How to get the series coefficient of a function defined by a differential equation [duplicate]

Question

I need a Taylor series approximation for $x(t)$, which is defined by the following differential equation.

$\frac{dx}{dt}=-k_1 x+(1-x)k_2 e^{-k_3 t}$, $x(0)=0$

     {
     D[x[t], t] == -k1 x[t] + (1 - x[t]) k2 Exp[-k2 t]
     , x[0] == 0
     }

How do I get the coefficients $x^{''}(0)$, $x^{'''}(0)$ from Mathematica?

Answer 1

Search the documentation for Taylor, first hit is Series.

Solve Differential equation,

sol = x[t] /. First@DSolve[
    {
     D[x[t], t] == -k1 x[t] + (1 - x[t]) k2 Exp[-k2 t]
     , x[0] == 0
     }
    , x[t], t];

You can get the series by Series[sol, {t, 0, 3}] but you are interested only in the 2nd and 3rd order coefficients, so you see at the bottom of the documentation the related functions and you find SeriesCoefficient

The Series 2nd order coefficient is:

SeriesCoefficient[sol, {t, 0, 2}]

-((k1 k2)/2) - k2^2

Third order:

SeriesCoefficient[sol, {t, 0, 3}]

1/6 (k1^2 k2 + 3 k1 k2^2 + 5 k2^3)

Answer 2

Looks like homework, but anyway. The idea is to take derivatives, substituting from lower derivatives to rewrite e.g. x'' in terms of x' (which you can already rewrite in terms of x). Below I show this for x''[0]. Similar substitutuins, keeping track of past derivative rewrites, can be used to get higher derivatives to evaluate at t=0.

xprime[t] := -k1 x[t] + (1 - x[t]) k2 Exp[-k2 t]
substrule = x'[t] -> xprime[t];
x[0] = 0;

In[203]:= D[xprime[t], t] /. substrule /. t -> 0

(* Out[203]= -k1 k2 - 2 k2^2 *)