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**davesface** Find a two-term expansion of the solution of

$\displaystyle \frac{dv}{dt}+\epsilon v^2+v=0$ for 0<t, $\displaystyle v(0)=1$.

I tried using v as a power series in $\displaystyle \epsilon$, ie $\displaystyle v(t)\sim v_0(t)+\epsilon v_1(t)+\epsilon^2 v_2(t)+...$, plugging it in, and then separating out terms by powers of $\displaystyle \epsilon$ and got

$\displaystyle O(1): v_0=Ae^{-t}$

$\displaystyle O(\epsilon): v_1=Be^{-t}+A^2e^{-2t}$

If I just continue through, I get $\displaystyle v(t) \sim Ae^{-t}+\epsilon (Be^{-t}+A^2e^{-2t})$, and then applying the BC I get $\displaystyle v(0) \sim A+\epsilon(B+A^2)=1$. Is it enough to say that B doesn't matter so long as $\displaystyle \epsilon$ is sufficiently small and thus A=1? Was my approach totally wrong?