Hi !

Note that X'+Y'+Z'=0 hence X+Y+Z=C

Bringing back Y=C-X-Z into the first and tbe third equation reduces to a system of two equations where the unknown are X and Z, easier to solve. Various methods are allowed (Matricial, or Laplace transform, or substitution)

For example, sustitution leads to

X''+(a+b+c+d)X'+(ab+ad+cd)X=Ccd

Solving this ODE leads to a general solution on the form :

X(t) = Xoo+C1*exp(k1*t)+C2*exp(k2*t)

Xoo is a constant term.

The coeffients C1 and C2 depend on the initial condition, but it doesn't matter.

k1 and k2 are rather simple functions of a, b, c, d which can be easily expressed.

It is then possible to show that k1 and k2 are negative as far as a, b, c, d are positive. So, when t tends to infinity, the exponential terms tends to 0. Only the constant term remains, which is the limit as t approches the infinity.