Q5.3-25E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Using the Runge–Kutta algorithm for systems with h = 0.05, approximate the solution to the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{t}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$ at t=1.

Step-by-Step Solution

Answer

The result can get by the Runge-Kutta method and the result is y(1)=1.25958

Transform the equation

Here the equation is $\mathrm{y}\text{'}\text{'}\text{'}+\mathrm{y}\text{'}\text{'}+{\mathrm{y}}^2=\mathrm{t}.$

 

The system can be written as:

$$\begin{gathered} {\mathrm{x}}_1=\mathrm{y}\left(\mathrm{t}\right) \\ {\mathrm{x}}_2=\mathrm{y}\text{'}=\mathrm{x} \\ \mathrm{y}\text{'}\text{'}={\mathrm{x}}_3 \end{gathered}$$

The transform equation is:

$$\begin{gathered} \mathrm{x}{\text{'}}_1={\mathrm{x}}_2 \\ \mathrm{x}{\text{'}}_2={\mathrm{x}}_3 \\ \mathrm{x}{\text{'}}_3=\mathrm{t}-{\mathrm{x}}_3-{{\mathrm{x}}^2}_1 \end{gathered}$$

The initial conditions are:

$$\begin{gathered} {\mathrm{x}}_1\left(0\right)=1 \\ {\mathrm{x}}_2\left(0\right)=1 \\ {\mathrm{x}}_3\left(0\right)=1 \end{gathered}$$

Apply the Runge-Kutta method

For h=1

$$\begin{gathered} {\mathrm{k}}_{1,1}={\mathrm{hx}}_2 \\ {\mathrm{k}}_{2,1}={\mathrm{hx}}_3 \\ {\mathrm{k}}_{3,1}=\mathrm{h}(\mathrm{t}-{\mathrm{x}}_3-{{\mathrm{x}}^2}_1) \\ {\mathrm{k}}_{1,2}=\mathrm{h}\left({\mathrm{x}}_2+\frac{{\mathrm{k}}_{2,1}}{2}\right) \\ {\mathrm{k}}_{2,2}=\mathrm{h}\left({\mathrm{x}}_3+\frac{{\mathrm{k}}_{3,1}}{2}\right) \\ {\mathrm{k}}_{3,2}=\mathrm{h}\left(\mathrm{t}+\frac{\mathrm{h}}{2}-{\mathrm{x}}_3-\frac{{\mathrm{k}}_{3,1}}{2}-{\left({\mathrm{x}}_1+\frac{{\mathrm{k}}_{1,1}}{\mathrm{w}}\right)}^2\right) \\ {\mathrm{k}}_{1,3}=\mathrm{h}\left({\mathrm{x}}_3+\frac{{\mathrm{k}}_{2,2}}{2}\right) \\ {\mathrm{k}}_{2,3}=\mathrm{h}\left({\mathrm{x}}_3+\frac{{\mathrm{k}}_{3,2}}{2}\right) \\ {\mathrm{k}}_{3,3}=\mathrm{h}\left(\mathrm{t}+\frac{\mathrm{h}}{2}-{\mathrm{x}}_3-\frac{{\mathrm{k}}_{3,2}}{2}-{\left({\mathrm{x}}_1+\frac{{\mathrm{k}}_{1,2}}{\mathrm{w}}\right)}^2\right) \\ {\mathrm{k}}_{1,4}=\mathrm{h}({\mathrm{x}}_2+{\mathrm{k}}_{2,3}) \\ {\mathrm{k}}_{2,4}=\mathrm{h}({\mathrm{x}}_3+{\mathrm{k}}_{3,3}) \\ {\mathrm{k}}_{3,4}=\mathrm{h}(\mathrm{t}+\mathrm{h}-{\mathrm{x}}_3-{\mathrm{k}}_{3,3}-({\mathrm{x}}_1+{\mathrm{k}}_{1,3}{)}^2) \end{gathered}$$

For,

$$\begin{gathered} {\mathrm{t}}_0=0 \\ {\mathrm{a}}_1=0 \\ {\mathrm{a}}_2=0 \\ {\mathrm{a}}_3=1 \\ {\mathrm{x}}_1(1,1)=1.29167 \\ {\mathrm{x}}_2(1,1)=0.28125 \end{gathered}$$

Repeating the same procedure for $\mathrm{h}=2^{-1,},2^{-2},2^{-3}$.

n	h	${\mathrm{x}}_1$	${\mathrm{x}}_2$	${\mathrm{x}}_3$$x_3$
0	1.0	1.29167	0.28125	0.03125
1	0.5	1.26039	0.34509	-0.06642
2	0.25	1.25960	0.346996	-0.06957
3	0.125	1.25958	0.34704	-0.06971

Hence, y (1) =1.25958

 

This is the required result.