first order differential equation I needed help with this Differential Equation, below:
$$dy/dt  = t + y, \text{ with } y(0) = -1$$
I tried $dy/(t+y) = dt$ and integrated both sides, but it looks like the $u$-substitution does not work out. 

This equation is not separable.  In other words, you can't write it as $f(y)\;dy=g(t)\;dt$.  A differential equation like this can be solved by integrating factors.  First, rewrite the equation as:
$$\frac{dy}{dt}-y=t$$
Now we multiply the equation by an integrating factor so we can use the product rule, $d(uv)=udv+vdu.$  For this problem, that integrating factor would be $e^{-t}$.
$$e^{-t}\frac{dy}{dt}-e^{-t}y=\frac d{dt}(e^{-t}y)=te^{-t}$$
$$e^{-t}y=\int te^{-t}dt=-te^{-t}+\int e^{-t}dt=-te^{-t}-e^{-t}+C$$
$$y=Ce^t-t-1$$
For this specific problem, we could also follow Iasafro's suggestion.
$$z=y+t,\frac{dz}{dt}=\frac{dy}{dt}+1,\frac{dy}{dt}=\frac{dz}{dt}-1$$
$$\frac{dz}{dt}-1=z,\frac{dz}{dt}=z+1,\frac{dz}{z+1}=dt$$
As you can see, this substitution resulted in a separable equation, allowing you to integrate both sides.
