Taylor Expansion involving $e^y$

Question

I'm asked to find the first few nonzero terms in the Taylor expansion about $x=0$ of the IVP $y''=e^y, y(0)=1, y'(0)=-1$

Is this 2nd derivative calculated as $y''=e^{y(0)}=e^1=e$ ?

If so, then $y'''=(e^y)(y')=(e^{y(0)})(-1)=(e^1)(-1)=-e$

and so on? Guess my question is really about how to calculate the 2nd derivative.

Thanks.

Answer 1

Note that, since $y(0)=1$, $y'(0)=-1$, and $y''(0)=\mathrm e^{y(0)}=\mathrm e$, its Taylor's expansion begins with $\;y(x)=1-x+\frac{\mathrm e}2 x^2+\dotsm$

Then differentiate both sides of the differential equation; you obtain $$y'''=\mathrm e^y\cdot y',\enspace\text{whence}\quad y'''(0)=\mathrm e\cdot -1=-\mathrm e,$$ so $$y=1-x+\frac{\mathrm e}2 x^2-\frac{\mathrm e}6 x^3+\dotsm$$ We'd obtain similarly $\;y^{(4)}(0)=\mathrm e(1+\mathrm e)$, &c.