calculus - Power series solution for ODE

The ODE I have is $$y'(x)+e^{y(x)}+\frac{e^x-e^{-x}}{4}=0, \hspace{0.2cm} y(0)=0$$

I want to determine the first five terms (coefficients $a_0,\ldots, a_5$) of the power series solution $$y(x)=\sum_{k=0}^{\infty} a_kx^k$$ So far, I know that $$y'(x)=\sum_{k=1}^{\infty} a_kkx^{k-1}$$
Now I plug these back into the equation and get:
$$\sum_{k=1}^{\infty} a_kkx^{k-1} + e^{\sum_{k=0}^{\infty} a_kx^k} + \frac{e^x-e^{-x}}{4}=0$$ . Now I'm not sure how to continue with this. Please help.

Answer

Since you just need a few terms, setting $$y=\sum_{n=1}^6 a_i x^i$$ (because of the condition $y(0)=0$) you could develop $e^y$ as a Taylor series around $x=0$ and get
$$e^y=1+a_1 x+\frac{1}{2} \left(a_1^2+2 a_2\right) x^2+\frac{1}{6} \left(a_1^3+6 a_2 a_1+6 a_3\right) x^3+$$ $$\frac{1}{24} \left(a_1^4+12 a_2 a_1^2+24 a_3 a_1+12 a_2^2+24 a_4\right) x^4+$$ $$\frac{1}{120} \left(a_1^5+20 a_2 a_1^3+60 a_3 a_1^2+60 a_2^2 a_1+120 a_4 a_1+120 a_2 a_3+120 a_5\right) x^5+$$ $$\frac{1}{720} \left(a_1^6+30 a_2 a_1^4+120 a_3 a_1^3+180 a_2^2 a_1^2+360 a_4 a_1^2+720 a_2 a_3 a_1+720 a_5 a_1+120 a_2^3+360 a_3^2+720 a_2 a_4+720 a_6\right) x^6$$

Expanding the term $\frac{e^x-e^{-x}}{4}$ as a Taylor series too, the differential equation then write
$$(a_1+1)+\left(a_1+2 a_2+\frac{1}{2}\right) x+\frac{1}{2} \left(a_1^2+2 a_2+6 a_3\right) x^2+\frac{1}{12} \left(2 a_1^3+12 a_2 a_1+12 a_3+48 a_4+1\right) x^3+$$ $$\frac{1}{24} \left(a_1^4+12 a_2 a_1^2+24 a_3 a_1+12 a_2^2+24 a_4+120 a_5\right) x^4+$$ $$\frac{1}{240} \left(2 a_1^5+40 a_2 a_1^3+120 a_3 a_1^2+120 a_2^2 a_1+240 a_4 a_1+240 a_2 a_3+240 a_5+1440 a_6+1\right) x^5+$$ $$\frac{1}{720} \left(a_1^6+30 a_2 a_1^4+120 a_3 a_1^3+180 a_2^2 a_1^2+360 a_4 a_1^2+720 a_2 a_3 a_1+720 a_5 a_1+120 a_2^3+360 a_3^2+720 a_2 a_4+720 a_6\right) x^6=0$$

Cancelling the coefficients lead to $$a_1=-1\qquad a_2=\frac{1}{4}\qquad a_3=-\frac{1}{4}\qquad a_4=\frac{7}{48}\qquad a_5=-\frac{19}{160}$$

I hope I did not make any mistake since my results do not coincide with DaveNine's answer.