The MATLAB function to solve implicit equation

Hi all,

Thanks once again to answer my questions every time.

I would like to request you tell me the best way to solve an Implicit equation in MATLAB.

Thank you very much

[Dyuman Joshi]

Please specify which equation you want to solve. It is difficult to suggest a proper working solution without adequate information.

[Teja Reddy]

Hi, thank you,

The equation is

m^2 / 3(a+a0)^2 * (2 * k1 + 3 * k2) = a0 * (2/v1 + 5/v2)

The unknown term to be calculated is "a0"

Thank you

[Torsten]

What is the correct form:

(m^2 / 3) * (a+a0)^2 * (2 * k1 + 3 * k2) = a0 * (2/v1 + 5/v2)

m^2 / (3 * (a+a0)^2) * (2 * k1 + 3 * k2) = a0 * (2/v1 + 5/v2)

m^2 / (3 * (a+a0)^2 * (2 * k1 + 3 * k2)) = a0 * (2/v1 + 5/v2)

?

[Teja Reddy]

Oh, I am very sorry, the correct form is

[m^2 /( 3(a+a0)^2)] * (2 * k1 + 3 * k2)] = a0 * (2/v1 + 5/v2)

Thank you

[Torsten]

You open one time [ and close two times ] ] . Still unbalanced parenthesis.

[Teja Reddy]

I am extremely sorry for the confusion created

[m^2 /( 3(a+a0)^2)] * (2 * k1 + 3 * k2) = a0 * (2/v1 + 5/v2)

this is the correct one.

Thank you

Answer 1

[Steven Lord]

@John D'Errico posted how to solve this equation symbolically, if you don't have values for the various symbols used in the code. If you do have values for the symbols (except for a0, the symbol for which you're trying to solve) write a function to evaluate f(a0) and use fzero.

m = 1;
a = 2;
k1 = 3;
k2 = 4;
v1 = 5;
v2 = 6;
EQ = @(a0) [m^2 /( 3*(a+a0)^2)] * (2 * k1 + 3 * k2) - (a0 * (2/v1 + 5/v2));
y = fzero(EQ, 7)

y = 0.6782

check = EQ(y) % should be small

check = -1.1102e-16

I changed your EQ slightly. Instead of being [one expression] == [a second expression] I made it f(a0) = [one expression] - [a second expression]. fzero solves for where that f(a0) is equal to 0 which is equivalent to your formulation.

Answer 2

[John D'Errico]

Assuming your question is how to solve the equation

[m^2 /( 3(a+a0)^2)] * (2 * k1 + 3 * k2) = a0 * (2/v1 + 5/v2)

I removed the second closing ], since it seems to have been the spurious one. Even if not, the solution is still little different. Just use solve. This is just a simple polynomial equation at heart. And, you NEED to use * to multiply. MATLAB does not allow implicit multiplication. That is, 3(a+a0) is NOT seen as 3*(a + a0) in MATLAB. It is just a syntax error.

syms m a a0 k1 k2 v1 v2
EQ = [m^2 /( 3*(a+a0)^2)] * (2 * k1 + 3 * k2) == a0 * (2/v1 + 5/v2)

EQ = 

a0sol = solve(EQ,a0,'maxdegree',3)

a0sol = 

Yes, the rsult is messy. But what do you expect from what is effectively a cubic polynomial? Not all problems have a nice simple solution. In fact, most implicit equations are written in that form because they do not have a simple solution, and worse yet, most implicit equations will have no algebraic solution at all.

[Teja Reddy]

I understood your explanation in a very good manner. Yes, I am expecting the value of "a0".

Thank you