ordinary differential equations
If $y(x)$ is the solution of differential equation$\frac {dy}{dx}=2(1+y)(\sqrt y)$ satisfying $y(0)=0,y(\frac {\pi}{2})=1 $ then the largest interval (to the right of origin ) on which the solution exists is
(a)$[0,\frac {3\pi}{4}]$
(b)$[0,\pi)$
(c)$[0,2\pi)$
(d) )$[0,\frac {2\pi}{3}]$
Exact image of question
This question was asked in GATE Mathematics exam
My try
I solved the givrn differential equation and the solution is
$${tan^{-1}}(\sqrt y)=x+c$$
Can anyone give me hint in these question?
I find it useful to plot the function after solving for the constant from the ICs.
We have:
$$y(x) = -\dfrac{8}{e^{4 x}-2}$$
A plot of $y(x)$ shows:
Of course, analytically, we care about where that denominator is $0$, so we have:
$$e^{4 x}-2 = 0 \rightarrow x = \dfrac{\ln 2}{4} \approx 0.173286795139986$$
From the plot, you can also see what happens as $x \rightarrow \pm \infty$, but it is not bad to derive that analytically either.
One last helpful thing you may have not gotten to yet, is a phase plot, which looks like:
This is a linear differential equation. Therefore it is defined on the largest intervals for which the functions used in the differential equation are all defined.
Hence solutions are defined on intervals for which $\cos x$ doesn't vanish. If you want a solution defined at $0$, the interval is $(-\frac{\pi}{2},\frac{\pi}{2})$
Best Answer
The curve $$y=\tan^2 x\qquad(x\geq0)\tag{1}$$ and all its horizontal translates are solution curves of this ODE. To this family of curves its envelope $y(x)\equiv0$ has to be added, since it is a solution as well. But this is not all: Since the ODE does not satisfy a crucial assumption of the existence and uniqueness theorem in the points $(x,0)$ we have to accept that for such initial points there are several solutions. Intuitively: We may choose an arbitrary $p\in{\mathbb R}$, then follow the envelope as long as $x\leq p$, and for $x\geq p$ go along a translate of the curve $(1)$.
Now you want the solution passing through the point $\bigl({\pi\over2},1\bigr)$. It so happens that the following "splined" function does the trick: $$y(x)=\left\{\eqalign{0\qquad\qquad&\bigl(x\leq{\pi\over4}\bigr)\cr \tan^2\bigl(x-{\pi\over4}\bigr)\qquad&\bigl({\pi\over4}\leq x<{3\pi\over4}\bigr)\ .\cr}\right.$$ This solution lives in the $x$-interval $\ \bigl]-\infty,{3\pi\over4}\bigr[\ $. It follows that answer (a) comes nearest.