TITLE: Approximate solution of differential equation
QUESTION [6 upvotes]: 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

REPLY [4 votes]: The solution is easier than you may think at first glance.
Just expand the solution of the differential equation into Taylor series.
$$y＝y(0)+\frac{y'(0)}{1!}x+\frac{y''(0)}{2!}x^{2}+...$$  
The coefficients $y'(0),y''(0)$, etc. we can evaluate directly from the differential equation by differentiating it. Some of the first coefficients: 
$$y(0)=1-\lambda$$  
$$y'(0)=\lambda e^{1-\lambda}$$  
$$y''(0)=1+\lambda^{2} e^{2(1-\lambda)}$$
$$y'''(0)=\lambda e^{1-\lambda}\left [1+2\lambda^{2} e^{2(1-\lambda)}\right ]$$  
Now, after you've put these results into the Taylor series, all you need is to regroup the series terms to separate terms that contain only $\lambda,\lambda^{2},\lambda^{3}$ etc.