Solving a differential equation

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

[Chunru]

Are k and N constants?

[Simone]

k is a constant of known value, also N is a constant but of unknown value

[Chunru]

Then you have an ODE of and the solution can be written as:

[Simone]

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]

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

Answer 1

[Bjorn Gustavsson]

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

[Simone]

Thank you for your answer, one problem I'm getting is that C and V are vectors made of 351 rows, N has to be a vector of 351 rows but by doing this code I get that it is a Matrix 351x351.

Writing

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

doesn't help

[Davide Masiello]

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]

@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));

Answer 2

[Davide Masiello]

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.

[Simone]

Thank you for your answer, I've tried changing the assumption with a real value of my dataset but I get the following error

"Error using fit>iFit

X must be a matrix with one or two columns.

Error in fit (line 116)

[fitobj, goodness, output, convmsg] = iFit( xdatain, ydatain, fittypeobj, ..."

[Davide Masiello]

Maybe your data was already arranged in columns, in which case you need to delete the apostrophe after C and V in the fit function.