Show D.E. is exact Show that the differential equation $$(y^3 + \cos(t))y'= 2+ y\sin(t)~,~~~~ y(0)=-1$$ is exact and solve the i.v.p.
So far I have $$(y^3 +\cos(t))y'- y\sin(t)-2=0$$ where $M_y=N_t,$ making it exact. So then I have 
$$\frac{dh}{dt}=-y\sin(t)-2~,~~~~~~~~ \frac{dh}{dy}=y^3tcos(t)$$
Integrate $dh/dy$ which would give $y^4/4+y\cos(t)+g(t)$.
Integrate $dh/dt$ which would give me $y\cos(t)-2t+g(y)$, and then I get stuck.

Hint: What term does $\int h_y$ have that $\int h_t$ doesn't? What term does $\int h_t$ have that $\int h_y$ doesn't? Do you see that each one has the extra term that the other doesn't (i.e., the $g(t)$ and the $g(y)$), and so you simply need to combine the two to form your solution $y(t)$. This is referred to as the method of "inspection" because you antidifferentiate both partials, and form your solution by combining the two antiderivatives, paying heed to the missing "constant of integration" that comes from each partial. 
