This ODE system can't be solved symbolically with the given information.
First let's define the differential equations:
dg1 = y1'[t] == -k1 y1[t] - k2 y1[t]
dg2 = y2'[t] == k2 y1[t] - k3 y2[t]
dg3 = y3'[t] == k1 y1[t] + k3 y2[t] - k4 y3[t]
dg4 = y4'[t] == k4 y3[t] - k5 y2[t] y4[t] + k6 y5[t]
dg5 = y5'[t] == k5 y2[t] y4[t] - k6 y5[t]
Now the first three differential equations can be solved, because these are independent of the last two.
pSol = DSolve[{dg1, dg2, dg3}, {y1[t], y2[t], y3[t]}, t]
{{y1[t] -> E^((-k1 - k2) t) C[1],
y2[t] -> (E^(-k3 t) (-1 + E^((-k1 - k2) t + k3 t)) k2 C[1])/(-k1 - k2 + k3) + E^(-k3 t) C[2],
y3[t] -> (E^(-k3 t - k4 t) (E^(k3 t) k1^2 k3 -
E^((-k1 - k2) t + k3 t + k4 t) k1^2 k3 +
2 E^(k3 t) k1 k2 k3 - E^(k4 t) k1 k2 k3 -
E^((-k1 - k2) t + k3 t + k4 t) k1 k2 k3 + E^(k3 t) k2^2 k3 -
E^(k4 t) k2^2 k3 - E^(k3 t) k1 k3^2 +
E^((-k1 - k2) t + k3 t + k4 t) k1 k3^2 - E^(k3 t) k2 k3^2 +
E^((-k1 - k2) t + k3 t + k4 t) k2 k3^2 - E^(k3 t) k1^2 k4 +
E^((-k1 - k2) t + k3 t + k4 t) k1^2 k4 - E^(k3 t) k1 k2 k4 +
E^((-k1 - k2) t + k3 t + k4 t) k1 k2 k4 + E^(k3 t) k1 k3 k4 -
E^((-k1 - k2) t + k3 t + k4 t) k1 k3 k4 +
E^(k4 t) k2 k3 k4 -
E^((-k1 - k2) t + k3 t + k4 t) k2 k3 k4) C[1])/((k1 + k2 - k3) (k1 + k2 - k4) (k3 - k4)) -
(E^(-k3 t - k4 t) (-E^(k3 t) + E^(k4 t)) k3 C[2])/(k3 - k4) +
E^(-k4 t) C[3]
}}
From the last differential equation we get
Solve[dg5, y5[t]]
{{y5[t] -> (k5 y2[t] y4[t] - y5'[t])/k6}}
This (or even easier to see k6 y5[t]
) can be inserted into the fourth differential equation
dg4 /. (First@%)
y4'[t] == k4 y3[t] - y5'[t]
(Note that there is no y5[t]
left.)
Inserting also pSol
and applying Simplify
provides the remaining equation
% /. (First@pSol) // Simplify
(E^(-(k3 + k4) t) k4 (E^((-k1 - k2 + k3 + k4) t) (k1 + k2) (k1 - k3) (k3 - k4) C[1] -
E^(k4 t) k3 (-k1 - k2 + k4) ((k1 - k3) C[2] + k2 (C[1] + C[2])) -
E^(k3 t) (k1 + k2 - k3) (k4 (k4 C[3] - k3 (C[2] + C[3])) +
k2 (-k4 C[3] + k3 (C[1] + C[2] + C[3])) +
k1 (-k4 (C[1] + C[3]) + k3 (C[1] + C[2] + C[3])))))/((k1 + k2 - k3) (k1 + k2 - k4) (k3 - k4)) +
y4'[t] + y5'[t] == 0
If one tries to solve this equation using
DSolve[%, {y4[t], y5[t]}, {t}]
the following message is received
DSolve::underdet: There are more dependent variables than equations, so the system is underdetermined. >>
The remaining equation can be solved for y4[t]
, though.
DSolve[%%, {y4[t]}, {t}]
{{y4[t] ->
C[4] + (E^(-(k1 + k2) t) (k1 - k3) (k3 - k4) k4 C[1] +
E^(-(k3 + k4) t) (-E^(k4 t) k4 (-k1 - k2 + k4) ((k1 - k3) C[2] + k2 (C[1] + C[2])) -
E^(k3 t) (k1 + k2 - k3) (k4 (k4 C[3] - k3 (C[2] + C[3])) +
k2 (-k4 C[3] + k3 (C[1] + C[2] + C[3])) +
k1 (-k4 (C[1] + C[3]) +
k3 (C[1] + C[2] + C[3])))) - (k1 + k2 - k3) (k1 + k2 - k4) (k3 - k4) y5[t])/((k1 + k2 - k3) (k1 + k2 - k4) (k3 - k4))
}}
The ODE system can also be solved, e.g. if y5'[t]
is a constant.
Additional note:
With the solutions of the first three equations (pSol
), the remaining equations can be rewritten as
y4'[t] == k6 y5[t] - a[t] y4[t] + b[t]
y5'[t] == a[t] y4[t] - k6 y5[t]
Therefore the remaining problem is to solve this non-homogeneous first-order linear differential equation with variable coefficients.
DSolve
? It would be good if you could provide the equations in Mathematica input form. It would let potential answerers try their solutions. $\endgroup$