## Classical Solution Technique to Solve a First Order ODE

I have a audiovisual digital lecture on YouTube that shows the use of Euler’s method to solve a first order ordinary differential equation (ODE).  To show the accuracy of Euler’s method,  I compare the approximate answer to the exact answer.  A YouTube viewer asked me: How did I get the exact answer?

In this blog, I use the classical solution technique to find the exact answer to the ODE.  In a previous blog, I showed how to find the exact answer to the ODE by the integrating factor method.

The pdf file of the solution is also available.

## Example: Solving First Order Linear ODE by Integrating Factor

I have a audiovisual digital lecture on YouTube that shows the use of Euler’s method to solve a first order ordinary differential equation (ODE).  To show the accuracy of Euler’s method,  I compare the approximate answer to the exact answer.  A YouTube viewer asked me: How did I get the exact answer?

In this blog, I use the integrating factor method to find the exact answer, because that is the method the viewer was using to solve the ODE exactly.  So here it is and in two future blogs, I will show the same example being solved by 1) Laplace transforms and 2) the classical (complementary + particular) solution techniques.

The pdf file of the solution is also available.

This post is brought to you by

## How do I numerically solve an ODE in MATLAB?

The other day a student came to ask me for help in solving a second order ordinary differential equation using the ode45 routine of MATLAB.  To use ode45, one needs to be familiar with how the inputs are required by MATLAB.  The understanding of these inputs is important to use ode45 successfully in problems that are more complex than solving a second order ODE.

The ordinary differential equation was
2y”+3y’+5y=7 exp(-x), y(0)=11, dy/dx(0)=13
This has to put in the state variable form by reducing it by using
y’=z
That gives
y’=z with the corresponding initial conditions as y(0)=11
Then
2y”+3y’+5y=7 exp(-x)
reduces to
2z’ + 3z+5y=7exp(-x)
z’ =(7exp(-x)-3z-5y)/2 with the corresponding initial conditions as z(0)=13

So as needed by MATLAB, call y as y(1) and z as y(2)
dy(1)=y(2), y(1) at x=0 is 11
dy(2)=(7exp(-x)-3y(2)-5y(1))/2, y(2) at x=0 is 13

These equations are now put in a MATLAB function we call odestate.m
dy=zeros(2,1);
dy(1)=y(2);
dy(2)=(7*exp(-x)-3*y(2)-5*y(1))/2;

To solve the ODE, the
The inputs are
1) the function odestate
2) The outputs are required between x=0 and x=17,
hence entered as [0 17]
3) The initial conditions are y(0)=11 and dy/dx(0)=13,
hence entered as [11  13]

The outputs are
1) X= array of x values between 0 and 17
2) Y= matrix of 2 columns;
first column is the y(x)
second column is dy/dx(x)
The MATLAB code then is
[X,Y]=ode45(@odestate,[0  17],[11 13]);

Click the links for the MATLAB mfiles for the function odestate.m and the ODE solver odetest.m

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This post is brought to you by Holistic Numerical Methods: Numerical Methods for the STEM undergraduate at http://numericalmethods.eng.usf.edu, the textbook on Numerical Methods with Applications available from the lulu storefront, and the YouTube video lectures available at http://numericalmethods.eng.usf.edu/videos

## How do I solve a boundary value ODE in MATLAB?

Many students ask me how do I do this or that in MATLAB.  So I thought why not have a small series of my next few blogs do that.  In this blog, I show you how to solve a boundary value ordinary differential equation.

• The MATLAB program link is here.
• The HTML version of the MATLAB program is here.
• DO NOT COPY AND PASTE THE PROGRAM BELOW BECAUSE THE SINGLE QUOTES DO NOT TRANSLATE TO THE CORRECT SINGLE QUOTES IN MATLAB EDITOR.  DOWNLOAD THE MATLAB PROGRAM INSTEAD

%% HOW DO I DO THAT IN MATLAB SERIES?
% In this series, I am answering questions that students have asked
% me about MATLAB.  Most of the questions relate to a mathematical
% procedure.

%% TOPIC
% How do I solve a boundary value ordinary differential equation?

