Hi Kamal,

Is this what you mean?

Solve[?x == D[u, {x, 1}], x]
(* {{x -> ?x/4}} *)

Gianluca Gorni,

How can I apply (ReplaceAll) for replacing more than one function like replacing both [Epsilon]x and Epsilon y altogether in one simple command? should i repeat [/.] for each function or use brackets?

Thank you

Solve[[Sigma]x == 2 G [Epsilon]x + [Lambda] ([Epsilon]x + [Epsilon]y), \ [Sigma]x] /. Solve[[Epsilon]x == D[u, {x, 1}], [Epsilon]x] Solve[[Epsilon]y == D[v, {y, 1}], [Epsilon]y] Solve[[Gamma]xy == D[u, {y, 1}] + D[v, {x, 1}], [Gamma]xy] Solve[[Sigma]y == 2 G [Epsilon]y + [Lambda] ([Epsilon]x + [Epsilon]y), [Sigma]y] Solve[[Tau]xy == G [Gamma]xy, [Tau]xy]

Attachments:

CH. 3 Query.nb

Hi Rohit Namjoshi,

Thanks for your reply. I have segma is function of epsilon, epsilon is function of neu and neu is function of x. i want to find segma in terms of x.

When I SOLVE epsilon i get it in terms of x (which is OK). Then i SOLVE Segma and get it in terms of epsilon ( i want it in terms of x)

is there a simple way to do so?

u = 2 x^2; Solve[[Epsilon]x == D[u, {x, 1}], [Epsilon]x] Solve[[Sigma]x == 2 G [Epsilon]x + [Lambda] ([Epsilon]x + [Epsilon]y), [Sigma]x]

i also attached the file

Many thanks for your help

Attachments:

CH. 3(query).nb