# Asymptotic expansion on 3 nonlinear ordinary differential equations

The 3 nonlinear differential equations are as follows $$\epsilon \frac{dc}{dt}=\alpha I + \ c (-K_F - K_D-K_N s-K_P(1-q)), \nonumber$$
$$\frac{ds}{dt}= \lambda_b P_C \ \epsilon \ c (1-s)- \lambda_r (1-q) \ s, \nonumber$$ $$\frac{dq}{dt}= K_P (1-q) \frac{P_C}{P_Q} \ \ c - \gamma \ q, \nonumber$$ I want to use asymptotic expansion on $c, s$ and $q$. And values of parameters are: $K_F = 6.7 \times 10^{-2},$
$K_N = 6.03 \times 10^{-1}$ $K_P = 2.92 \times 10^{-2}$, $K_D = 4.94 \times 10^{-2}$, $\lambda_b= 0.0087$, $I=1200$ $P_C = 3 \times 10^{11}$ $P_Q = 2.304 \times 10^{9}$ $\gamma=2.74$ $\lambda_{b}=0.0087$ $\lambda_{r}= 835$ $\alpha=1.14437 \times 10^{-3}$ For initial conditions: $$c_0(0)= c(0) = 0.25 \nonumber$$ $$s_0(0)= cs(0) = 0.02 \nonumber \nonumber$$ $$q_0(0)=q(0) = 0.98 \nonumber \nonumber$$ and $$c_i(0)= 0, \ i>0\nonumber$$ $$s_i(0)= 0, \ i>0 \nonumber \nonumber$$ $$q_i(0)=0, i>0. \nonumber \nonumber$$ => i started with the expansions : $$c= c_0+ \epsilon c_1 + \epsilon^2 c_2+......... \nonumber$$ $$s= s_0+ \epsilon s_1 + \epsilon^2 s_2+......... \nonumber$$ $$q= q_0+ \epsilon q_1 + \epsilon^2 q_2+......... \nonumber$$ we are only interseted in up to fisrt power of $\epsilon$. so, we should get total 6 approximate differential equations to get answer for $\frac{dc_0}{dt}, \frac{ds_0}{dt}, \frac{dq_0}{dt}, \frac{dc_1}{dt}, \frac{ds_1}{dt}$ and $\frac{dq_1}{dt}$ but i think $\frac{dc_1}{dt}$ will disappear while expanding and equating the up to first power of $\epsilon$, do i need to go further up to $\epsilon{^2}$ because $\frac{dc_1}{dt}$ is very important to find and we need 6 approximate differetial equations in total. what can i do? please some one help me.

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Hi! Is this a question about the software system Mathematica? – Michael E2 Jul 25 '14 at 22:03
not really but you can solve it using software once you got 6 approximate differential equations – Manjushree Jul 25 '14 at 23:14
Then it would be convenient if you posted your code, so those who might help don't have to type it in themselves. It reduces needless errors; moreover, it's rather expected on this site. – Michael E2 Jul 26 '14 at 0:56