%% SUMMARY

% Language : Matlab 2008a;
% Authors : Autar Kaw;
% Mfile available at
% http://numericalmethods.eng.usf.edu/blog/ode_boundaryvalue.m;
% Last Revised : May 26, 2009;
% Abstract: This program shows you how to solve an
%           boundary value ordinary differential equation.
clc
clear all

%% INTRODUCTION

disp(‘ABSTRACT’)
disp(‘   This program shows you how to solve’)
disp(‘   a boundary value ordinary differential equation’)
disp(‘ ‘)
disp(‘AUTHOR’)
disp(‘   Autar K Kaw of https://autarkaw.wordpress.com&#8217;)
disp(‘ ‘)
disp(‘MFILE SOURCE’)
disp(‘   http://numericalmethods.eng.usf.edu/blog/ode_boundaryvalue.m&#8217;)
disp(‘ ‘)
disp(‘LAST REVISED’)
disp(‘   May 26, 2009’)
disp(‘ ‘)

%% INPUTS
% Solve the ordinary differential equation
% y”-0.0000002y=0.000000075*x*(75-x)
% Define x as a symbol
syms x
%The ODE
ode_eqn=’D2y-0.000002*y=0.000000075*x*(75-x)’;
% The initial conditions
iv_1=’y(0)=0′;
iv_2=’y(75)=0′;
% The value at which y is sought at
xval=37.5;
%% DISPLAYING INPUTS

disp(‘INPUTS’)
func=[‘  The ODE to be solved is ‘ ode_eqn];
disp(func)
iv_explain=[‘  The boundary conditions are ‘ iv_1 ‘    ‘ iv_2];
disp(iv_explain)
fprintf(‘  The value of y is sought at x=%g’,xval)
disp(‘  ‘)

%% THE CODE

% Finding the solution of the ordinary differential equation
soln=dsolve(ode_eqn,iv_1,iv_2,’x’);
% vpa below uses variable-precision arithmetic (VPA) to compute each
% element of soln to 5 decimal digits of accuracy
soln=vpa(soln,5);

%% DISPLAYING OUTPUTS
disp(‘  ‘)
disp(‘OUTPUTS’)
output=[‘  The solution to the ODE is ‘ char(soln)];
disp(output)
value=subs(soln,x,xval);
fprintf(‘  The value of y at x=%g is %g’,xval,value)
disp(‘  ‘)

This post is brought to you by Holistic Numerical Methods: Numerical Methods for the STEM undergraduate at http://numericalmethods.eng.usf.edu, the textbook on Numerical Methods with Applications available from the lulu storefront, and the YouTube video lectures available at http://numericalmethods.eng.usf.edu/videos

## How do I solve an initial value ODE in MATLAB?

Many students ask me how do I do this or that in MATLAB.  So I thought why not have a small series of my next few blogs do that.  In this blog, I show you how to solve an initial value ordinary differential equation.

• The MATLAB program link is here.
• The HTML version of the MATLAB program is here.
• DO NOT COPY AND PASTE THE PROGRAM BELOW BECAUSE THE SINGLE QUOTES DO NOT TRANSLATE TO THE CORRECT SINGLE QUOTES IN MATLAB EDITOR.  DOWNLOAD THE MATLAB PROGRAM INSTEAD

%% HOW DO I DO THAT IN MATLAB SERIES?
% In this series, I am answering questions that students have asked
% me about MATLAB.  Most of the questions relate to a mathematical
% procedure.

%% TOPIC
% How do I solve an initial value ordinary differential equation?

%% SUMMARY

% Language : Matlab 2008a;
% Authors : Autar Kaw;
% Mfile available at
% http://numericalmethods.eng.usf.edu/blog/ode_initial.m;
% Last Revised : May 14, 2009;
% Abstract: This program shows you how to solve an
%           initial value ordinary differential equation.
clc
clear all

%% INTRODUCTION

disp(‘ABSTRACT’)
disp(‘   This program shows you how to solve’)
disp(‘   an initial value ordinary differential equation’)
disp(‘ ‘)
disp(‘AUTHOR’)
disp(‘   Autar K Kaw of https://autarkaw.wordpress.com&#8217;)
disp(‘ ‘)
disp(‘MFILE SOURCE’)
disp(‘   http://numericalmethods.eng.usf.edu/blog/ode_initial.m&#8217;)
disp(‘ ‘)
disp(‘LAST REVISED’)
disp(‘   May 14, 2009’)
disp(‘ ‘)

