Transform expression into a certain standard form for an ODE

Question

I'm using Mathmatica 11.1 on Windows.

Given an expression containing a differntial equation of the form

$\qquad A\, x'(z) = C\, x''(z) - B\, x(z)$

Or

A x'[z] == -B x[z] + C x''[z]

I want to transform the expression into a standard form for an ODE, namely,

$\qquad x''(z)-\frac{A}{C}x'(z)-\frac{B}{C}x(z)=0$

Or

x''[z] - A/C x'[z] - B/C x[z] == 0

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A more generalized version of this format may be represented as

$\qquad y^n(x)-a_{1}\cdot y^{n-1}(x)\cdots a_{n-1}\cdot y'(x)-a_{n}\cdot y(x)=0$

for linear ODEs, or

$\qquad y^n(x)-a_{1}\cdot y^{n-1}(x)\cdots a_{n-1}\cdot y'(x)-a_{n}\cdot y(x)=p(x)$

for non-linear ODES, where $p(x)$ may be some non-linear function on x, such as $sin(x)$ or $1/x$.

---

I have attempted to address this by using StandardForm and NonlinearStateSpaceModel, amongst other ways, and have not yet been able to convert the expression.

I would prefer a solution that is generalized and would function for any given ODE however, this may or may not be practical.

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Answer 1

A little clunky, but this should work pretty generally:

standardForm[eqn_Equal] := Module[{equation = # - eqn[[1]] & /@ eqn},
  equation = Collect[#, _''[_]] & /@ equation;
  #/First@Cases[equation, a_ _''[_] :> a, Infinity] & /@ equation // Apart // Expand
 ]

Note: as of V11.3, # - eqn[[1]] & /@ eqn can be replaced with SubtractSides[eqn, First@eqn].

Then,

standardForm[a x'[z] == -c x''[z] - b x[z] + d x''[z]]
(* 0 == (b*x[z])/(c - d) + (a*Derivative[1][x][z])/(c - d) + Derivative[2][x][z] *)

---

Explanation:

In the first step,

eqn = a x'[z] == -c x''[z] - b x[z] + d x''[z];
equation = # - eqn[[1]] & /@ eqn
(* 0 == -b x[z] - a x'[z] - c x''[z] + d x''[z] *)

we move everything to the right-hand side. In the second step,

equation = Collect[#, _''[_]] & /@ equation
(* 0 == -b x[z] - a x'[z] + (-c + d) x''[z] *)

we collect all the instances of x''[z] so that we can extract the coefficient in the next step,

First@Cases[equation, a_ _''[_] :> a, Infinity]
(* -c + d *)

in order to then divide both sides of the equation by this value:

#/First@Cases[equation, a_ _''[_] :> a, Infinity] & /@ equation // Apart // Expand
(* 0 == b x[z]/(c - d) + a x'[z]/(c - d) + x''[z] *)

The // Apart // Expand is there to make sure that the cancellation occurs in the coefficient of the x''[z] term.

Answer 2

Another way, easily generalized for linear ODEs:

depvars = NestList[D[#, z] &, x[z], 2]; (* {x[z], x'[z], x''[z]} *)
ca = CoefficientArrays[
   Subtract @@ Solve[a x'[z] == -b x[z] + c x''[z] + p[z], x''[z]][[1, 1]],
   depvars];
[email protected] == -First@ca
(*  -((b x[z])/c) - (a x'[z])/c + x''[z] == -(p[z]/c)  *)