Chapter 27, Problem 27P

The following nonlinear, parasitic ODE was suggested by Hornbeck (1975):

$\frac{d{y}_1}{dt}=5\left(y_1-t^2\right)$

If the initial condition is $y_1\left(0\right)=0.08$, obtain a solution from $t=0\text{ to 5}$:

(a) Analytically

(b) Using the fourth-order RK method with a constant step size of 0.03125.

(c) Using the MATLAB function ode45.

(d) Using the MATLAB function ode23s.

(e) Using the MATLAB function ode23tb.

Present your results in graphical form.

To determine

To calculate: The analytical solution of the nonlinear, parasitic ordinary differential equation, $\frac{d{y}_1}{dt}=5(y_1-t^2)$, where the initial condition is $y_1\left(0\right)=0.08$ and $t=0\text{ to }5$.

Answer to Problem 27P

Solution:

The analytical solution of the differential equation is $y_1\left(t\right)=t^2+0.4t+0.08$.

Explanation of Solution

Given Information:

A nonlinear, parasitic ordinary differential equation, $\frac{d{y}_1}{dt}=5(y_1-t^2)$ with initial condition $y_1\left(0\right)=0.08$ and $t=0\text{ to }5$.

Formula used:

The general linear differential equation is,

$\frac{d{y}_1}{dt}+P\left(t\right)y_1=Q\left(t\right)$

Calculation:

Consider the nonlinear ordinary differential equation, $\frac{d{y}_1}{dt}=5(y_1-t^2)$.

Rearrange the above differential equationto get,

$\frac{d{y}_1}{dt}-5y_1=-5t^2$

Now, compare the above differential equation with the general linear differential equation

$\frac{d{y}_1}{dt}+P\left(t\right)y_1=Q\left(t\right)$

Thus,

$P\left(t\right)=-5\text{ and }Q\left(t\right)=-5t^2$

Now, find the integrating factor (I. F.) as shown below,

$\begin{array}{c}I.F=e^{{\displaystyle \int -5dt}}\\ =e^{-5t}\end{array}$

Therefore, the solution of the linear differential equation is given as,

$y_1\left(I.F.\right)={\displaystyle \int Q\left(t\right)}\left(I.F.\right)dt+c$

Substitute the value of integrating factor (I. F.) in above equation,

$\begin{array}{c}{y}_1\left(I.F.\right)={\displaystyle \int Q\left(t\right)}\left(I.F.\right)dt+c\\ y_1\cdot e^{-5t}={\displaystyle \int -5t^2}\cdot e^{-5t}dt+c\end{array}$

Now, integrate the right-hand-side of the above equation,

$\begin{array}{c}{y}_1\cdot e^{-5t}={\displaystyle \int -5t^2}\cdot e^{-5t}dt+c\\ y_1\cdot e^{-5t}=\frac{1}{25}{e}^{-5t}\left(25t^2+10t+2\right)+c\end{array}$

Solve further, to get

$\begin{array}{c}{y}_1=\frac{1}{25}{e}^{5t}{e}^{-5t}\left(25t^2+10t+2\right)+c{e}^{5t}\\ =\frac{1}{25}\left(25t^2+10t+2\right)+c{e}^{5t}\\ =t^2+0.4t+0.08+c{e}^{5t}\end{array}$

$$

Thus, the solution is $y_1\left(t\right)=t^2+0.4t+0.08+c{e}^{5t}$.

Now, to determine the constant c, use the initial condition $y_1\left(0\right)=0.08$. Thus,

$\begin{array}{c}{y}_1\left(0\right)={\left(0\right)}^2+0.4\left(0\right)+0.08+c{e}^{5\left(0\right)}\\ 0.08=0+0+0.08+c\\ c=0\end{array}$

Substitute $c=0$ in the solution obtained above, that is, $y_1\left(t\right)$.

Therefore,

$\begin{array}{l}{y}_1\left(t\right)=t^2+0.4t+0.08+c{e}^{5t}\\ y_1\left(t\right)=t^2+0.4t+0.08+\left(0\right)e^{5t}\\ y_1\left(t\right)=t^2+0.4t+0.08\end{array}$

Hence, the solution of the differential equation is $y_1\left(t\right)=t^2+0.4t+0.08$.

