Can anyone help to write a code for plotting the following equation with time please?

    x = exp( (-B/omega) * cos(omega * t) ) ... 
       ./ ( (B/A)*(integral(exp( (-B/omega)* cos(omega * t) ))))

Where

    A= 1;
    B= 10;
    omega= 1;
    x0 =0.1;
    t = 0 :0.0001:1000;

[Avan Al-Saffar]

The above result is for solving the following system analytically:

dx/dt = (B*sin*(omega*t) * x) - (x^2 * B / A)

[Roger Stafford]

As it stands, the integral in your expression is an indefinite integral and therefore has an arbitrary constant of integration. You need to specify what that constant is in order to successfully plot x as a function of t.

[Avan Al-Saffar]

We can find the constant at x(t=0).

Answer 1

[per isakson]

With a little bit of guessing

    A= 1;
    B= 10;
    omega= 1;
    x0 =0.1;
    t = 0 :0.0001:1000;
    fi = @(ti) exp( (-B/omega).*cos(omega*ti) );
    fx = @(tj) exp( (-B/omega) .* cos(omega*tj) ) ... 
       ./ ( (B/A).*(integral( fi, 0, tj )));
    ezplot( fx, 0:1e-3:12*pi )

produces this

&nbsp

I don't use x0 =0.1; and I get a warning

    Warning: Function failed to evaluate on array inputs; [...]

[Avan Al-Saffar]

Many thanks, this is really useful. If I want to find Max(x) and Mean(x), can you help me to find those as well please?

[Torsten]

Note that x(0)=Infinity since you get a division by zero there for your quotient.

Best wishes

Torsten.

[Avan Al-Saffar]

But I have cos(omega*t) in the denominator so if t =0, I will get 1 .

[Torsten]

If lower limit and upper limit of an integral are identical (t=0 in this case), its value is zero - independent of the function to be integrated.

Best wishes

Torsten.