Calculus

Solve the following initial value problem and determine the interval where the solution exists.

dy/dt = 2ty^2
y(0) = y_0

Textbook answer is:

When y_0 <> 0, then y = 1/(1/y_0 - t^2).
When y_0 = 0, then y = 0

When y_0 > 0, the interval where solution exists is |t| < 1/sqrt(y_0)
When y_0 <= 0, the interval where solution exists is -infinity < t < +infinity


My work:

The equation is separable:

dy/y^2 = 2t dt
-1/y = t^2 + C

When y_0 = 0, I use intuition to come to the same solution that the textbook shows.

When y_0 <> 0
C = -1/y_0
y = 1/(1/y_0 - t^2).
The interval where the solution exists is where the denominator expression does not equal zero. Algebraically, that simplifies to t <> 1/sqrt(y_0)

The textbook answer is different. What did I do wrong or omit?

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asked by Mark

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