To determine

To calculate: Thesolution of the nonlinear, parasitic ordinary differential equation, $\frac{d{y}_1}{dt}=5(y_1-t^2)$, using the fourth-order RK method with a constant step size of 0.03125.

Answer to Problem 27P

Solution:

The graph of the solution of the differential equation is,

Explanation of Solution

Given Information:

A nonlinear, parasitic ordinary differential equation, $\frac{d{y}_1}{dt}=5(y_1-t^2)$ step size of 0.03125.

Calculation:

Consider the nonlinear ordinary differential equation, $\frac{d{y}_1}{dt}=5(y_1-t^2)$.

The VBA code to solve the above differential equation with fourth-order RK method with a constant step size of 0.03125 is given below,

OptionExplicit

Subfind()

'Define dimension of variables

Dim t AsDouble, y1 AsDouble, h AsDouble,hhAsDouble

'Reset the values

t =0

y1 =0.08

h =0.03125

'move to the cell b15

Range("b15").Select

'Assign name to columns

ActiveCell. Value="4th order RK"

ActiveCell. Offset(1,0).Select

ActiveCell. Value="t"

ActiveCell. Offset(0,1).Select

ActiveCell. Value="y1"

'determine y1 at different values of t using RK method of 4th order

hh= RK4(t, y1, h)

EndSub

'define function derivative

Functiondrive(t, y1)

'define dimension of variable temp

Dim temp AsDouble

'evaluate the value of derivative

temp =5*(y1 -(t * t))

drive = temp

EndFunction

'define function

Function RK4(t, y1, h)

'define dimension of variables

Dim k1 AsDouble, k2 AsDouble, k3 AsDouble, k4 AsDouble,yanalAsDouble

Dim cm AsDouble, cm1 AsDouble,ceAsDouble,slpAsDouble,cnewAsDouble

Dim j AsInteger

'evaluate k1, k2, k3 and k4

For j =1To160

'Move to the specified cell b16

Range("b16").Select

ActiveCell. Offset(j,0).Select

ActiveCell. Value= t

ActiveCell. Offset(0,1).Select

ActiveCell. Value= y1

'Call drive function to find k1

k1 =drive(t, y1)

cm = y1 +(k1 *(h /2))

'Call drive function to find k2

k2 =drive(t +(h /2), cm)

cm1 = y1 +(k2 *(h /2))

'Call drive function to find k3

k3 =drive(t +(h /2), cm1)

ce= y1 + k3 * h

'Call drive function to find k4

k4 =drive(t + h,ce)

slp=(k1 +2*(k2 + k3)+ k4)/6

cnew= y1 +(slp* h)

yanal=yanalfun(t)

ActiveCell. Offset(0,1).Select

ActiveCell. Value=yanal

t = t + h

y1 =cnew

Next

EndFunction

'define function

Functionyanalfun(t)

'define dimension of variable temp1

Dim temp1 AsDouble

'evaluate y1 analytically

temp1 =(t * t +0.4* t +0.08)

yanalfun= temp1

EndFunction

The output given below is obtained in the Excel after the execution of the above code:

