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$\textbf{Question:}$ In the following question choose the appropriate domain of definition of the given function and corresponding initial condition , find the maximal interval $(a,b)$ of existence of the solution and find its limit as $x$ approaches to $a$ and $b$ $$\frac{dy}{dx}=\frac{1}{1+y^2}$$

I consider the ODE with initial condition $$\frac{dy}{dx}=\frac{1}{1+y^2}, y(0)=0.$$ If we solve it then we have $$3y+y^3=3x,$$ which is implicitly defined function. Now i am stuck how to find maximal interval of existence of solution $y$ and its limit $x$ approaches to $a$ and $b.$ Please help. Thanks.

neelkanth
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    The maximal interval is the largest $x$-interval for which you can solve the equation $y^3 + 3y = 3x$ for $y$. Plot the function $g(y) = y^3 + 3y - a$ for different $a$ values. For which $a$ does this equation have a solution? Now replace $a$ with $-3x$. For which $x$ does this have a solution? And what happens to this solution if $x$ is very large (positive or negative)? – Hans Engler Mar 25 '19 at 16:20
  • @HansEngler Thanks sir for reply... i will try with this hint...Thanks... – neelkanth Mar 25 '19 at 16:23
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    The implicit function theorem does the trick. Note that $(3y+y^3-3x)'_y \ne 0, \forall y$. – PierreCarre Mar 25 '19 at 16:27

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Since $0 < 1/(1+y^2) \le 1$, any solution must satisfy $y(0) \le y(x) \le y(0) + x$ for $x > 0$ and $y(0) + x \le y(x) \le y(0)$ for $x < 0$. Thus $a=-\infty$ and $b = \infty$.

Robert Israel
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