Question: Why does dsolve return a solution with a singularity?

I don't understand why the solution of sys_2 isn't those of sys_1 when M__p=1 and M__a=0 ?

Traces of the computation seem to indicate that dsolve proceeds exactly the same for sys_2 and sys_1 .

Please note that sol_1 contains a term of the form t*cos(t) that sol_2 doesn't, thus the question: "Is sol_2 correct?"

Could you help me to fix this?
TIA

restart

infolevel[dsolve] := 4;

4

(1)

sys_1 := {diff(x(t), t$2)=sin(t)-x(t), x(0)=0, D(x)(0)=0};
sol_1 := dsolve(sys_1)

{diff(diff(x(t), t), t) = sin(t)-x(t), x(0) = 0, (D(x))(0) = 0}

 

Methods for second order ODEs:
--- Trying classification methods ---
trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
-> Try solving first the homogeneous part of the ODE
   checking if the LODE has constant coefficients
   <- constant coefficients successful
   -> Determining now a particular solution to the non-homogeneous ODE
      building a particular solution using variation of parameters
<- solving first the homogeneous part of the ODE successful

 

x(t) = (1/2)*sin(t)-(1/2)*cos(t)*t

(2)

sys_2 := {(M__p+M__a)*diff(x(t), t$2)=M__p*sin(t)-x(t), x(0)=0, D(x)(0)=0};
sol_2 := dsolve(sys_2)

{(M__p+M__a)*(diff(diff(x(t), t), t)) = M__p*sin(t)-x(t), x(0) = 0, (D(x))(0) = 0}

 

Methods for second order ODEs:
--- Trying classification methods ---
trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
-> Try solving first the homogeneous part of the ODE
   checking if the LODE has constant coefficients
   <- constant coefficients successful
   -> Determining now a particular solution to the non-homogeneous ODE
      building a particular solution using variation of parameters
<- solving first the homogeneous part of the ODE successful

 

x(t) = sin(t/(M__p+M__a)^(1/2))*M__p*(M__p+M__a)^(1/2)/(M__p+M__a-1)-M__p*sin(t)/(M__p+M__a-1)

(3)

eval(sol_2, [M__p=1, M__a=0])

Error, numeric exception: division by zero

 

 

Download SomethingWrong.mw

PS: Already, in the following case, dsolve doesn't return the solution of sys_1.

sys_3 := {(A+B)*diff(x(t), t$2)=(A+B)*sin(t)-x(t), x(0)=0, D(x)(0)=0};
sol_3 := dsolve(sys_3)

If I do this

sys_4 := {(A+B)*diff(v(t), t)=(A+B)*sin(t)-x(t), diff(x(t), t)=v(t), x(0)=0, v(0)=0}:
sol_4 := dsolve(sys_4)

I get a very complex solution wich contains a piecewise function which separates the cases A+B=1 and A+B<>1.
Evaluating sol_4 for A+B=1 gives the same expression than sys_1:

simplify(eval(sol_4, A=1-B), trig)
       /       1                  1          1         \ 
      { v(t) = - sin(t) t, x(t) = - sin(t) - - cos(t) t }
       \       2                  2          2         / 

Here is a workaround to get the correct solution of sys_2:

sys_5 := {(M__P+M__A)*diff(v(t), t)=(M__P+C)*sin(t)-x(t), diff(x(t), t)=v(t), x(0)=0, v(0)=0}:
sol_5 := dsolve(sys_5):
simplify(eval(sol_5, [M__P=1, M__A=0, C=0]), trig)
       /       1                  1          1         \ 
      { v(t) = - sin(t) t, x(t) = - sin(t) - - cos(t) t }
       \       2                  2          2         / 

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