Solving a differential equation in mechanics

Question

I have tried to solve a second order differential equation, but unfortunately the answer is too long and different from the desired one. Can you help me? Here is the code:

    DSolve[D[Subscript[X, 1, 1][Subscript[T, 0]], {Subscript[T, 0], 2}] + 
   Subscript[\[Omega], 
    1]^2 Subscript[X, 1, 1][Subscript[T, 0]] == -Subscript[c, 1, 
     2] (E^(2 I  Subscript[\[Omega], 1] Subscript[T, 0]) Subscript[A, 
       1]^2 + Subscript[A, 1] Subscript[
\!\(\*OverscriptBox[\(A\), \(_\)]\), 1]) - 
   Subscript[c, 1, 
    3] (E^(I (Subscript[\[Omega], 1] + Subscript[\[Omega], 
          2]) Subscript[T, 0]) Subscript[A, 1] Subscript[A, 2] + 
      E^(I (Subscript[\[Omega], 1] - Subscript[\[Omega], 
          2]) Subscript[T, 0]) Subscript[A, 1] Subscript[
\!\(\*OverscriptBox[\(A\), \(_\)]\), 2]) - 
   0.5 Subscript[c, 1, 1] Subscript[A, 
    1] (E^(I (Subscript[\[Omega], 1] + p) Subscript[T, 0]) + E^(
      I (Subscript[\[Omega], 1] - p) Subscript[T, 0])) - 
   Subscript[c, 1, 
    4] (E^(2 I  Subscript[\[Omega], 2] Subscript[T, 0]) Subscript[A, 
       2]^2 + Subscript[A, 2] Subscript[
\!\(\*OverscriptBox[\(A\), \(_\)]\), 2]), 
 Subscript[X, 1, 1][Subscript[T, 0]], Subscript[T, 0]] 

Answer 1

Clear["Global`*"]

Format[A[n_]] := Subscript[A, n]
Format[AO[n_]] := Subscript[\!\(\*OverscriptBox[\(A\), \(_\)]\), n]
Format[c[m_, n_]] := Subscript[c, m, n]
Format[T[n_]] := Subscript[T, n]
Format[ω[n_]] := Subscript[ω, n]
Format[X[m_, n_]] := Subscript[X, m, n]

eqn = D[X[1, 1][T[0]], {T[0], 2}] +
    ω[1]^2 X[1, 1][
      T[0]] == -c[1, 2] (E^(2 I ω[1] T[0]) A[1]^2 + A[1] AO[1]) - 
    c[1, 3] (E^(I (ω[1] + ω[2]) T[0]) A[1] A[2] + 
       E^(I (ω[1] - ω[2]) T[0]) A[1] AO[2]) - 
    1/2 c[1, 1] A[
      1] (E^(I (ω[1] + p) T[0]) + E^(I (ω[1] - p) T[0])) - 
    c[1, 4] (E^(2 I ω[2] T[0]) A[2]^2 + A[2] AO[2]);

sol = DSolve[eqn, X[1, 1][T[0]], T[0]][[1]] // FullSimplify

Verifying that the solution satisfies the equation,

eqn /. (NestList[D[#, T[0]] &, sol, 2] // Flatten) // Simplify

(* True *)

Using ExpToTrig

sol // ExpToTrig // Simplify

Answer 2

Calling $A_k = \rho_k e^{i \phi_k}, \bar A_k = \rho_k e^{-i \phi_k}$ and $X(t) = X_r(t) + i X_i(t)$ and using the Laplace transform with the initial conditions $X_{r,i}(0), X'_{r,i}(0)$ according with the script which follows.

