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Find the solution of the initial value problem $$ \frac{dy}{dt}=1-y^2,\;y(0)=y_0 $$ (i) in the case $y_0=1$; (ii) $y_0=-1$; (iii) in all other cases.

Are there any $y_0$ such that the solution blows up in finite time

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I understand by integrating the $1-y^2$ on the left and integrating the 1 on the right to give $$\arctan(y)=t+c$$ then not really sure where to go from here.

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    $\begingroup$ If you don't give us the ODE we can't help you. $\endgroup$ – Frieder Jäckel May 10 '17 at 9:30
  • $\begingroup$ Press on the title it should give a link to the question. $\endgroup$ – Ben Jones May 10 '17 at 9:33
  • $\begingroup$ Try to use $\frac{1}{1-y^2}=\frac{1}{2}\left(\frac{1}{1-y}+\frac{1}{1+y}\right).$ $\endgroup$ – Frieder Jäckel May 10 '17 at 9:36
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Well, we have:

$$\text{y}'\left(t\right)=1-\text{y}\left(t\right)^2\space\Longleftrightarrow\space\int\frac{\text{y}'\left(t\right)}{1-\text{y}\left(t\right)^2}\space\text{d}t=\int1\space\text{d}t\tag1$$

Now, substitute $\text{u}=\text{y}\left(t\right)$:

$$\int\frac{\text{y}'\left(t\right)}{1-\text{y}\left(t\right)^2}\space\text{d}t=\int\frac{1}{1-\text{u}^2}\space\text{d}\text{u}=\frac{\ln\left|1+\text{u}\right|-\ln\left|1-\text{u}\right|}{2}+\text{C}_1\tag2$$

So, we get:

$$\frac{\ln\left|1+\text{y}\left(t\right)\right|-\ln\left|1-\text{y}\left(t\right)\right|}{2}+\text{C}_1=t+\text{C}_2\space\Longleftrightarrow\space$$ $$\ln\left|1+\text{y}\left(t\right)\right|-\ln\left|1-\text{y}\left(t\right)\right|=2t+\text{C}\tag3$$

Take the $\exp$ of both sides:

$$\left|\frac{1+\text{y}\left(t\right)}{1-\text{y}\left(t\right)}\right|=\text{K}\cdot e^{2t}\tag4$$

So, when $t=0$:

$$\left|\frac{1+\text{y}\left(0\right)}{1-\text{y}\left(0\right)}\right|=\text{K}\cdot e^{2\cdot0}=\text{K}=\left|\frac{1+\text{y}_0}{1-\text{y}_0}\right|\tag5$$

So, the solution is:

$$\left|\frac{1+\text{y}\left(t\right)}{1-\text{y}\left(t\right)}\right|=\left|\frac{1+\text{y}_0}{1-\text{y}_0}\right|\cdot e^{2t}\tag6$$

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    $\begingroup$ After you insert the initial point, the sign is fixed, so you can remove the absolute values in the last equation as it has to be valid at $t=0$. $\endgroup$ – LutzL May 10 '17 at 10:18
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I understand by integrating the $1-y^2$ on the left and integrating the $1$ on the right to give $\arctan(y)=t+c$ then not really sure where to go from here.

Actually $$ \arctan' x = \frac{1}{1+x^2} $$ so your solution is wrong.

For the determination of the integration constant you just apply the initial condition $y(0) = y_0$ to the general solution, thus $y_0 = 1$ and so on for the sub tasks.

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