Which is right?, separation of variables, differential equations

Question

I've recently found two explanations on how to solve this IVP, One is wrong , one is not. Which one is which? and more importantly why ?

There's free treasure for getting the right answer ^^

For $X'(t)= (X^2)t,$

and Initial Conditions $X(o)=X_o,$

Does $X(t)= -2/(t^2+c) \Longrightarrow X(t)= -2/(t^2-2/X_o)$ as $C=-2/X_o$

Or does $X(t)= -2/(t^2+2c) \Longrightarrow X(t)=-2/(t^2-2X_o)$ as $C=-X_o$?

on second thoughts , treasure that is free isn't very valuable , so an imaginary high five should suffice as a prize.

Answer 1

Easy way to check:

1. Make sure the answer satisfies the initial condition.
2. Make sure the answer satisfies the differential equation.

In your question, (1) will fail for one of the answers -- which one?

As for your treasure comments, I consider them impolite.

UPDATE

So you have $$\frac{dx}{dt} = x^2 t$$ which separates into $x^{-2}dx = tdt$ and integrates out to $$-x^{-1} = \frac{t^2}{2} + C$$ which simplifies to $$ x(t) = - \frac{1}{t^2/2+C} = \frac{-2}{t^2+2C} $$ and the initial condition implies $$ x_0 = x(0) = \frac{-2}{2C} = -1/C $$ so $C = -1/x_0$ and the final form would be $$ x(t) = \frac{-2}{t^2-2/x_0} = \frac{2x_0}{2-x_0t^2} $$