4th order RK
t	y1
0	0.08	0.08
0.03125	0.093476	0.093477
0.0625	0.108906	0.108906
0.09375	0.126289	0.126289
0.125	0.145625	0.145625
0.15625	0.166914	0.166914
0.1875	0.190156	0.190156
0.21875	0.215351	0.215352
0.25	0.242499	0.2425
0.28125	0.2716	0.271602
0.3125	0.302655	0.302656
0.34375	0.335662	0.335664
0.375	0.370622	0.370625
0.40625	0.407536	0.407539
0.4375	0.446403	0.446406
0.46875	0.487222	0.487227
0.5	0.529995	0.53
0.53125	0.57472	0.574727
0.5625	0.621399	0.621406
0.59375	0.670031	0.670039
0.625	0.720615	0.720625
0.65625	0.773152	0.773164
0.6875	0.827642	0.827656
0.71875	0.884085	0.884102
0.75	0.942481	0.9425
0.78125	1.002829	1.002852
0.8125	1.06513	1.065156
0.84375	1.129383	1.129414
0.875	1.195589	1.195625
0.90625	1.263747	1.263789
0.9375	1.333857	1.333906
0.96875	1.405919	1.405977
1	1.479932	1.48
1.03125	1.555897	1.555977
1.0625	1.633814	1.633906
1.09375	1.713681	1.713789
1.125	1.795498	1.795625
1.15625	1.879266	1.879414
1.1875	1.964983	1.965156
1.21875	2.052649	2.052852
1.25	2.142263	2.1425
1.28125	2.233824	2.234102
1.3125	2.327332	2.327656
1.34375	2.422785	2.423164
1.375	2.520181	2.520625
1.40625	2.61952	2.620039
1.4375	2.7208	2.721406
1.46875	2.824017	2.824727
1.5	2.929171	2.93
1.53125	3.036257	3.037227
1.5625	3.145273	3.146406
1.59375	3.256214	3.257539
1.625	3.369075	3.370625
1.65625	3.483852	3.485664
1.6875	3.600538	3.602656
1.71875	3.719125	3.721602
1.75	3.839605	3.8425
1.78125	3.961966	3.965352
1.8125	4.086198	4.090156
1.84375	4.212287	4.216914
1.875	4.340215	4.345625
1.90625	4.469964	4.476289
1.9375	4.601512	4.608906
1.96875	4.734831	4.743477
2	4.869893	4.88
2.03125	5.00666	5.018477
2.0625	5.145091	5.158906
2.09375	5.285137	5.301289
2.125	5.426742	5.445625
2.15625	5.569837	5.591914
2.1875	5.714346	5.740156
2.21875	5.860176	5.890352
2.25	6.007221	6.0425
2.28125	6.155356	6.196602
2.3125	6.304436	6.352656
2.34375	6.454288	6.510664
2.375	6.604715	6.670625
2.40625	6.755482	6.832539
2.4375	6.906318	6.996406
2.46875	7.056903	7.162227
2.5	7.206864	7.33
2.53125	7.355766	7.499727
2.5625	7.503099	7.671406
2.59375	7.648268	7.845039
2.625	7.790577	8.020625
2.65625	7.92921	8.198164
2.6875	8.063218	8.377656
2.71875	8.191486	8.559102
2.75	8.312714	8.7425
2.78125	8.425381	8.927852
2.8125	8.527709	9.115156
2.84375	8.617619	9.304414
2.875	8.69268	9.495625
2.90625	8.750052	9.688789
2.9375	8.786412	9.883906
2.96875	8.797877	10.08098
3	8.779905	10.28
3.03125	8.727189	10.48098
3.0625	8.633523	10.68391
3.09375	8.491649	10.88879
3.125	8.293087	11.09563
3.15625	8.027917	11.30441
3.1875	7.684546	11.51516
3.21875	7.249417	11.72785
3.25	6.706683	11.9425
3.28125	6.037815	12.1591
3.3125	5.221152	12.37766
3.34375	4.231369	12.59816
3.375	3.038857	12.82063
3.40625	1.609002	13.04504
3.4375	-0.09867	13.27141
3.46875	-2.13146	13.49973
3.5	-4.5447	13.73
3.53125	-7.40305	13.96223
3.5625	-10.7821	14.19641
3.59375	-14.7703	14.43254
3.625	-19.4709	14.67063
3.65625	-25.0048	14.91066
3.6875	-31.5132	15.15266
3.71875	-39.1613	15.3966
3.75	-48.1421	15.6425
3.78125	-58.6814	15.89035
3.8125	-71.043	16.14016
3.84375	-85.5354	16.39191
3.875	-102.519	16.64563
3.90625	-122.417	16.90129
3.9375	-145.72	17.15891
3.96875	-173.006	17.41848
4	-204.949	17.68
4.03125	-242.336	17.94348
4.0625	-286.088	18.20891
4.09375	-337.283	18.47629
4.125	-397.179	18.74563
4.15625	-467.248	19.01691
4.1875	-549.211	19.29016
4.21875	-645.079	19.56535
4.25	-757.205	19.8425
4.28125	-888.338	20.1216
4.3125	-1041.69	20.40266
4.34375	-1221.03	20.68566
4.375	-1430.74	20.97063
4.40625	-1675.96	21.25754
4.4375	-1962.71	21.54641
4.46875	-2297.99	21.83723
4.5	-2690.02	22.13
4.53125	-3148.39	22.42473
4.5625	-3684.34	22.72141
4.59375	-4310.97	23.02004
4.625	-5043.62	23.32063
4.65625	-5900.23	23.62316
4.6875	-6901.75	23.92766
4.71875	-8072.7	24.2341
4.75	-9441.73	24.5425
4.78125	-11042.3	24.85285
4.8125	-12913.7	25.16516
4.84375	-15101.5	25.47941
4.875	-17659.5	25.79563
4.90625	-20650.1	26.11379
4.9375	-24146.4	26.43391
4.96875	-28234.2	26.75598