%% INPUTS
% Solve the ordinary differential equation 3y”+5y’+7y=11exp(-x)
% Define x as a symbol
syms x
%The ODE
ode_eqn=’3*D2y+5*Dy+7*y=11*exp(-13*x)’;
% The initial conditions
iv_1=’Dy(0)=17′;
iv_2=’y(0)=19′;
% The value at which y is sought at
xval=23.0;
%% DISPLAYING INPUTS

disp(‘INPUTS’)
func=[‘  The ODE to be solved is ‘ ode_eqn];
disp(func)
iv_explain=[‘  The initial conditions are ‘ iv_1 ‘    ‘ iv_2];
disp(iv_explain)
fprintf(‘  The value of y is sought at x=%g’,xval)
disp(‘  ‘)

%% THE CODE

% Finding the solution of the ordinary differential equation
soln=dsolve(ode_eqn,iv_1,iv_2,’x’);
% vpa below uses variable-precision arithmetic (VPA) to compute each
% element of soln to 5 decimal digits of accuracy
soln=vpa(soln,5);

%% DISPLAYING OUTPUTS
disp(‘  ‘)
disp(‘OUTPUTS’)
output=[‘  The solution to the ODE is ‘ char(soln)];
disp(output)
value=subs(soln,x,xval);
fprintf(‘  The value of y at x=%g is %g’,xval,value)
disp(‘  ‘)

This post is brought to you by Holistic Numerical Methods: Numerical Methods for the STEM undergraduate at http://numericalmethods.eng.usf.edu, the textbook on Numerical Methods with Applications available from the lulu storefront, and the YouTube video lectures available at http://numericalmethods.eng.usf.edu/videos

## Runge-Kutta 2nd order equations derived

In my class, I present the 2nd order Runge-Kutta method equations without proof. Although I do discuss where the equations come from, there are still students who want to see the proof. So here it is.

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This post is brought to you by Holistic Numerical Methods: Numerical Methods for the STEM undergraduate at http://numericalmethods.eng.usf.edu

## A Matlab program for comparing Runge-Kutta methods

In a previous post, we compared the results from various 2nd order Runge-Kutta methods to solve a first order ordinary differential equation. In this post, I am posting the matlab program. It is better to download the program as single quotes in the pasted version do not translate properly when pasted into a mfile editor of MATLAB or see the html version for clarity and sample output .

Do your own testing on a different ODE, a different value of step size, a different initial condition, etc. See the inputs section below that is colored in bold brick.

% Simulation : Comparing the Runge Kutta 2nd order method of
% solving ODEs
% Language : Matlab 2007a
% Authors : Autar Kaw
% Last Revised : July 12, 2008
% Abstract: This program compares results from the
% exact solution to 2nd order Runge-Kutta methods
% of Heun’s method, Ralston’s method, Improved Polygon
% method, and directly using the three terms of Taylor series
clc
clear all
clc
clf
disp(‘This program compares results from the’)
disp(‘exact solution to 2nd order Runge-Kutta methods’)
disp(‘of Heuns method, Ralstons method, Improved Polygon’)
disp(‘ method, and directly using the three terms of Taylor series’)

%INPUTS. If you want to experiment these are only things
% you should and can change. Be sure that the ode has an exact
% solution
% Enter the rhs of the ode of form dy/dx=f(x,y)
fcnstr=’sin(5*x)-0.4*y’ ;
% Initial value of x
x0=0;
% Initial value of y
y0=5;
% Final value of y
xf=5.5;
% number of steps to go from x0 to xf.
% This determines step size h=(xf-x0)/n
n=10;

%REST OF PROGRAM
%Converting the input function to that can be used
f=inline(fcnstr) ;

% EXACT SOLUTION
syms x
eqn=[‘Dy=’ fcnstr]
% exact solution of the ode
exact_solution=dsolve(eqn,’y(0)=5′,’x’)
% geting points for plotting the exact solution
xx=x0:(xf-x0)/100:xf;
yy=subs(exact_solution,x,xx);
yexact=subs(exact_solution,x,xf);
plot(xx,yy,’.’)
hold on

