Chapter 4.3, Problem 1P

(a)

Program Plan Intro

Program Description: Purpose of the problem is to obtain the approximate values of $x\left(0.2\right)\text{ and }y\left(0.2\right)$ by Euler’s method.

Explanation of Solution

Given information:

The differential problem is $x^{\prime }=x+2y,x\left(0\right)=0\text{ and }{y}^{\prime }=2x+y,y\left(0\right)=2$ .

The exact solution of the differential equation is $x\left(t\right)=e^{3t}-e^{-t}$ and $y\left(t\right)=e^{3t}+e^{-t}$ .

The two step size is $0.1$ .

Calculation:

The differential problem can be represented as,

  $\begin{array}{c}f\left(x,y\right)=x+2y,\\ g\left(x,y\right)=2x+y\end{array}$

The Euler’s formula for $x_{n+1}$ is shown below.

  $x_{n+1}=x_n+h\left(x_n+2y_n\right)$ …… (1)

The Euler’s formula for $y_{n+1}$ is shown below.

  $y_{n+1}=y_n+h\left(2x_n+y_n\right)$ …… (2)

Substitute 0 for $n$ in equation (1)

  $x_{0+1}=x_0+h\left(x_0+2y_0\right)$ …… (3)

Substitute 0 for $n$ in equation (2)

  $y_{0+1}=y_0+h\left(2x_0+y_0\right)$ …… (4)

Substitute 0 for $x_0$ , $0.1$ for h and $2$ for $y_0$ in equation (3).

  $\begin{array}{c}{x}_{0+1}=0+0.1\left(0+2\left(2\right)\right)\\ =0.4\end{array}$

Substitute 0 for $x_0$ , $0.1$ for h and $2$ for $y_0$ in equation (4).

  $\begin{array}{c}{y}_{0+1}=2+0.1\left(2\left(0\right)+2\right)\\ =2+0.2\\ =2.2\end{array}$

Substitute 1 for $n$ in equation (1)

  $x_2=x_1+h\left(x_1+2y_1\right)$ …… (5)

Substitute 1 for $n$ in equation (2)

  $y_2=y_1+h\left(2x_1+y_1\right)$ …… (6)

Substitute $0.4$ for $x_1$ , $0.1$ for h and $2.2$ for $y_1$ in equation (5).

  $\begin{array}{c}{x}_2=0.4+0.1\left(0.4+2\left(2.2\right)\right)\\ =0.4+0.48\\ =0.88\end{array}$

Substitute $0.4$ for $x_1$ , $0.1$ for h and $2.2$ for $y_1$ in equation (6).

  $\begin{array}{c}{y}_2=2.2+0.1\left(2\left(0.4\right)+2.2\right)\\ =2.2+0.3\\ =2.5\end{array}$

Therefore, the value of $x_2\text{ and }{y}_2$ is $0.88\text{ and 2}\text{.5}$

Conclusion:

Thus, the approximate values of $x\left(0.2\right)\text{ and }y\left(0.2\right)$ by Euler’s method is $0.8800\text{ and 2}\text{.5000}$ .

(b)

Program Plan Intro

Program Description: Purpose of the problem is to find the approximate values of $x\left(0.2\right)\text{ and }y\left(0.2\right)$ by improved Euler’s method.

Explanation of Solution

Given information:

The differential problem is $x^{\prime }=x+2y,x\left(0\right)=0\text{ and }{y}^{\prime }=2x+y,y\left(0\right)=2$ .

The exact solution of the differential equation is $x\left(t\right)=e^{3t}-e^{-t}$ and $y\left(t\right)=e^{3t}+e^{-t}$ .

The two step size is $0.2$ .

Calculation:

The improved Euler’s formula for predicators $u_{n+1}$ is shown below.

  $u_{n+1}=x_n+h\left(x_n+2y_n\right)$ …… (7)

The improved Euler’s formula for predicators $v_{n+1}$ is shown below.

  $v_{n+1}=y_n+h\left(2x_n+y_n\right)$ …… (8)

Substitute 0 for $n$ in equation (7)

  $u_{0+1}=x_0+h\left(x_0+2y_0\right)$ …… (9)

Substitute 0 for $n$ in equation (8)

  $v_{0+1}=y_0+h\left(2x_0+y_0\right)$ …… (10)

Substitute 0 for $x_0$ , $0.2$ for h and $2$ for $y_0$ in equation (9).

  $\begin{array}{c}{u}_{0+1}=0+0.2\left(0+2\left(2\right)\right)\\ =0.8\end{array}$

Substitute 0 for $x_0$ , $0.2$ for h and $2$ for $y_0$ in equation (10).

  $\begin{array}{c}{v}_{0+1}=2+0.2\left(2\left(0\right)+2\right)\\ =2+0.2\\ =2.4\end{array}$

The improved Euler’s formula for correctors is shown below.