ode = \!\(
\*SubsuperscriptBox[\(\[Omega]\), \(1\), \(2\)]\ \((
\(\*SubscriptBox[\(X\), \(r\)]\)[t] + I 
\(\*SubscriptBox[\(X\), \(i\)]\)[t]\ )\)\) + Subscript[X, r]''[t] + 
  I Subscript[X, i]''[t] - (-(1/2) E^( I Subscript[\[Phi], 1]) (E^(I t (-p + Subscript[\[Omega], 1])) +
 E^(I t (p + Subscript[\[Omega], 1]))) Subscript[c, 11]
  Subscript[\[Rho], 1] - Subscript[c, 12] (\!\(
  \*SubsuperscriptBox[\(\[Rho]\), \(1\), \(2\)] + \(
\*SuperscriptBox[\(E\), \(2\ I\ 
\*SubscriptBox[\(\[Phi]\), \(1\)] + 2\ I\ t\ 
\*SubscriptBox[\(\[Omega]\), \(1\)]\)]\ 
\*SubsuperscriptBox[\(\[Rho]\), \(1\), \(2\)]\)\)) - 
Subscript[c, 13] (E^(I Subscript[\[Phi], 1] - I Subscript[\[Phi], 2] + 
I t (Subscript[\[Omega], 1] - Subscript[\[Omega], 2]))
Subscript[\[Rho], 1] Subscript[\[Rho], 2] + 
E^(I Subscript[\[Phi], 1] + I Subscript[\[Phi], 2] + 
I t (Subscript[\[Omega], 1] + Subscript[\[Omega], 2]))
Subscript[\[Rho], 1] Subscript[\[Rho], 2]) - 
Subscript[c, 14] (\!\(
\*SubsuperscriptBox[\(\[Rho]\), \(2\), \(2\)] + \(
\*SuperscriptBox[\(E\), \(2\ I\ 
\*SubscriptBox[\(\[Phi]\), \(2\)] + 2\ I\ t\ 
\*SubscriptBox[\(\[Omega]\), \(2\)]\)]\ 
\*SubsuperscriptBox[\(\[Rho]\), \(2\), \(2\)]\)\)));

Lode = LaplaceTransform[ode, t, s]
odei = ComplexExpand[Lode]
odere = (odei + (odei /. {I -> -I}))/2 // Expand
odeim = (odei - (odei /. {I -> -I}))/(2 I) // Expand
sol = Solve[{odere == 0, odeim == 0}, {LaplaceTransform[Subscript[X, r][t], t, s], LaplaceTransform[Subscript[X, i][t], t, s]}][[1]]
x1t = InverseLaplaceTransform[Last[sol[[1]]], s, t] // FullSimplify
x2t = InverseLaplaceTransform[Last[sol[[2]]], s, t] // FullSimplify

We will obtain the following results.

