How to solve an initial value problem if my function is implicit? I have the differential equation:
$\dfrac{dy}{dt} = \dfrac{1}{2y+3}$ 
with the initial value of $y(0) = 1$
So I solve the diffeq via separation of variables:
$y^2 + 3y = t + c$
But from here, how do I solve for the function? Am I "allowed" to use the initial value information with an implicit function such as this, and give an implicit function as my answer, as below?
$y^2 + 3y = t + 4$
As far as I can tell, it should be perfectly fine to leave my answer as an implicit function. And I don't see a way to solve for $y$ here. TO be honest, I used a diffeq solver to check my work, and that tool did come up with an explicit solution, though when I tried to see where it came from I had no idea. 
 A: Yes, you should use your known initial condition to solve for $c$. 
Proceeding to solve for $y(t)$, we write $y^2+3y+\frac{9}{4}-\frac{9}{4}=\left(y+\frac{3}{2}\right)^2-\frac{9}{4}$ by completing the square.
Now, we write $\left(y+\frac{3}{2}\right)^2-\frac{9}{4}=t+4.$ and solve for $y$ in terms of $t$.
Adding $\frac{9}{4}$, taking the square root, and subtracting $\frac{9}{2}$, we get: $$y(t)=\sqrt{t+\frac{25}{4}}-\frac{3}{2}$$
EDIT: I had misplaced a $3$.
Edit: To address your comment:
Note that we can write $y(t)=\sqrt{t+\frac{25}{4}}-\frac{3}{2}=\sqrt{\frac{1}{4}(4t+25)}-\frac{3}{2}=\frac{1}{2}\sqrt{(4t+25)}-\frac{3}{2}=\frac{1}{2}(\sqrt{(4t+25)}-3).$
The reason that $c$ is different is because $c$ is some arbitrary constant. In this case, we have already solved for it. 
A: From where you arrived,
$y^2+3y=t+c$
Substituting $y(0)=1$, we get $c=4$
Hence, we get $y^2+3y-4=t$
Proceed by completing the square for $y^2+3y-4=(y+\frac{3}{2})^2-\frac{7}{4}$
$(y+\frac{3}{2})^2-\frac{7}{4}=t$
Then we get $y=\frac{1}{2}(3±\sqrt{4t+7})$