  $x_{n+1}=x_n+\frac{h}{2}\left(\left(x_n+2y_n\right)+\left(u_{n+1}+2v_{n+1}\right)\right)$ …… (11)

The improved Euler’s formula for correctors $x_1$ is shown below.

  $y_{n+1}=y_n+\frac{h}{2}\left(\left(2x_n+y_n\right)+\left(2u_{n+1}+v_{n+1}\right)\right)$ …… (12)

Substitute 0 for $n$ in equation (11)

  $x_1=x_0+\frac{h}{2}\left(\left(x_0+2y_0\right)+\left(u_1+2v_1\right)\right)$ …… (13)

Substitute 0 for $n$ in equation (12)

  $y_1=y_0+\frac{h}{2}\left(\left(2x_0+y_0\right)+\left(2u_1+v_1\right)\right)$ …… (14)

Substitute 0 for $x_0$ , $0.2$ for h , $0.8$ for $u_1$ , $2.4$ for $v_1$ and $2$ for $y_0$ in equation (13).

  $\begin{array}{c}{x}_1=0+\frac{0.2}{2}\left(\left(0+2\left(2\right)\right)+\left(0.8+2\left(2.4\right)\right)\right)\\ =0.1\left(9.6\right)\\ =0.96\end{array}$

Substitute 0 for $x_0$ , $0.2$ for h , $0.8$ for $u_1$ , $2.4$ for $v_1$ and $2$ for $y_0$ in equation (14).

  $\begin{array}{c}{y}_1=2+\frac{0.2}{2}\left(\left(2\left(0\right)+2\right)+\left(2\left(0.8\right)+2.4\right)\right)\\ =2+0.2+0.4\\ =2.6\end{array}$

Therefore, the value of $x_1\text{ and }{y}_1$ is $0.96\text{ and 2}\text{.6}$

Conclusion:

Thus, the approximate values of $x\left(0.2\right)\text{ and }y\left(0.2\right)$ by improved Euler’s method is $0.9600\text{ and 2}\text{.6000}$ .

(c)

Program Plan Intro

Program Description: Purpose of the problem is to find the approximate values of $x\left(0.2\right)\text{ and }y\left(0.2\right)$ by Runge-Kutta method and also compare the result with all the obtained results.

Explanation of Solution

Given information:

The differential problem is $x^{\prime }=x+2y,x\left(0\right)=0\text{ and }{y}^{\prime }=2x+y,y\left(0\right)=2$ .

The exact solution of the differential equation is $x\left(t\right)=e^{3t}-e^{-t}$ and $y\left(t\right)=e^{3t}+e^{-t}$ .

The two step size is $0.2$ .

Calculation:

The Runge-Kutta iteration formulas are shown below.

  $x_{n+1}=x_n+\frac{h}{6}\left(F_1+2F_2+2F_3+F_4\right)$ …… (15)

The Runge-Kutta iteration formulas are shown below

  $y_{n+1}=y_n+\frac{h}{6}\left(G_1+2G_2+2G_3+G_4\right)$ …… (16)

The value of $F_1,F_2,F_3\text{ and }{F}_4$ is shown below.

  $\begin{array}{c}{F}_1=x_0+2y_0\\ F_2=\left(x_0+\frac{h}{2}{F}_1\right)+2\left(y_0+\frac{h}{2}{G}_1\right)\\ F_3=\left(x_0+\frac{h}{2}{F}_2\right)+2\left(y_0+\frac{h}{2}{G}_2\right)\\ F_4=\left(x_0+h{F}_3\right)+2\left(y_0+h{G}_3\right)\end{array}$

The value of $G_1,G_2,G_3\text{ and }{G}_4$ is shown below.

  $\begin{array}{c}{G}_1=2x_0+y_0\\ G_2=2\left(x_0+\frac{h}{2}{F}_1\right)+\left(y_0+\frac{h}{2}{G}_1\right)\\ G_3=2\left(x_0+\frac{h}{2}{F}_2\right)+\left(y_0+\frac{h}{2}{G}_2\right)\\ G_4=2\left(x_0+h{F}_3\right)+\left(y_0+h{G}_3\right)\end{array}$

Substitute 0 for $x_0$ ,and $2$ for $y_0$ in $F_1=x_0+2y_0$ as,

  $\begin{array}{c}{F}_1=0+4\\ =4\end{array}$

Substitute 0 for $x_0$ , and $2$ for $y_0$ in $G_1=2x_0+y_0$ as,

  $\begin{array}{c}{G}_1=2\left(0\right)+2\\ =2\end{array}$

Substitute 0 for $x_0$ , $0.2$ for h , $4$ for $F_1$ , $2$ for $G_1$ and $2$ for $y_0$ in $F_2=\left(x_0+\frac{h}{2}{F}_1\right)+2\left(y_0+\frac{h}{2}{G}_1\right)$ as,

