MT, P1

[Victor Ivrii]

Solve the following initial value problem:
\begin{equation*}
y'(t) = 1- y^2(t),\qquad y(0)=0.
\end{equation*}

[Yangming Cai]

answer to p1

[Xiaozeng Yu]

1

[Xuewen Yang]

For Xiaozeng Yu,

Is there a problem with the sign change? from "plus" (2nd row) to "minus" (3rd row) ?

[Xiaozeng Yu]

For Xiaozeng Yu,

Is there a problem with the sign change? from "plus" (2nd row) to "minus" (3rd row) ?

no mistake. $d(\ln(1-y))/dt = -1/(1-y)$.

[Xuewen Yang]

For Xiaozeng Yu,

Is there a problem with the sign change? from "plus" (2nd row) to "minus" (3rd row) ?

no mistake. d(ln(1-y))/dt = -1/1-y

You are right! My bad 

[Victor Ivrii]

And where this solution is defined? $\DeclareMathOperator{\arctanh}{arctanh}$

BTW, when collecting MT I have seen this problem and in many papers integration was atrocious.

As long as $|y|<1$ we have $\int \frac{dy}{1-y^2}=\arctanh (y)$ (inverse hyperbolic function $\tanh$) and then
$y=\tanh (t)=\frac{{e^t}-e^{-t}}{e^{t}+e^{-t}}$ (similar to $\int \frac{dy}{1+y^2}=\arctan(y)$).

[Xiaozeng Yu]

And where this solution is defined? $\DeclareMathOperator{\arctanh}{arctanh}$

BTW, when collecting MT I have seen this problem and in many papers integration was atrocious.

As long as $|y|<1$ we have $\int \frac{dy}{1-y^2}=\arctanh (y)$ (inverse hyperbolic function $\tanh$) and then
$y=\tanh (t)=\frac{{e^t}-e^{-t}}{e^{t}+e^{-t}}$ (similar to $\int \frac{dy}{1+y^2}=\arctan(y)$).

hmm...yea...-1<y<1  T.T i feel i'm screwed...

[Victor Ivrii]

And where this solution is defined?
hmm...yea...-1<y<1 

This is a range of $y$ (what values it takes), I asked about domain--for which $t$ it is defined.

So, my question about domain is pending

[Xiaozeng Yu]

t is defined everywhere...

[Victor Ivrii]

$t$ is defined everywhere...

You think right but write incorrectly: Solution $y(t)$ is defined everywhere.

[Xiaozeng Yu]

thank u prof.