How do I solve an ODE of the form y'=ay^3 +by^2 +cy +d?

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Gabriel Baranoski 2022-1-25

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https://ww2.mathworks.cn/matlabcentral/answers/1636020-how-do-i-solve-an-ode-of-the-form-y-ay-3-by-2-cy-d

评论： Jon 2022-1-26

I have a vector, for example "pn" of size (20,1). Each point of data describes a coeffecient of y to the power of 20 minus that data point indices (made from the polyfit function in matlab).

For example, pn(15,1) = 5 translates to 5y^5

This defines a large polynomial which looks similar to the title example.

I know that the ODE I have is of the form y'=ay^19 + by^18 +cy^17 +...+gy +h

how can i solve this ode to make a plot of y as a function of t?

I know about the ODE functions like ode45 etc, but I'm not sure how to use them with my ode form.

Thanks!

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James Tursa 2022-1-25

编辑：James Tursa 2022-1-25

Can you verify that the form has y's in the polynomial and not x's or t's? And do you have numeric initial conditions?

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Answer 1

Jon 2022-1-25

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The ode solvers, e.g. ode45 require a function handle which will evaluate the current value of the derivative given the current state. So define your function for example as:

fun = @(y)polyval(pn,y)

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Answer 2

Gabriel Baranoski 2022-1-26

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https://ww2.mathworks.cn/matlabcentral/answers/1636020-how-do-i-solve-an-ode-of-the-form-y-ay-3-by-2-cy-d#answer_882335

Hey so I ended up figuring it out. This is my code to solve the ODE below:

FK=polyfit(MFPs(:,1),MFPs(:,2),20);

tspan = [0 :0.05:50];

x0 = 0;

[t,v]=ode45(@(t,v) ((FK(1)*v^20) + (FK(2)*v^19) + (FK(3)*v^18) + (FK(4)*v^17) + (FK(5)*v^16) + (FK(6)*v^15) + (FK(7)*v^14) + (FK(8)*v^13) + (FK(9)*v^12) + (FK(10)*v^11) + (FK(11)*v^10) + (FK(12)*v^9) + (FK(13)*v^8) + (FK(14)*v^7) + (FK(15)*v^6) + (FK(16)*v^5) + (FK(17)*v^4) + (FK(18)*v^3) + (FK(19)*v^2) + (FK(20)*v) + (FK(21)) - FR - (Y*(v^2)))/M ,tspan,x0);

plot (app.UIAxes,t,v*3.6);

There are a few terms included there that change the ODE from how i originally stated, but this solves the ode accurately.

Maybe theres a way to reduce the ode solve command but I'm not to sure how.