DSolve + basis solution

Question

DSolve can't solve this system of differential equation by DSolve. But the basis solution is known. How I can help DSolve to use this basis solution?

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WM code:

DSolve[{
x1'[t] == x2[t], 
x2'[t] == 2Cos[t]/(3+2Sin[t])x1[t]-x2[t](3-2Cos[t]+2Sin[t])/(3+2 Sin[t])},
{x1[t], x2[t]},t]

---

System of equations:

$x_1'(t)=x_2(t)$

$x_2'(t)=\frac{2cos(t)}{3+2sin(t)}x_1(t)-x_2(t)\frac{3-2cos(t)+2sin(t)} {3+2sin(t)}$

Basis solution:

$e^{-t}\{-1,1\}\ and\ \{3-cos(t)+sin(t),sin(t)+cos(t)\}$

Answer 1

Here's a way to automate the solution of a system that, like the OP's, is a linear system whose coefficient matrix happens to have a constant eigenvector (constant in that it is independent of $t$). If $\dot x = Ax$ and $A$ is an $n \times n$ matrix that has some constant eigenvectors $v_1,v_2,\dots,v_k$, then completing them to a basis $v_1,v_2,\dots,v_k,v_{k+1},\dots,v_n$, we can form the transition matrix $$P = \left( v1 \mid v_2 \mid \cdots \mid v_n \right) \,.$$ Then with $x = Pu$, we get the transformed ODE $\dot u = P^{-1}AP\,u$, which hopefully DSolve can deal with. It should be able to find the first $k$ solutions, but it might fail to find the last $n-k$.

vars = {x1, x2};
ode = {x1'[t] == x2[t], 
   x2'[t] == 2 Cos[t]/(3 + 2 Sin[t]) x1[t] - 
     x2[t] (3 - 2 Cos[t] + 2 Sin[t])/(3 + 2 Sin[t])};
rhs = D[Through[vars[t]], t] /. First@Solve[ode, D[Through[vars[t]], t]];
If[AllTrue[rhs, Internal`LinearQ[#, Through[vars[t]]] &],   (* check linearity *)
 rhsA = CoefficientArrays[rhs, Through[vars[t]]][[2]],
 rhsA = Failure["nonlinear", <||>]
 ]
(*  SparseArray[<<4>>]    < check: was linear > *)

ev = Cases[Eigenvectors[rhsA], v_ /; FreeQ[v, t]]
If[Length[ev] < 1, Print["Bummer. No constant eigenvectors."]]
(*  {{-1, 1}}    < check: got a constant eigenvector > *)

pp = Transpose@Join[ev, NullSpace[ev]];  (* transition matrix *)
aa = Inverse[pp].rhsA.pp // Simplify;    (* transform coefficient matrix *)
dsol = DSolve[{u'[t], v'[t]} == aa.{u[t], v[t]}, {u, v}, t]  (* check DSolve works *)
(*  check: DSolve found a solution
  {{u -> Function[{t}, E^-t C[2] + C[1] (-3 + 2 Cos[t])], 
    v -> Function[{t}, C[1] (3 + 2 Sin[t])]}}
*)

pp.{u[t], v[t]} /. First[dsol] // Simplify;  (* pp.{u, v} yields {x1, x2} *)
xsol = Thread[vars -> (Function @@ {t, #} & /@ %)]
(*
  {x1 -> Function[t, 6 C[1] - E^-t C[2] - 2 C[1] Cos[t] + 2 C[1] Sin[t]], 
   x2 -> Function[t, E^-t C[2] + 2 C[1] Cos[t] + 2 C[1] Sin[t]]}
*)

ode /. xsol // Simplify    (* check solution *)    
(*  {True, True}  *)

Answer 2

An alternative approach is to solve the system of ODEs with DSolve without knowing a particular solution. As a first try, eliminate x2 to obtain and attempt to solve the resulting single ODE.

DSolve[x1''[t] == x1[t] 2 Cos[t]/(3 + 2 Sin[t]) - 
    x1'[t] (3 - 2 Cos[t] + 2 Sin[t])/(3 + 2 Sin[t]), x1[t], t]

Unfortunately, it returns unevaluated. Next, reorder terms in the ODE and again attempt to solve it.

DSolve[x1''[t] + x1'[t])/(x1'[t] + x1[t]) == (2 Cos[t])/(3 + 2 Sin[t]), x1[t], t]

It, too, returns unevaluated, even though the ODE has a now obvious first integral, x1'[t] + x1[t] == C[1] (3 + 2 Sin[t]). Only when the auxiliary variable, v[t], is introduced can DSolve obtain a solution.

DSolve[{v'[t]/v[t] == (2 Cos[t])/(3 + 2 Sin[t]), 
    x1'[t] + x1[t] == v[t]}, {x1[t], v[t]}, t]
(* {{v[t] -> C[1] (3 + 2 Sin[t]), 
     x1[t] -> E^-t C[2] + C[1] (3 - Cos[t] + Sin[t])}} *)

and x2[t] follows immediately from x2[t] == x1'[t]. Incidentally, introducing v[t] directly into the original ODEs also yields this result.

Simplify[{x1'[t] == x2[t], x2'[t] == 2 Cos[t]/(3 + 2 Sin[t]) x1[t] - 
  x2[t] (3 - 2 Cos[t] + 2 Sin[t])/(3 + 2 Sin[t])} /. x2 -> Function[t, -x1[t] + v[t]]]
(* {v[t] == x1[t] + x1'[t], ((-3 + 2 Cos[t] - 2 Sin[t]) v[t])/
    (3 + 2 Sin[t]) + x1[t] + x1'[t] == v'[t]} *)
DSolve[%, {x1[t], v[t]}, t]

Answer 3

I am also not fully understanding the question. Since you already know the basis solutions. But May be you are looking for reduction of order method?

If one basis solution is known, a second one can be found using reduction of order. For example, let $x_1=-e^{-t}$ be one basis solution. The second one can be found by assuming the solution of the ODE has the form $x_2(t)= x_1(t) v(t)$ and substituting this back into the ODE and solving for $v(t)$.

ClearAll[u,x1,x2,v,x,t]

(*this is the orginal ODE. Assume we know only one solution for it *)
ode =x''[t]==2*(Cos[t]/(3+2*Sin[t]))*x[t]-x'[t]*((3-2*Cos[t]+2*Sin[t])/(3+2*Sin[t])); 

x1[t_]:=-Exp[-t]; (*this is one solution we know *)
u[t_]:=x1[t]*v[t];
newODE=ode/.{x[t]->u[t],x'[t]->u'[t],x''[t]->u''[t]}//Simplify

Now find v[t]

sol = v[t]/.First@DSolve[newODE,v[t],t]

Therefore the second solution is

   x2[t_] = sol*x1[t]

Verify

 ode/.{x''[t]->x2''[t],x'[t]->x2'[t],x[t]->x2[t]}//Simplify

This way, you now have both solutions. so you kinda helped Mathematica, since it could not find both solutions before, but it can find one solution, given the other.

If this is not what you want, then I will delete this.