Now, plot the following chart using the data obtained in Excel.

In the above plot, the series1 represent the numerical solution whereas the series2 represent the exact solution.

To determine

The solution of the nonlinear, parasitic ordinary differential equation, $\frac{d{y}_1}{dt}=5(y_1-t^2)$, using the MATLAB function ode45.

Answer to Problem 27P

Solution:

The graph of the solution of the differential equation is,

Explanation of Solution

Given Information:

A nonlinear, parasitic ordinary differential equation, $\frac{d{y}_1}{dt}=5(y_1-t^2)$.

Consider the nonlinear ordinary differential equation, $\frac{d{y}_1}{dt}=5(y_1-t^2)$.

Use the MATLAB function ODE45 to solve the above differential equation. The result is obtained in graphical form.

Write the code given below in MATLAB editor window and save it.

%create a function yp

functionyp=dy(t,y)

%calculate the value of yp

yp=5*(y-t^2);

Now, write the following code in MATLAB command window

Following graph is obtained after the execution of the above MATLAB code.

To determine

The solution of the nonlinear, parasitic ordinary differential equation, $\frac{d{y}_1}{dt}=5(y_1-t^2)$, using the MATLAB function ode23s.

Answer to Problem 27P

Solution:

The graph of the solution of the differential equation is,

Explanation of Solution

Given Information:

A nonlinear, parasitic ordinary differential equation, $\frac{d{y}_1}{dt}=5(y_1-t^2)$.

Consider the nonlinear ordinary differential equation, $\frac{d{y}_1}{dt}=5(y_1-t^2)$.

Use the MATLAB function ODE23s to solve the above differential equation. The result is obtained in graphical form.

Write the code given below in MATLAB editor window and save it.

%create a function yp

functionyp=dy(t,y)

%calculate the value of yp

yp=5*(y-t^2);

Now, write the following code in MATLAB command window

Following graph is obtained after the execution of the above MATLAB code.

To determine

The solution of the nonlinear, parasitic ordinary differential equation, $\frac{d{y}_1}{dt}=5(y_1-t^2)$, using the MATLAB function ode23s.

Answer to Problem 27P

Solution:

The graph of the solution of the differential equation is,

Explanation of Solution

Given Information:

A nonlinear, parasitic ordinary differential equation, $\frac{d{y}_1}{dt}=5(y_1-t^2)$.

Consider the nonlinear ordinary differential equation, $\frac{d{y}_1}{dt}=5(y_1-t^2)$.

Use the MATLAB function ODE23s to solve the above differential equation. The result is obtained in graphical form.

Write the code given below in MATLAB editor window and save it.

%create a function yp

functionyp=dy(t,y)

%calculate the value of yp

yp=5*(y-t^2);

Now, write the following code in MATLAB command window

Following graph is obtained after the execution of the above MATLAB code.