Question: pde system

I have a set of PDEs & I tried to solve it. First of all I defined some functions as:
H := (diff(a(t, y), t))/a(t, y), Hy := (diff(a(t, y), y))/a(t, y), H1 := diff(H(t, y), t), a1 := diff(a(t, y), t), a2 := diff(a1, t), Hy1 := diff(Hy(t, y), y), ay1 := diff(a(t, y), y), ay2 := diff(ay1, y), n1 := diff(n(t, y), t), n2 := diff(n1, t), ny1 := diff(n(t, y), y), ny2 := diff(ny1, y), pef := -2*n(t, y)^2*Hy1-3*n(t, y)^2*Hy^2-n(t, y)*ny2-2*n(t, y)*Hy*ny1-2*H(t, y)*n1/n(t, y), `ρef` := 3*n(t, y)^2*Hy1+6*n(t, y)^2*Hy^2, `ωef` := pef/`ρef`,
then I wrote my equations as:
pde1 := 3*H(t, y)^2-`ρef` = 0, pde2 := 2*H1+3*H(t, y)^2+pef = 0, pde3 := Hy^2+Hy*ny1/n(t, y)-H1/n(t, y)^2-2*H(t, y)^2/n(t, y)^2+H(t, y)*n1/n(t, y)^3 = 0,
I just want to solve two of them as:
sys := {pde1, pde3}, ics := {a(0, 0) = 1, n(0, 0) = -2, (D[1](a))(0, 0) = 1, (D[1](n))(0, 0) = 1, (D[2](a))(0, 0) = 1, (D[2](n))(0, 0) = 1}, sol := pdsolve(sys, {ics}, numeric)
after this I have an error like this:"Error, (in pdsolve/numeric/process_PDEs) number of dependent variables and number of PDE must be the same"
please help me. I really need to solve it. thanks a lot.