The first-order ODE \[\frac{\dd y}{\dd t}=\frac{-3 t^2+4 t + 3}{2y}\] can be rewritten in differential form as \[2y\,\dd y=(-3 t^2+4 t + 3)\,\dd t\] The left- and right-hand sides can be replaced by \(\dd(y^2)\) and \(\dd(-t^3+2 t^2+3 t)\), respectively. After all, \(y^2\) and \(-t^3+2 t^2+3 t\) are antiderivatives of \(y\) and \(-3 t^2+4 t + 3\), respectively. So we have an equality between two differentials: \[\dd(y^2)=\dd(-t^3+2 t^2+3 t)\] The functions ‘behind the d's’ are thus equal to each other up to a constant: \[y^2=-t^3+2 t^2+3 t+c\] for some constant \(c\). The explicit solution is: \[y=\sqrt{-t^3+2 t^2+3 t+c}\quad \text{or} \quad y=-\sqrt{-t^3+2 t^2+3 t+c}\]

For those who find the final steps in the processing of differentials too formal or are not familar with differentials: you can also add to the left- and right-hand side of the \[2y\,\dd y=(-3 t^2+4 t + 3)\,\dd t\] the integral symbol: \[\int 2y\,\dd y=\int (-3 t^2+4 t + 3)\,\dd t\] Now you just continue with computing these integrals. So: \[y^2=-t^3+2 t^2+3 t+c\] where we have combined the integration constants into a single constant \(c\) and have placed it on the right-hand side of the equation.