# How to define a dependent variable for differentiaion?

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Abinav il 7 Ott 2020
Commentato: Ameer Hamza il 7 Ott 2020
Hi,
I have a variable A which is dependent on x, but I don't know the expression yet.
I have a differential equation dA/dx which is given as
dA_x = (T*b-E*Q*diff(C_s,x))/(E*S);
All the entities except C_s are constants. I can get A by integrating this equation.
A = int(dA_x,x);
But the problem lies with the differential expression of C_s which is given by
C_s = (M-2*A*E*Q)/(2*E*I);
Since A is not yet defined as a function or dependent variable of x, the differentiation results in 0.
How do I solve this?
I'm not looking for a numeriacal answer, I just need a symbolic expression.
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### Risposta accettata

Ameer Hamza il 7 Ott 2020
Define symbolic variable A like this
syms A(x)
It will tell MATLAB that a depends on x and will not become zero on differentiation.
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Abinav il 7 Ott 2020
syms L b h S I Q E u F M T c theta w k
syms x A(x) dA_x C_s(x)
S = b*h;
I = b*h^3/3;
Q = b*h^2/2;
M = -((F*L^2)/2)+(F*(L*x-x^2/2));
C_s = (M-2*A*E*Q)/(2*E*I);
dA_x = (T*b-E*Q*diff(C_s,x))/(E*S);
A = int(dA_x,x);
At this point, the diff(C_s,x) becomes C_s. And matlab makes a substitution from the above equation resulting in A(x) on both sides of equation. Which is mathematically correct.
C_s = subs(C_s,A)
But when I substitute A back into my C_s, there is still an A term in the equation. This could be avoided if matlab retains keeps the term as C_s while integrating instead of substituing the equation in there.
In either case, the problem can be solved by rearranging the terms but I can't figure out a way to do that in Matlab.
Ameer Hamza il 7 Ott 2020
Symbolic variables can be a bit confusing sometimes. For this case, the following code will work
syms L b h S I Q E u F M T c theta w k
syms x A(x) dA_x(x) C_s(x)
S = b*h;
I = b*h^3/3;
Q = b*h^2/2;
M = -((F*L^2)/2)+(F*(L*x-x^2/2));
C_s = (M-2*A*E*Q)/(2*E*I);
dA_x = (T*b-E*Q*diff(C_s,x))/(E*S);
A = int(dA_x,x);
C_s = subs(C_s);

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