I.V.P. $y'=1+y^2$ $y(0)=0$

Question

I.V.P. $y'=1+y^2$, $y(0)=0$.

a) find the maximum interval there exist an unique solution for the given I.V.P.

b) find the solution of the given I.V.P.

My try:

a)

I should to find M,L and h.

Such that :

1) $M=\max |f(x,y)|$.

2) $L$ is Lipschitz constant: $$|f(x,y_2)-f(x,y_1)|\leq L|y_2-y_1|.$$

3) $h=\min(a,{b\over M})$.

So

1) $M=|f(x,y)|=|1+y^2|\leq 1+|y|^2$

$M=1+b^2$

2) $L=|{\partial f\over \partial y}|=|2y|$

$\leq 2|y|=2b$

$L=2b$

3) $h=min(a,{b\over M})$

$=min(a,{b\over {1+b^2}})$

$b=1$ [by use the First Derivative Test]

So

$M=2$ , $L=2$ and $h={1\over 2}$.

True ?

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For b)

$y'=1+y^2$

${dy\over dx}=1+y^2$

$dy=(1+y^2) dx$

${1\over {(1+y^2)}} dy= dx$

$\tan^{-1} (y)=x+c$

From condition

$x=0, y=0$

$\tan^{-1} (0)=0+c$

$c=0$

So

$\tan^{-1} (y)=x$

$y= \tan (x)$

True ?

Thanks.

Answer 1

True.

So how far can $y = \tan(x)$ go until it blows up?