yn
1.2	1.48
1.4	2.24788
1.6	3.6518
1.8	6.88733

Step 1: 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

 

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

 

Step 2: Solve for x 1 and y 1

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

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

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

 

 

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

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

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

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

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

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

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

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

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

Hence the solution is 

xn	yn
1.2	1.48
1.4	2.24788
1.6	3.6518
1.8	6.88733