solving trascendental equations, proper setting

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PatrizioGraziosi 2020-6-14

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https://ww2.mathworks.cn/matlabcentral/answers/547923-solving-trascendental-equations-proper-setting

评论： Ameer Hamza 2020-6-17

采纳的回答： Ameer Hamza

Hello everybody,

I'd like to solve for y = y(x) the following equation

d log( y ) / d x + y = 1 + f

with f = f(x).

f is a 1D numerically known array, I don't know its nalytical form.

I cannot set properly solve or fzero.

Can you help me, please?

Patrizio

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Answer 1

Ameer Hamza 2020-6-14

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https://ww2.mathworks.cn/matlabcentral/answers/547923-solving-trascendental-equations-proper-setting#answer_450990

编辑：Ameer Hamza 2020-6-14

在 MATLAB Online 中打开

This is a differential equation and you can use symbolic toolbox to find an anayltical solution

syms y(x) f
eq = diff(log(y), x) + y == 1 + f;
sol = dsolve(eq);

Result

>> sol
sol =
 (exp((C1 + x)*(f + 1))*(f + 1))/(exp((C1 + x)*(f + 1)) + 1)
                                                       f + 1

Following shows how to get a numerical solution using ode45

syms y(x) f
eq = diff(log(y), x) + y == 1 + f;
sol = dsolve(eq);
odeFun = matlabFunction(odeToVectorField(eq), 'Vars', {'t', 'Y', 'f'});
tspan = [0 10]; % time span for numerical solution
ic = 1; % initial condition: y(0)==1
fv = 1; % numerical solution for f=1
[t, y] = ode45(@(t, y) odeFun(t, y, fv), tspan, ic);
plot(t, y);

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PatrizioGraziosi 2020-6-14

编辑：PatrizioGraziosi 2020-6-14

在 MATLAB Online 中打开

Thank you Ameer!

In the system I'm stuying, for each 'x' value there is one value for 'f' and one value for 'y'.

I mean, there is a f(x) that I know and a y(x) I don't know but for each 'x' value there are only one 'f' and one 'y' . I know f(x) and the relation between f(x) and y(x), and should find y(x).

As far as I understand, the tspan plays the role of the 'x' variable and I get a y(x) for each 'fv' but this not actually the case: for each 'x' value I have also a different 'fv' value.

Hope this makes my problem clearer. If I have misunderstood your answer, sorry.

About the symbolic toolbok lines you wrote, it looks to be the case for a scalar f.

if I input

syms y(x) f(x)

I get this

sol =
 
exp(int(f(x) + 1, x, 'IgnoreSpecialCases', true, 'IgnoreAnalyticConstraints', true))/(C6 - int(-exp(int(f(x) + 1, x, 'IgnoreSpecialCases', true, 'IgnoreAnalyticConstraints', true)), x, 'IgnoreSpecialCases', true, 'IgnoreAnalyticConstraints', true))