$$ X_r(t) = \frac{c_{11} \rho _1 \cos \left(p t+t \omega _1+\phi _1\right)}{2 p^2+4 p \omega _1}+\frac{c_{11} \rho _1 \cos \left(\phi _1-t \omega _1\right)}{4 \omega _1^2-p^2}+\frac{c_{11} \rho _1 \cos \left(t \left(\omega _1-p\right)+\phi _1\right)}{2 p \left(p-2 \omega _1\right)}-\frac{c_{12} \rho _1^2}{\omega _1^2}-\frac{c_{14} \rho _2^2}{\omega _1^2}-\frac{c_{13} \rho _1 \rho _2 \cos \left(t \left(\omega _1-\omega _2\right)+\phi _1-\phi _2\right)}{\left(2 \omega _1-\omega _2\right) \omega _2}+\frac{c_{13} \rho _1 \rho _2 \cos \left(t \left(\omega _1+\omega _2\right)+\phi _1+\phi _2\right)}{\omega _2 \left(2 \omega _1+\omega _2\right)}-\frac{c_{14} \rho _2^2 \cos \left(2 \left(t \omega _2+\phi _2\right)\right)}{\omega _1^2-4 \omega _2^2}+\frac{c_{12} \rho _1^2 \cos \left(2 \left(t \omega _1+\phi _1\right)\right)}{3 \omega _1^2}+\frac{c_{12} \rho _1^2 \left(6 \cos \left(t \omega _1\right)+\cos \left(2 \phi _1-t \omega _1\right)-3 \cos \left(t \omega _1+2 \phi _1\right)\right)}{6 \omega _1^2}+\frac{2 c_{14} \rho _2^2 \left(-2 \omega _2^2 \cos \left(t \omega _1\right)-\omega _2 \omega _1 \sin \left(2 \phi _2\right) \sin \left(t \omega _1\right)+\omega _1^2 \cos ^2\left(\phi _2\right) \cos \left(t \omega _1\right)\right)}{\omega _1^4-4 \omega _1^2 \omega _2^2}+\frac{2 c_{13} \rho _1 \rho _2 \left(2 \omega _1^2 \sin \left(\phi _2\right) \sin \left(t \omega _1+\phi _1\right)+\omega _2 \omega _1 \cos \left(\phi _2\right) \cos \left(\phi _1-t \omega _1\right)-\omega _2^2 \sin \left(\phi _2\right) \cos \left(\phi _1\right) \sin \left(t \omega _1\right)\right)}{4 \omega _1^3 \omega _2-\omega _1 \omega _2^3}+\frac{p^2 X_r(0) \cos \left(t \omega _1\right)}{p^2-4 \omega _1^2}-\frac{4 \omega _1^2 X_r(0) \cos \left(t \omega _1\right)}{p^2-4 \omega _1^2}+\frac{X_r'(0) \sin \left(t \omega _1\right)}{\omega _1}\\ X_i(t) = -\frac{c_{11} \rho _1 \sin \left(p t-t \omega _1-\phi _1\right)}{2 p^2-4 p \omega _1}+\frac{c_{11} \rho _1 \sin \left(p t+t \omega _1+\phi _1\right)}{2 p^2+4 p \omega _1}+\frac{c_{11} \rho _1 \sin \left(\phi _1-t \omega _1\right)}{4 \omega _1^2-p^2}-\frac{c_{13} \rho _1 \rho _2 \sin \left(t \left(\omega _1-\omega _2\right)+\phi _1-\phi _2\right)}{\left(2 \omega _1-\omega _2\right) \omega _2}+\frac{c_{13} \rho _1 \rho _2 \sin \left(t \left(\omega _1+\omega _2\right)+\phi _1+\phi _2\right)}{\omega _2 \left(2 \omega _1+\omega _2\right)}-\frac{c_{14} \rho _2^2 \sin \left(2 \left(t \omega _2+\phi _2\right)\right)}{\omega _1^2-4 \omega _2^2}+\frac{c_{12} \rho _1^2 \sin \left(2 \left(t \omega _1+\phi _1\right)\right)}{3 \omega _1^2}+\frac{c_{12} \rho _1^2 \left(\sin \left(2 \phi _1-t \omega _1\right)-3 \sin \left(t \omega _1+2 \phi _1\right)\right)}{6 \omega _1^2}+\frac{2 c_{14} \rho _2^2 \left(\omega _1 \sin \left(\phi _2\right) \cos \left(\phi _2\right) \cos \left(t \omega _1\right)+\omega _2 \cos \left(2 \phi _2\right) \sin \left(t \omega _1\right)\right)}{\omega _1^3-4 \omega _1 \omega _2^2}-\frac{2 c_{13} \rho _1 \rho _2 \left(\omega _2^2 \sin \left(\phi _1\right) \sin \left(\phi _2\right) \sin \left(t \omega _1\right)+2 \omega _1^2 \sin \left(\phi _2\right) \cos \left(t \omega _1+\phi _1\right)-\omega _2 \omega _1 \cos \left(\phi _2\right) \sin \left(\phi _1-t \omega _1\right)\right)}{4 \omega _1^3 \omega _2-\omega _1 \omega _2^3}+\frac{p^2 X_i(0) \cos \left(t \omega _1\right)}{p^2-4 \omega _1^2}-\frac{4 \omega _1^2 X_i(0) \cos \left(t \omega _1\right)}{p^2-4 \omega _1^2}+\frac{X_i'(0) \sin \left(t \omega _1\right)}{\omega _1} $$

Answer 3

I have taken your equation and tidied it up by getting rid of the subscripts. (I do not have the ability to do Greek letters on this computer - can someone edit and improve further ?)

