Approximate solution of differential equation My task: find approximate solution as $$y = y_0(x) + y_1(x)\lambda + y_2(x)\lambda^2 + y_3(x)\lambda^3$$
of differential equation 
$$y' = \sin x + \lambda e^y, y(0)=1-\lambda.   \ \ \ \   (*)$$
My attempt :
Let $$y(x,\lambda) = y_0(x) + y_1(x)\lambda + y_2(x)\lambda^2 + y_3(x)\lambda^3.$$
 Then $$ y_0(x) = y(x,0)$$ $$y_1(x) =\frac{ \partial y(x,\lambda)}{\partial \lambda}_{\lambda = 0}$$ $$y_2(x) =\frac{1}{2}\frac{ \partial^2y(x,\lambda)}{\partial \lambda^2}_{\lambda = 0}$$
$$y_3(x) =\frac{1}{6}\frac{ \partial^3y(x,\lambda)}{\partial \lambda^3}_{\lambda = 0}$$
And 
$$ y_0'(x) = y'_x(x,0)$$ $$y_1'(x) =\frac{ \partial}{\partial \lambda}y'_x(x,\lambda)_{\lambda = 0}$$ $$y_2'(x) =\frac{1}{2}\frac{ \partial^2}{\partial \lambda^2}y'_x(x,\lambda)_{\lambda = 0}$$
$$y_3'(x) =\frac{1}{6}\frac{ \partial^3}{\partial \lambda^3}y'_x(x,\lambda)_{\lambda = 0}$$
After that, using (*), i got
$$y_0'(x)= sin x$$
$$y_1'(x)=e^{(y_0)}$$
$$y_2'(x)=e^{(y_0)}y_1;$$
$$y_3'(x)=e^{(y_0)}(y_1^2+y_2)$$
And i stopped here. There isn't a solution in these integrals. Maybe, there is another solution of my task? Thanks

I don't know if you prefer the exact solution instead of approximates. Infortunately, the analytic expression includes an integral $\int e^{-\cos(x)}dx$ which has no standard closed form :

Note that $y(x)=-\cos(x)-\ln(f(x)+C)$
where $f(x)$ is the function as shown above. This is consistent with the previous answer from grdgfgr.
For numerical calculus, there is no difficulty to compute accurate values of the integral, then accurate numerical values of the whole function $y(x)$. 
If you want good approximates on the form of formula, it is possible to express $\int_0^x e^{-\cos(t)}dt$ on the form of finite series.
Ultimately, the integral can be remplaced by the next Fourier series : 
$$\int_0^{x} e^{-\cos(t)}dt=a_0x+\displaystyle\sum_{n=1} \frac{a_n}{n}\sin(nx)$$
$a_0=1.266066$
$a_1=-1.130318$
$a_2=0.271495$
$a_3=-0.0443368$
$a_4=0.00547424$
$a_5=-0.000542926$
It is sufficient to make negligible the deviations.
