Solving a differential equation

Question

Simone 2022-11-28

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评论： Bjorn Gustavsson 2022-11-28

Hello everybody, I can't seem to find a way to solve the following differential equation.

I have an array for V (which is my indipendent variable) and C (which is the dependent one), I need to find a punctual value of N for each value of C and V.

Looking online I've tried to use this line of code

ode = y == ((C).^3)/(k)*diff(V,C)

But I get the error : Error using diff Difference order N must be a positive integer scalar

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Simone 2022-11-28

Thank you for your answer, I get the mathematical aspect of the problem, what I don't understand is how to get the value of N in matlab. I don't understand if it's possible to get the answer

Simone 2022-11-28

I was thinking about setting up a line that goes like

So If I know the value of the angular coefficient by my dataset, I could get the punctual value of N

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Answer 1

Bjorn Gustavsson 2022-11-28

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According to your description you don't have a differential equation if you have the values of C and the known dependent values in V, and you want the values of N according to your equation. The best you can do from this is simply:

N = C^3/k*gradient(V,C);

HTH

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Davide Masiello 2022-11-28

编辑：Davide Masiello 2022-11-28

In your original response to @Chunru you said N is a constant, but now you say it must be a 351X1 column vector. Which one is it?

If it oughta be a vector, that's eaily obtained with @Bjorn Gustavsson's method, just write

N = C.^3/k.*gradient(V,C);

Bjorn Gustavsson 2022-11-28

@Simone, you have to keep track of when to use the element-wise operations ( ./ .* and .^ ) and when to use the matrix operations ( * / \ and ^). I also notice that I goofed on that one. The solution should be as @Davide Masiello gave above:

N = C.^3/k.*gradient(V,C);

Here we use the elementwise power in the first factor. Then, since k is a scalar constant it doesn't matter if we use / (right array divide) or ./. Then we take the gradient. In your variant you tried the right array divide with k*gradient(V,C) in the denominator which doesn't match the equation in your question. Your answer might therefore also be:

N = C.^3./(k.*gradient(V,C));

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Answer 2

Davide Masiello 2022-11-28

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Since you have data for V and C you could fit them with the analytical solution of the differential equation.

Let's assume this is your dataset

k = 10;
C = [1 2 3 4 5 6 7 8 9 10];
V = [0.0158 11.3471 13.4291 14.1110 14.4800 14.5975 14.7361 14.8572 14.8940 14.9459];

Let us also assume that the initial condition is v(c=1) = 0.

You can find the analytical solution to the differential equation for a generic N.

syms x y(x) N

myode = diff(y,x) == N*10/x^3;

sol = dsolve(myode,y(1) == 0)

sol =

Then, turn the symbolic solution into an anonymous function.

f = matlabFunction(sol)
f = function_handle with value:
    @(N,x)N.*1.0./x.^2.*(x.^2-1.0).*5.0

And finally fit the function to the data

fitobj = fit(C',V',f,'StartPoint',1)
fitobj = 
     General model:
     fitobj(x) = N.*1.0./x.^2.*(x.^2-1.0).*5.0
     Coefficients (with 95% confidence bounds):
       N =       3.015  (3.01, 3.02)

The result is N = 3.015, which is good considering that I had produced the fake C and V data by using N = 3 and adding a bit of random noise.