1. Another differential equation

Explicitly find the maximal interval I about 0 on which we can solve the following differential equation:

I'm not sure how to approach. I know that if it was

then I can divide both sides and integrate and get
, but there's a square term.

Thank you.

2. Originally Posted by Paperwings
Explicitly find the maximal interval I about 0 on which we can solve the following differential equation:

I'm not sure how to approach. I know that if it was

then I can divide both sides and integrate and get
, but there's a square term.

Thank you.

3. Ah, ok. Thanks mathdude.

1.
2. Integrating yields

3. Solving for f(x) gives

However, substituting the initial condition gives

I don't understand this part. Wouldn't this function be undefined at zero?

4. Originally Posted by Paperwings
Ah, ok. Thanks mathdude.

1.
2. Integrating yields

3. Solving for f(x) gives

However, substituting the initial condition gives

I don't understand this part. Wouldn't this function be undefined at zero?

Why not do the last transformation the easy way:

$\f(x) = \frac{\-1}{x+C}\\-1 = \frac{\-1}{0+C}\\C = \1$

5. Assume there is a non-vanishing solution on an interval I about 0.
Thus, y' = (y)^2 this means y'/(y^2) = 1, integrating means (1/y) = x+k.
Since y(0)=1 it means 1/y(0) = k and so k =1.
Thus, the solution must be 1/y=x+1.
Thus, if -1 is not in I then we can take reciprocals to get y = 1/(x+1).
Therefore, I has the condition that -1 is not in it.
Hence, the largest interval is (-1, $\infty$).

6. Thus, y' = (y)^2 this means y'/(y^2) = 1, integrating means (1/y) = x+k.
Isn't it -1/y ?

7. It's a typo, that's all.