First order non-linear ordinary differential equation containing $e^x$ $$(t+e^y) + \left(\frac{t^2}{2} + 2te^y\right)y' = 0$$
How can i solve this? I've never seen a differential equation like that.

$$(t+e^y)+\left(\dfrac{t^2}2+2te^y\right)\dfrac{dy}{dt}=0$$
$$\Rightarrow 2(t+e^y)dt+(t^2+4te^y)dy=0$$
$$2tdt+2e^ydt+t^2dy+4te^ydy=0$$
$$$$
On multiplying throughout by $e^y$ (to convert the ODE into an exact ODE) and subsequently rearranging, we get 
$$$$
$$(2te^ydt+t^2e^ydy)+ 2(2te^{2y}dy+e^{2y}dt)=0$$ 
$$d(t^2e^y)+d(2e^{2y}t)=0$$
$$t^2e^y+2te^{2y}=C$$
$$$$
Edit: If you (like me) dislike messy looking terms, you can substitute $u=e^y$ in the first step and modify the equation.