% RUNGE-KUTTA METHODS
h=(xf-x0)/n;
% Heun’s method
a1=0.5;
a2=0.5;
p1=1;
q11=1;
xr=zeros(1,n+1);
yr=zeros(1,n+1);
%Initial values of x and y
xr(1)=x0;
yr(1)=y0;
for i=1:1:n
k1=f(xr(i),yr(i));
k2=f(xr(i)+p1*h,yr(i)+q11*k1*h);
yr(i+1)=yr(i)+(a1*k1+a2*k2)*h;
xr(i+1)=xr(i)+h;
end
%Value of y at x=xf
y_heun=yr(n+1);
% Absolute relative true error for value using Heun’s Method
et_heun=abs((y_heun-yexact)/yexact)*100;
hold on
xlabel(‘x’)
ylabel(‘y’)
title_name=[‘Comparing exact and Runge-Kutta methods with h=’ num2str(h)] ;
title(title_name)
plot(xr,yr, ‘color’,’magenta’,’LineWidth’,2)
% Midpoint Method (also called Improved Polygon Method)
a1=0;
a2=1;
p1=1/2;
q11=1/2;
%Initial values of x and y
xr(1)=x0;
yr(1)=y0;
for i=1:1:n
k1=f(xr(i),yr(i));
k2=f(xr(i)+p1*h,yr(i)+q11*k1*h);
yr(i+1)=yr(i)+(a1*k1+a2*k2)*h;
xr(i+1)=xr(i)+h;
end
%Value of y at x=xf
y_improved=yr(n+1);
% Absolute relative true error for value using Improved Polygon Method
et_improved=abs((y_improved-yexact)/yexact)*100;
hold on
plot(xr,yr,’color’,’red’,’LineWidth’,2)

% Ralston’s method
a1=1/3;
a2=2/3;
p1=3/4;
q11=3/4;
xr(1)=x0;
yr(1)=y0;
for i=1:1:n
k1=f(xr(i),yr(i));
k2=f(xr(i)+p1*h,yr(i)+q11*k1*h);
yr(i+1)=yr(i)+(a1*k1+a2*k2)*h;
xr(i+1)=xr(i)+h;
end
%Value of y at x=xf
y_ralston=yr(n+1);
% Absolute relative true error for value using Ralston’s Method
et_ralston=abs((y_ralston-yexact)/yexact)*100;
hold on
plot(xr,yr,’color’,’green’,’LineWidth’,2)

% Using first three terms of the Taylor series
syms x y;
fs=char(fcnstr);
% fsp=calculating f'(x,y) using chain rule
fsp=diff(fs,x)+diff(fs,y)*fs;
%Initial values of x and y
xr(1)=x0;
yr(1)=y0;
for i=1:1:n
k1=subs(fs,{x,y},{xr(i),yr(i)});
kk1=subs(fsp,{x,y},{xr(i),yr(i)});
yr(i+1)=yr(i)+k1*h+1/2*kk1*h^2;
xr(i+1)=xr(i)+h;
end
%Value of y at x=xf
y_taylor=yr(n+1);
% Absolute relative true error for value using Taylor series
et_taylor=abs((y_taylor-yexact)/yexact)*100;
hold on
plot(xr,yr,’color’,’black’,’LineWidth’,2)
hold off
legend(‘exact’,’heun’,’midpoint’,’ralston’,’taylor’,1)

% THE OUTPUT
fprintf(‘\nAt x = %g ‘,xf)
disp(‘ ‘)
disp(‘_________________________________________________________________’)
disp(‘Method Value Absolute Relative True Error’)
disp(‘_________________________________________________________________’)
fprintf(‘\nExact Solution %g’,yexact)
fprintf(‘\nHeuns Method %g %g ‘,y_heun,et_heun)
fprintf(‘\nImproved method %g %g ‘,y_improved,et_improved)
fprintf(‘\nRalston method %g %g ‘,y_ralston,et_ralston)
fprintf(‘\nTaylor method %g %g ‘,y_taylor,et_taylor)
disp( ‘ ‘)

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This post is brought to you by Holistic Numerical Methods: Numerical Methods for the STEM undergraduate at http://numericalmethods.eng.usf.edu