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The initial value problem $y' = 2\times\sqrt{y}$ , $y(0)=a$ has

  1. a unique solution for $a<0$

  2. no solution for $a>0$

  3. infinitely many solutions if $a=0$

  4. a unique solution if $a\ge 0$.

    My answer is 4.

    As if we solve the ode by separating $y'$ ,then we have a general solution as $\sqrt{y}=x+c$. Now using $y(0)=a$ , we have $c=\sqrt a$, which is valid if $a \ge 0$, and in that case we get a unique solution for that $a$.

    Now the answer is given to be 3, if $a=0$ ,then we get $y=x²$ ,which is unique , then how is it 3 ?

    Is my method correct and my answer also ? Or am I wrong ?

    Hoping for a help.Thank you.

hiren_garai
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2 Answers2

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You get $c^2=a$ thus $c=|\sqrt {a}|=\pm \sqrt{a} $ thus the equation is $y=(x\pm \sqrt {a})^2$. Now if $a\geq 0$ we have two solutions while for $a=0$ we have infinite solutions as the locus is parabola.

Archis Welankar
  • 15,910
  • I got it. Thanks for the solution. – hiren_garai Jun 16 '17 at 12:11
  • $y(x)=(x-\sqrt a)^2$ is not a solution to the differential equation for $a>0$, because $y'(x)=-2\sqrt{y(x)}$ for sufficiently small values of $x$. As a general observation, when facing an instance where gibberish methods of ODE fail, using another gibberish method to try to go around the mistake is pointless. –  Jun 16 '17 at 16:22
  • can someone please explain me the line "for a=0 we have infinite solutions as the locus is parabola"... not getting... – math student Jul 30 '23 at 10:28
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You are wrong. For instace, for any $\beta\ge0$ the function $$f_\beta(x)=\begin{cases}(x-\beta)^2&\text{if }x\ge \beta\\ 0&\text{if }x<\beta\end{cases}$$

satisfies $f'_\beta=2\sqrt{f_\beta}$ and $f_\beta(0)=0$.

The problem with your argument is that, keen as it is on working only algebraically, it "conveniently" skips all the crucial details of a proper proof, basically:

  • the fact that the differential equation $y'=2\sqrt{\lvert y\rvert}$ fails to satisfy the hypothesis of Cauchy-Lipschitz theorem in any neighbourhood of the line $\Bbb R\times \{0\}$

  • the fact that, while for $y(0)>0$ you can rewrite the problem as an integral equation $$\int_0^x \frac{y'(t)}{2\sqrt{y(t)}}\,dt=\int_0^x1\,ds,$$ this cannot be done when $y$ is constantly $\equiv 0$ in a neighbourhood of $0$.