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How do i translate this equation to mathematica language?

azade53

Posted 10 years ago

i need the translation because i need to solve this equation.

POSTED BY: azade53

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Nasser M. Abbasi

Nasser M. Abbasi, student

Posted 10 years ago

well, you can just write the 3 equations as is eq1 = r1'[t] == -k13 r1[t] + k21 r2[t]; eq2 = r2'[t] == -k21 r2[t] + k32 r3[t]; eq3 = r3'[t] == k13 r1[t] - k32 r3[t]; DSolve[{eq1, eq2, eq3}, {r1[t], r2[t], r3[t]}, t] or if you want little less typing ClearAll["Global`*"] state = {r1[t], r2[t], r3[t]}; mat = {{-k13, k21, 0}, {0, -k21, k32}, {k13, 0, -k32}} eq = D[state, t] == mat. state; DSolve[eq, state, t] See DSolve, there are many examples there

well, you can just write the 3 equations as is

eq1 = r1'[t] == -k13 r1[t] + k21 r2[t];
eq2 = r2'[t] == -k21 r2[t] + k32 r3[t];
eq3 = r3'[t] == k13 r1[t] - k32 r3[t];
DSolve[{eq1, eq2, eq3}, {r1[t], r2[t], r3[t]}, t]

or if you want little less typing

ClearAll["Global`*"]
state = {r1[t], r2[t], r3[t]};
mat = {{-k13, k21, 0}, {0, -k21, k32}, {k13, 0, -k32}}
eq = D[state, t] == mat. state;
DSolve[eq, state, t]

See DSolve, there are many examples there

POSTED BY: Nasser M. Abbasi

David Keith

Posted 10 years ago

This will produce the solution. It of course includes 3 integration constants, C[1],C[2], C[3]. eqs = {r1'[t], r2'[t], r3'[t]} == {{-k13, k21, 0}, {0, -k21, k32}, {k13, 0, -k32}}.{r1[ t], r2[t], r3[t]}; sol = DSolve[eqs, {r1, r2, r3}, t] // First; These could also be solved for by including them in the equation set. sol2 = DSolve[{eqs, {r1[0] == 1, r2[0] == 1, r3[0] == 1}}, {r1, r2, r3}, t] // First;

This will produce the solution. It of course includes 3 integration constants, C[1],C[2], C[3].

eqs = {r1'[t], r2'[t], 
    r3'[t]} == {{-k13, k21, 0}, {0, -k21, k32}, {k13, 0, -k32}}.{r1[
      t], r2[t], r3[t]};

sol = DSolve[eqs, {r1, r2, r3}, t] // First;

These could also be solved for by including them in the equation set.

sol2 = DSolve[{eqs, {r1[0] == 1, r2[0] == 1, r3[0] == 1}}, {r1, r2, 
     r3}, t] // First;

POSTED BY: David Keith

azade53

Posted 10 years ago

Thanks a lot for responses

POSTED BY: azade53

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