MATLAB: 1st order differential equation

1st orderode45

i am trying to solve this differential equation and the solution i get it is wrong so where is the error? and this my function dy/dt=k*y*ln(450/y) y(0)=40 and y(15)=95 and k is constant and this my MATLAB

tspan = [0 300];
y0 = 40;y15=95;
[t,y] = ode45(@(t,y) 'k'*y*exp(450/y), tspan, y0,y15);
plot(t,y,'-o')

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Hi , the plot is still wrong , k should be get from the initial condition and the solution is y(t)=450/( e^e^(-0.02948t+0.8839) so how can i get this by MATLAB

Related Solutions

MATLAB: Solving differential equation with ode45

func=@(t,y)(y-1)^2;
tspan = [0 10];
for x = 0.01:0.05:1
    y0 = x;
    [t,y] = ode45(func,tspan,y0);
    plot(t, y(:,1));
    hold on
end

MATLAB: ODE45, differential equation

Hi,

if you search a value for k that complies with the boundary conditions, you could use fzero to solve the problem numerically:

k = fzero(@calculate_k, 0.001);
disp(['k = ', num2str(k)])
[t,y] = ode45(@(t,y) k*y*exp(450/y), [0 300], 40);
plot(t,y,'r')
hold on
scatter(15,95,'ob')
text(23,96,'y(t=15) = 95','Color','b')
hold off
function k_value = calculate_k(x)
tspan = [0 15];
y0 = 40;
[~,y] = ode45(@(t,y) x*y*exp(450/y), tspan, y0);
k_value = y(end) - 95;
end

This will give you as result:

k = 0.00010435

or if you need it more accurate:

k =
       1.043495007807761e-04

and the plot which belongs to this result looks like the boundary conditions are met (To be precise, it keeps the boundary conditions. This is because fzero is a powerful tool and also because you can think of a very good initial value for x0 with just a few estimates...):

Best regards

Stephan

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MATLAB: Solving differential equation with ode45

MATLAB: ODE45, differential equation

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