Toronto Math Forum
MAT2442013F => MAT244 MathTests => MidTerm => Topic started by: Victor Ivrii on October 09, 2013, 07:20:21 PM

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

answer to p1

1

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

For Xiaozeng Yu,
Is there a problem with the sign change? from "plus" (2nd row) to "minus" (3rd row) ?
no mistake. $d(\ln(1y))/dt = 1/(1y)$.

For Xiaozeng Yu,
Is there a problem with the sign change? from "plus" (2nd row) to "minus" (3rd row) ?
no mistake. d(ln(1y))/dt = 1/1y
You are right! My bad :P

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 $\int \frac{dy}{1y^2}=\arctanh (y)$ (inverse hyperbolic function $\tanh$) and then
$y=\tanh (t)=\frac{{e^t}e^{t}}{e^{t}+e^{t}}$ (similar to $\int \frac{dy}{1+y^2}=\arctan(y)$).

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 $\int \frac{dy}{1y^2}=\arctanh (y)$ (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...

And where this solution is defined?
hmm...yea...1<y<1
This is a range of $y$ (what values it takes), I asked about domainfor which $t$ it is defined.
So, my question about domain is pending

t is defined everywhere...

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

thank u prof.