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I am given a system first order differential equations: $x'=y$ and $y'=6x^2-a/2$, where $a$ is a constant and $'$ denotes $t$-derivatives.

I then make the substitution $(x,y)=(x_1y_1,y_1)$.

This implies that $x_1=\frac{x}{y}$ and $y_1=y$.

The $t$-derivatives are $$x_{1}'=\left(\frac{x}{y}\right)'=\frac{yx'-xy'}{y^2}=1-x_{1}y_{1}^{-1}(6x_{1}^2y_{1}^2-a/2)$$ and $$y_{1}'=6x_{1}^2y_{1}^2-a/2.$$

I need to continue in this fashion, that is, in the next step, for example, we let $(x_{1},y_{1})=(x_{2}y_{2},y_{2})$ and then find $x_{2}'$, $y_{2}'$, etc.

Is there a way I can compute the derivatives in Mathematica? For instance, how would I find $x_{1}'$ in Mathematica.

Thanks, Radz.

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up vote 2 down vote accepted

I assume that while the replacement rule is to be selectable; the same rule is used in each iteration.

Format[x[n_Integer]] := Subscript[x, n];
Format[y[n_Integer]] := Subscript[y, n];

eqn[1][0] = eqn[2][0] =
   {x[0]'[t] == y[0][t], y[0]'[t] == 6 x[0][t]^2 - a/2};

Rule set for {x, y} -> {x1*y1, y1}

sub[1][n_Integer?Positive] := sub[1][n] =
   NestList[D[#, t] &,
     {x[n - 1][t] :> y[n][t]*x[n][t], y[n - 1][t] :> y[n][t]},
     1] // Flatten;

Rule set for {x, y} -> {x1, x1*y1}

sub[2][n_Integer?Positive] := sub[2][n] =
   NestList[D[#, t] &,
     {y[n - 1][t] :> x[n][t]*y[n][t], x[n - 1][t] :> x[n][t]},
     1] // Flatten;

eqn[ruleSet_Integer?(# == 1 || # == 2 &)][n_Integer?Positive] :=
  eqn[m][n] =
   Equal @@@ Solve[eqn[ruleSet][n - 1] //. sub[ruleSet][n],
      {x[n]'[t], y[n]'[t]}][[1]];

Column[
 Grid[#, Frame -> All] & /@
  Table[eqn[ruleSet][n], {ruleSet, 2}, {n, 0, 3}]]

enter image description here

share|improve this answer
    
Dear @Bob Hanlon. Thanks your help. How can I modify this if for each iteration I want to use two different coordinates? For example, for the 1st iteration, either $(x,y)=(x_1y_1,y_1)$ or $(x,y)=(x_1,x_1y_1)$. – Radz Jun 27 '14 at 16:30
1  
@Radz - modified answer to handle selectable replacement rules. – Bob Hanlon Jun 27 '14 at 17:39
    
Dear @Bob Hanlon, Thank you. – Radz Aug 12 '14 at 12:35

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