dy/dt = t/2y+1. Find the general solution.
I get that y^2 +y = 1/2t^2 + C. How do I solve for Y?
Complete the square.
$\displaystyle y^2 + y + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 = \frac{1}{2}t^2 + C$
$\displaystyle \left(y + \frac{1}{2}\right)^2 - \frac{1}{4} = \frac{1}{2}t^2 + C$
$\displaystyle \left(y + \frac{1}{2}\right)^2 = \frac{1}{2}t^2 + \frac{1}{4} + C$
$\displaystyle y + \frac{1}{2} = \sqrt{\frac{1}{2}t^2 + \frac{1}{4} +C}$
$\displaystyle y= \sqrt{\frac{1}{2}t^2 + \frac{1}{4} +C} - \frac{1}{2}$.
First, your equation for $\displaystyle y$ is incorrect.
$\displaystyle y = \pm\sqrt{\frac{1}{2}t^2 + \frac{1}{4} + C} - \frac{1}{2}$.
If $\displaystyle y(1) = \frac{3}{2}$, then
$\displaystyle \frac{3}{2} = \pm \sqrt{\frac{1}{2}(1)^2 + \frac{1}{4} +C} - \frac{1}{2}$
$\displaystyle 2 = \pm \sqrt{\frac{1}{2} + \frac{1}{4} + C}$
$\displaystyle 4 = \frac{3}{4} + C$
$\displaystyle \frac{13}{4} = C$.
I think you'll find the solution is NOT unique...
$\displaystyle y = \pm \sqrt{\frac{1}{2}t^2+ \frac{1}{4} + \frac{13}{4}} - \frac{1}{2}$
$\displaystyle = \pm \sqrt{\frac{1}{2}t^2 + \frac{7}{2}} - \frac{1}{2}$.