 eqn = D[Subscript[X, 1, 1][Subscript[T, 0]], {Subscript[T, 0], 2}] + 
      Subscript[ω, 
          1]^2 Subscript[X, 1, 1][Subscript[T, 0]] == -Subscript[c, 1, 

      2] (E^(2 I  Subscript[ω, 1] Subscript[T, 0]) Subscript[A, 
                1]^2 + Subscript[A, 1] Subscript[

\!\(\*OverscriptBox[\(A\), \(_\)]\), 1]) - 
      Subscript[c, 1, 
         3] (E^(I (Subscript[ω, 1] + Subscript[ω, 
                       2]) Subscript[T, 0]) Subscript[A, 1] Subscript[
        A, 2] + 
            E^(I (Subscript[ω, 1] - Subscript[ω, 
                       2]) Subscript[T, 0]) Subscript[A, 1] Subscript[

\!\(\*OverscriptBox[\(A\), \(_\)]\), 2]) - 
      0.5 Subscript[c, 1, 1] Subscript[A, 
         1] (E^(I (Subscript[ω, 1] + p) Subscript[T, 0]) + E^(
              I (Subscript[ω, 1] - p) Subscript[T, 0])) - 
      Subscript[c, 1, 
         4] (E^(2 I  Subscript[ω, 2] Subscript[T, 
           0]) Subscript[A, 
                2]^2 + Subscript[A, 2] Subscript[

\!\(\*OverscriptBox[\(A\), \(_\)]\), 2])

Here are some substitutions

    subs = {Subscript[X, 1, 1] -> x11, 
   Subscript[X, 1, 1]^′′ -> x11'', 
   Subscript[c, 1, 1] -> c11, Subscript[c, 1, 2] -> c12, 
   Subscript[c, 1, 3] -> c13, Subscript[c, 1, 4] -> c14,
   Subscript[ω, 1] -> ω1, 
   Subscript[ω, 2] -> ω2, Subscript[A, 1] -> a1, 
   Subscript[A, 2] -> a2, Subscript[
\!\(\*OverscriptBox[\(A\), \(_\)]\), 1] -> ab1, Subscript[
\!\(\*OverscriptBox[\(A\), \(_\)]\), 2] -> ab2, Subscript[T, 0] -> t};

Now we replace the 0.5 by rationalising and effect the substitutions.

eqn1 = Rationalize[ eqn] /. subs

giving:

ω1^2 x11[t] + (x11^′′)[
   t] == -c12 (a1 ab1 + a1^2 E^(2 I t ω1)) - 
  1/2 a1 c11 (E^(I t (-p + ω1)) + E^(I t (p + ω1))) - 
  c14 (a2 ab2 + a2^2 E^(2 I t ω2)) - 
  c13 (a1 ab2 E^(I t (ω1 - ω2)) + 
     a1 a2 E^(I t (ω1 + ω2)))

Which I can see is a second order differential equation with a complex forcing. We can solve and simplify by doing

 sol =Simplify[DSolve[eqn1, x11[t], t]]

The output is long

{{x11[t] -> (E^(-I t (p + 2 ω1 - ω2)) (4 a1^2 c12 E^(
         I t (p + 4 ω1 - ω2))
          p (p^2 - 4 ω1^2) ω2 (4 ω1^4 - 
           17 ω1^2 ω2^2 + 4 ω2^4) + 
        6 a1 (ω1^2 - 
           4 ω2^2) (2 a2 c13 E^(I t (p + 3 ω1))
             p ω1^2 (p^2 - 
              4 ω1^2) (2 ω1 - ω2) + (2 ω1 \
+ ω2) (-2 ab2 c13 E^(I t (p + 3 ω1 - 2 ω2))
                p ω1^2 (p^2 - 4 ω1^2) + 
              E^(I t (2 ω1 - ω2)) (c11 E^(
                  I t ω1) ω1^2 (p + E^(2 I p t) p + 
                    2 ω1 - 2 E^(2 I p t) ω1) - 
                 2 ab1 c12 E^(I p t)
                   p (p^2 - 
                    4 ω1^2)) (2 ω1 - ω2) \
ω2)) - 
        6 E^(I t (p + ω1 - ω2))
          p (p^2 - 
           4 ω1^2) ω2 (4 ω1^2 - ω2^2) (2 \
a2^2 c14 E^(I t (ω1 + 2 ω2)) ω1^2 + 
           2 a2 ab2 c14 E^(
            I t ω1) (ω1^2 - 
              4 ω2^2) - ω1^2 (ω1^2 - 
              4 ω2^2) ((1 + E^(2 I t ω1)) C[1] - 
              I (-1 + E^(2 I t ω1)) C[2]))))/(12 p (p - 
        2 ω1) ω1^2 (p + 2 ω1) (ω1 - 
        2 ω2) (2 ω1 - ω2) ω2 (2 ω1 \
+ ω2) (ω1 + 2 ω2))}}

