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 $y(t)$

$$\frac{3y^2-t^2}{y^5} \frac{dy}{dt} + \frac{t}{2y^4}=0$$
$$\frac{dt}{dy}=-2\frac{3y^2-t^2}{ty} $$
$$\frac{2tdt}{2ydy}=-2\frac{3y^2-t^2}{y^2} $$
$T=t^2\quad;\quad Y=y^2$
$$\frac{dT}{dY}=2\frac{T}{Y} -6$$
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 $t^2$.