  $\begin{array}{c}{F}_2=\left(0+\frac{0.2}{2}\left(4\right)\right)+2\left(2+\frac{0.2}{2}\left(2\right)\right)\\ =0.4+4.4\\ =4.8\end{array}$

Substitute 0 for $x_0$ , $0.2$ for h , $4$ for $F_1$ , $2$ for $G_1$ and $2$ for $y_0$ in $G_2=2\left(x_0+\frac{h}{2}{F}_1\right)+\left(y_0+\frac{h}{2}{G}_1\right)$ as,

  $\begin{array}{c}{G}_2=2\left(0+\frac{0.2}{2}\left(4\right)\right)+\left(2+\frac{0.2}{2}\left(2\right)\right)\\ =0.8+2.2\\ =3\end{array}$

Substitute 0 for $x_0$ , $0.2$ for h , $4.8$ for $F_2$ , $3$ for $G_2$ and $2$ for $y_0$ in $F_3=\left(x_0+\frac{h}{2}{F}_2\right)+2\left(y_0+\frac{h}{2}{G}_2\right)$ as,

  $\begin{array}{c}{F}_3=\left(0+\frac{0.2}{2}\left(4.8\right)\right)+2\left(2+\frac{0.2}{2}\left(3\right)\right)\\ =0.48+4.6\\ =5.08\end{array}$

Substitute 0 for $x_0$ , $0.2$ for h , $4.8$ for $F_2$ , $3$ for $G_2$ and $2$ for $y_0$ in $G_3=2\left(x_0+\frac{h}{2}{F}_2\right)+\left(y_0+\frac{h}{2}{G}_2\right)$ as,

  $\begin{array}{c}{G}_3=2\left(0+\frac{0.2}{2}\left(4.8\right)\right)+\left(2+\frac{0.2}{2}\left(3\right)\right)\\ =0.96+2.3\\ =3.26\end{array}$

Substitute 0 for $x_0$ , $0.2$ for h , $5.08$ for $F_3$ , $3.26$ for $G_3$ and $2$ for $y_0$ in equation $F_4=\left(x_0+h{F}_3\right)+2\left(y_0+h{G}_3\right)$ as,

  $\begin{array}{c}{F}_4=\left(0+0.2\left(5.08\right)\right)+2\left(2+0.2\left(3.26\right)\right)\\ =1.016+5.304\\ =6.32\end{array}$

Substitute 0 for $x_0$ , $0.2$ for h , $5.08$ for $F_3$ , $3.26$ for $G_3$ and $2$ for $y_0$ in $G_4=2\left(x_0+h{F}_3\right)+\left(y_0+h{G}_3\right)$ as,

  $\begin{array}{c}{G}_4=2\left(0+0.2\left(5.08\right)\right)+\left(2+0.2\left(3.26\right)\right)\\ =2.032+2.652\\ =4.684\end{array}$

Substitute $5.08$ for $F_3$ , 0 for $x_0$ , $0.2$ for h , $3$ for $G_2$ , $4$ for $F_1$ , and $6.32$ for $F_4$ in equation (15).

  $\begin{array}{c}{x}_{n+1}=0+\frac{0.2}{6}\left(4+2\left(4.8\right)+2\left(5.08\right)+6.32\right)\\ =\frac{0.2}{6}\left(30.08\right)\\ =1.0027\end{array}$

Substitute $2$ for $G_1$ , 2 for $y_0$ , $0.2$ for h , $4.8$ for $F_2$ , $3.26$ for $G_3$ and $4.684$ for $G_4$ in equation (16).

  $\begin{array}{c}{y}_{n+1}=2++\frac{0.2}{6}\left(2+2\left(3\right)+2\left(3.26\right)+4.684\right)\\ =2+\frac{0.2}{6}\left(19.204\right)\\ =2+0.6401\\ =2.64101\end{array}$

The exact solution $x\left(0.2\right)$ can be obtained as,

  $\begin{array}{c}x\left(0.2\right)=e^{3\left(0.2\right)}-e^{-0.2}\\ =1.0034\end{array}$

The exact solution $y\left(0.2\right)$ can be obtained as,

  $\begin{array}{c}y\left(0.2\right)=e^{3\left(0.2\right)}+e^{-0.2}\\ =2.6408\end{array}$

It can be observed that the Runge-Kutta method is more closes to the exact solution of the differential equation.

Conclusion:

Thus, the approximate values of $x\left(0.2\right)\text{ and }y\left(0.2\right)$ by Runge-Kutta method is $1.0027\text{ and 2}\text{.64101}$ .