We can make some progress by looking at the real and imaginary parts seperately

FullSimplify[ComplexExpand[Re[sol]]]

giving

{{x11[t] -> -((a1 ab1 c12 + a2 ab2 c14)/ω1^2) + 
    C[1] Cos[t ω1] + (a1^2 c12 Cos[2 t ω1])/(
    3 ω1^2) - (
    a2^2 c14 Cos[2 t ω2])/(ω1^2 - 4 ω2^2) + 
    C[2] Sin[t ω1] + 
    a1 (-((ab2 c13 Cos[t (ω1 - ω2)])/(
        2 ω1 ω2 - ω2^2)) + (
       a2 c13 Cos[t (ω1 + ω2)])/(
       2 ω1 ω2 + ω2^2) + (c11 (p Cos[p t] Cos[
              t ω1] + 
            2 ω1 Sin[p t] Sin[t ω1]))/(p^3 - 
          4 p ω1^2))}}

Similarly for the imaginary part

FullSimplify[ComplexExpand[Im[sol]]]

giving

{{0 -> -((2 a1 c11 ω1 Cos[t ω1] Sin[p t])/(
     p^3 - 4 p ω1^2)) + (a1 c11 Cos[p t] Sin[t ω1])/(
    p^2 - 4 ω1^2) + ((2 ω1 + ω2) (a1 \
(ω1^2 - 
             4 ω2^2) (a1 c12 (2 ω1 - ω2) \
ω2 Sin[2 t ω1] - 
             3 ab2 c13 ω1^2 Sin[t (ω1 - ω2)]) + 
          3 a2^2 c14 ω1^2 ω2 (-2 ω1 + ω2) \
Sin[2 t ω2]) + 
       3 a1 a2 c13 ω1^2 (ω1 - 
          2 ω2) (2 ω1 - ω2) (ω1 + 
          2 ω2) Sin[
         t (ω1 + ω2)])/(3 ω1^2 (4 ω1^4 \
ω2 - 17 ω1^2 ω2^3 + 4 ω2^5))}}

There is no reason for this to be more simple. Perhaps putting in some initial conditions would help. Perhaps you just want the steady state solution without the starting transients? I note that your forcing term is a sum of forces. Perhaps it would be worthwhile looking at the response due to each term taken separately. The solution could then be found by superposition.

I just had a go at doing a FullSimplify on the solution. This looks a bit better.

{{x11[t] -> -((a1 ab1 c12)/ω1^2) + (
    a1^2 c12 E^(2 I t ω1))/(3 ω1^2) + (
    a1 c13 E^(
     I t (ω1 - ω2)) (a2 E^(
        2 I t ω2) (-2 ω1 + ω2) + 
       ab2 (2 ω1 + ω2)))/(-4 ω1^2 ω2 + \
ω2^3) + 
    a2 c14 (-(ab2/ω1^2) - (
       a2 E^(2 I t ω2))/(ω1^2 - 4 ω2^2)) + 
    C[1] Cos[t ω1] + (
    a1 c11 E^(I t ω1) (p Cos[p t] - 2 I ω1 Sin[p t]))/(
    p^3 - 4 p ω1^2) + C[2] Sin[t ω1]}}