DIfferences on Sloped field using 'dsolve' and 'ode45'

I tried to create the slope field with 'dsolve' and I received the following results.

syms y(x)  % Define the symbolic function y(x)
% Define the differential equation
eq = diff(y, x) == sin(y);
% Solve the general solution of the differential equation
gsol = dsolve(eq);
% Define the initial condition and solve the particular solution
cond = y(0) == 1;
psol = dsolve(eq, cond);
% Plot the particular solution
fplot(psol, [-5 5]);
hold on;
% The slope field for the differential equation
[x, y] = meshgrid(-5:0.5:5, -1:0.5:5);
title('Particular Solution and Slope Field of the Differential Equation')
axis tight
m=sin(y);
L=sqrt(1+m.^2);
quiver(x,y,1./L,m./L)
hold off;

However when I used 'ode45' the results it were not what i wanted, as you can see bellow. What should I change to my code?

 %Define the function
f = @(u,v) sin(v);

% Solve the differential equation using ode45
[u,v] = ode45(f,[0:0.1:5],1);

% Plot the solution of the differential equation
plot(u, v, 'LineWidth', 2) % Increase line width for better visibility
hold on

% Create a meshgrid for the vector field
[x, y] = meshgrid(-5:0.5:5, -1:0.5:5);

% Calculate the vector field
m = sin(y);
L = sqrt(1 + m.^2);

% Plot the vector field using quiver
quiver(x, y, 1./L, m./L, 'r') % Use red color for vectors

% Set the axis limits to fit the data
axis tight

% Add title
title('Solution of the differential equation and vector field')

% Add a legend
legend('ode45 solution', 'Vector field')

% Turn off hold
hold off

% Improve the overall aesthetics
set(gca, 'FontSize', 7) % Set font size for readability
grid on % Add a grid for better readability of the plot

Answer 1

[Torsten]

The dsolve solution satisfies y(0) = 1:

cond = y(0) == 1;

the ode45 solution satisfies y(-5) = 1:

[u,v] = ode45(f,[-5 5],1);

The slopefield looks the same to me.

[Athanasios Paraskevopoulos]

@Torsten How can I adjust the initial condition to y(0)=1 for 'ode45' command?

[Torsten]

f = @(u,v) sin(v);
[ur,vr] = ode45(f,[0 5],1);
[ul,vl] = ode45(f,[0 -5],1);
u = [flipud(ul);ur];
v = [flipud(vl);vr];
plot(u, v, 'LineWidth', 2) 

[Athanasios Paraskevopoulos]

@Torsten Cool thank you very much

Answer 2

[Sam Chak]

Hi @Athanasios Paraskevopoulos

The reason for obtaining different results is due to the selection of an incorrect initial value for the ode45 run.

% Define the function
f = @(u,v) sin(v);

% Solve the differential equation using ode45
% [u, v] = ode45(f, [-5 5], 1);     % pick wrong initial value f(-5) = 1
[u, v] = ode45(f, [-5 5], 7.35825e-3);  % adjust initial value f(-5) so that f(0) = 1

% Plot the solution of the differential equation
plot(u, v, 'LineWidth', 2) % Increase line width for better visibility
hold on

% Create a meshgrid for the vector field
[x, y] = meshgrid(-5:0.5:5, -1:0.5:5);

% Calculate the vector field
m = sin(y);
L = sqrt(1 + m.^2);

% Plot the vector field using quiver
quiver(x, y, 1./L, m./L, 'r') % Use red color for vectors

% Set the axis limits to fit the data
axis tight

% Add title
title('Solution of the differential equation and vector field')

% Add a legend
legend('ode45 solution', 'Vector field')

% Turn off hold
hold off

% Improve the overall aesthetics
set(gca, 'FontSize', 7) % Set font size for readability
grid on % Add a grid for better readability of the plot

[Athanasios Paraskevopoulos]

Hello @Sam Chak! Thank you for your answer. So I need to adjust the initial conditions when I use 'ode45' right?

[Sam Chak]

Yes, @Athanasios Paraskevopoulos. If you can actually analytically find the initial value at u = -5 because the differential equation has an analytical solution.

format long
v   = @(u) 2*acot(exp(-u)*cot(1/2));    % analytical solution
u   = -5:1e-4:5;
iv  = v(-5)     % initial value at u = -5

iv = 
   0.007361881194388

plot(u, v(u)), hold on
plot(0, v(0), 'p', 'markersize', 20), grid on
xlabel u, ylabel v

[Sam Chak]

format long
syms y(x)  % Define the symbolic function y(x)

% Define the differential equation
eq = diff(y, x) == sin(y);

% Define the initial condition and solve the particular solution
cond    = y(0) == 1;
ySol(x) = dsolve(eq, cond)

ySol(x) = 

% find initial value at x = -5
iv  = double(subs(ySol, x, -5))

iv = 
   0.007361881194388

[Athanasios Paraskevopoulos]

@Sam Chak That was really helpful