$\mathbf {(1)} $ The solution of the differential equation, $$y'=1+y^2,\;\;\;y(0)=1$$ exists on the interval-
(a) $ \displaystyle|x|< \frac {\pi}{2}$
(b) $-\displaystyle\frac {\pi}{2}<x< \pi$
(c) $-\displaystyle\frac{3\pi}{4} <x< \frac{\pi}{4}$
(d) $ |x|< \pi$
My Attempt:
The given DE is in variable separable form, on solving we have, $$\tan^{-1}y=x+c\implies y=\tan(x+c)$$ using the initial condition we obtain, $c=\displaystyle \frac{\pi}{4}$
Thus, $ y=\tan\left(x+\displaystyle \frac{\pi}{4}\right)$
For the solution to exist, $$-\frac {\pi}{2}<x+ \frac{\pi}{4}< \frac {\pi}{2}$$ (because only then will $y$ be defined)
$$\implies -\frac{3\pi}{4} <x<\frac{\pi}{4}$$
So, (c) is true.
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(2) If $y(x)$ is the solution of the DE, $$y'=2(1+y)\sqrt y \;\; ,y\left( \frac{\pi}{2}\right)=1$$ then the largest interval of(to the right of origin) on which the solution exists is-
(a) $\;\;\;\left[0, \displaystyle \frac {3\pi}{4}\right)$
(b) $\;\;\;[0, \pi)$
(c)$\;\;\;[0, 2\pi)$
(d)$\;\;\;\left[0, \displaystyle \frac {2\pi}{3}\right)$
My Attempt:
The DE is again in variable separable from, on using the substitution $\sqrt y =t$ and solving we have, $$y=(\tan(x+c))^2$$ (I hope I have not made any mistakes while solving)
Since we are dealing with only values from $\Bbb R$,I feel $\tan(x+c)$ must take non-negative values.
Now from the initial conditions we get, $c=\displaystyle - \frac {\pi}{4}$
Thus, $y=\left(\tan \left (x\displaystyle - \frac {\pi}{4}\right) \right)^2$
Beyond this point I'm finding it difficult to proceed.
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I want to know if I've got the first problem right and whether or not all my arguments are correct in the first problem.
For the second problem, Hints Please!
Yes, that is all correct. Now proceed as in the first case starting from $$ -\frac\pi2< x-\frac\pi4<\frac\pi2 $$ and reduce to the positive part of the interval.