Getting wrong answer with exact equation with initial condition. Solve the initial value problem $(4y+2t-5)dt+(6y+4t-1)dy=0, y(-1)=2$
This is an exact equation with $M(t,y)=\frac{\partial f}{\partial y}=4y+2t-5$ and $N(t,y)=\frac{\partial f}{\partial t}=6y+4t-1$
Integrating the first one $f = \int 4y+2t-5 dy = 2y^2+2ty-5y+g(t)$
Then differentiating with respect to the opposite variable $\frac{\partial f}{\partial t} = 2y + g'(t)$
But this is equivalent to $N(t, y)$ so $6y+4t-1 = 2y + g'(t)$ rewriting gives $g'(t)=rt-1+4y$ and integrating with respect to $t$ gives $g(t)=2t^2-t+4yt$
The solution to the ODE is $4y^2+6ty-5y+2t^2-t=C$
Applying $y(-1)=2$ gives $4(2)^2+6(-1)(2)-5(2)+2(-1)^2+1=-1$
But the answer key gives $4ty+t^2-5t+3y^2-y=8$
Where did I go wrong? 

You have the roles of the two pieces reversed.
With 
$$
M(t,y) = 4y + 2t - 5; \quad N(t,y) = 6y + 4t - 1,
$$
you check exactness by verifying that $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, but then you proceed with
$$
\frac{\partial f}{\partial t} = M; \quad \frac{\partial f}{\partial y} = N.
$$
It looks like you are doing everything else correctly; if you restart from the above line, you will arrive at the correct answer.
