Please solve the system Solve the system dxdt zabt1 Xmiddo

Please solve the system:

Solve the system: dx/dt = -z(a/bt+1) - Xmiddotg - B dy/dt = n y (a/bt+1) + 2.5 y/[1-k^2(1-e^-c/mt)]^2 [c middot e^-c/mt]

Solution

Given dx/dt and dy/dt

let Y\'=dy/dt, X\'=dx/dt

X\'= -Z(a/bz+1) -Xg - B

Y\'=ny(a/bz+1) + 2.5y[c.e^-cz/m]/(1-k^2(1-e^(-c/mz)))

X\'\' differentiating with respect to t,

X\'\' = - g (1) = -g

Y\'\' = n.(a/bz+1) + 2.5[c.e^-cz/m]^2/(1-k^2(1-e^(-c/mz)))

differentiation dy/dt(y) = 1 so we get

=   n.(a/bz+1) + 2.5[c.e^-cz/m]^2/(1-k^2(1-e^(-c/mz)))

from this we can write characteristic equation

c1 - c2 =0

x(t) = e^t.

for solving y we have  

n.(a/bz+1) + 2.5[c.e^-cz/m]^2/(1-k^2(1-e^(-c/mz)))=0

From this all are constants so writing characteristic equation for this gives

(1-k^2(1-e^(-c/mz)))n.(a/bz+1) + 2.5[c.e^-cz/m]^2 =0

(1-k^2(1-e^(-c/mz)))n.(a/bz) + (1-k^2(1-e^(-c/mz)))n +2.5[c.e^-cz/m]^2 =0

taking characteristic equation we get

c1 + c2 =1

c1 - c2 =1

c1 = 1 and c2=0

x(t) = e^t , y(t) =e^t