Runge Kutta solving differential equations

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Mj
Mj 2020-11-6
编辑： Samuele Sandrini 2020-11-7
Hello everyone!
I have to solve this second order differential equation by using the Runge-Kutta method in matlab:
can anyone help me please? and how can i plot the figure?(a against e)
d2a/de2=(((((2+c2)*(Fu^2))/(1+c2))+1)*(a^c2)-((2+c2/1+c2)*(Fu^2/a))-a^(2+(2*c2)))/(((2+c2)*Fu^2)/(1+c2)*(3+c2));
Fu=1
c2=0 , 0.5 , 1 (there are 3 values for c2)
initial conditions are: a=0.8 , d_a=0
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Mj
Mj 2020-11-7
is it possible for you to send me the code? because i am beginner in matlab.
Alan Stevens
Alan Stevens 2020-11-7
I have not generated any code. If you look at the ode45 help file you will see explanations of what to do.

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Samuele Sandrini
Samuele Sandrini 2020-11-7
编辑：Samuele Sandrini 2020-11-7
Hi, i'm sorry but your differential equation is unreadable in that way.
However i try to give you all element that you need; as already told by @Alan Stevens, first of all you have to pass from a second order diff. eq. to a system of two first order diff. equation, in this way:
  1. starting from your second order equation, in general like this: , you take two auxiliary variable and .
  2. After that you can rewrite the first equation like:
Note that the first equation in the system is always the same, the second one corrispond to y'' and is equal to the starting equation where y and y' is replaced by e .
Once you obtained the system, programmaticaly you can do something like this:
f = @(z,t) [z(2); ...(the second eqn in the system using z(1) end z(2)) ];
[t,y] = ode45(f,[t0 tF],[y(0);y'(0)]) %where y(0) and y'(0) are the initial condition (in your case [0.8;0])
Note that y will be a matrix with two column that corrispond respectively to and .
If you want an example here you can find in Examples paragraph, the second one is the solution of the van der Pol equation that is similar to this procedure.

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