Solving y(t) from a differential equation Given the initial problem
$$\frac{3y^2-t^2}{y^5} \frac{dy}{dt} + \frac{t}{2y^4}=0, y(1) =3 $$
I got the solution to be
$$\frac{t^2}{4y^4}-\frac{3}{2y^2} = -\frac{53}{324} $$
But how do I write it in terms of $y(t)$
 A: Letting $P:=\frac{t}{2y^4}$ and $Q:=\frac{-t^2+3y^2}{t^5}$ then $P_y=Q_t$ so the non linear DE is exact and we have the implicit solution $\frac{-3}{2y^2}+\frac{t^2}{4y^4}=k$ for an arbitrary constant $k$. We have for $y,k\not=0$ that  $-\frac{3}{2y^2}+\frac{t^2}{4y^4}=k$ give $t^2-6u-4ku^2=0$ with $u:=y^2$. Completing square we have $(u+\frac{3}{4k})^2=\frac{1}{4}(\frac{t^2}{k}+\frac{9}{4k^2})$. Squaring both sides we find $u+\frac{3}{4k}=\pm\frac{1}{2}\sqrt{\frac{4kt^2+9}{4k^2}}$ and then $y^2=\pm\frac{\sqrt{4kt^2+9}}{4k}-\frac{3}{4k}$. Hence $y=\pm\frac{1}{2}\sqrt{\frac{-3}{k}\pm\frac{\sqrt{4kt^2+9}}{k}}$. Hence, we have the four cases for to study. Invoking the condition $y(1)=3$ we find only $k=\frac{-53}{324}$ with the combination the sign $(+,-)$. I think that you can complete the details.
A: $$\frac{3y^2-t^2}{y^5} \frac{dy}{dt} + \frac{t}{2y^4}=0$$
$$\frac{dt}{dy}=-2\frac{3y^2-t^2}{ty} $$
$$\frac{2tdt}{2ydy}=-2\frac{3y^2-t^2}{y^2} $$
$T=t^2\quad;\quad Y=y^2$
$$\frac{dT}{dY}=2\frac{T}{Y} -6$$
First order linear ODE easy to solve :
$$T=6Y+c\:Y^2$$
Condition : $y(1)=3\quad\implies\quad Y(1^2)=3^2$
$1^2=6*(3^2)+c\:3^4\quad\implies\quad c=-\frac{53}{81}$
$$T=6Y-\frac{53}{81}Y^2$$
Solve it for $Y$ . Then replace $Y$ by $y^2$ and replace $T$ by $t^2$.
