Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

consider the IVP $$\dot{y}=\sqrt{|y|},y(x_0)=y_0$$ then it has two solutions namely y=0 and y= \begin{cases} \frac{x^2}{4}, & \text{if $x\ge0$} \\ -\frac{x^2}{4}, & \text{if $x\lt0$} \\ \end{cases} but on solving i found that it has solution $y=0$ and y=\begin{cases} \frac{x^2}{4}, & \text{if $y\ge0$} \\ -\frac{x^2}{4}, & \text{if $y\lt0$} \\ \end{cases} i found it as $|y|=y$ if $y \ge 0$ and $|y|=-y$ if $y<0$ and doing integration can someone tell me where I am wrong how the case become at $x$ not on $y$. $\bf EDIT$if we changes the initial condition as $y(0)=0$ then how to solve the problem?the actuall problem i wanted to ask is with these initials condition

share|cite|improve this question
Are you asking which is correct? That is, using $x \ge 0$ or $y \ge 0$, for example. Regards – Amzoti Oct 4 '13 at 5:49
yes i want to ask the same – abc Oct 4 '13 at 6:07
please someone response – abc Oct 4 '13 at 6:14
up vote 0 down vote accepted

Your solutions need to be functions of $x_0$ and $y(x_0)$.

Since $\dot{y}(x) \ge 0$, it is clear that $y$ is non-decreasing for $x \ge x_0$.

First I will deal with $y(x_0) >0$. Then it is clear that $\dot{y}(x) > 0$, hence $y(x) \ge y(x_0) $ for all $x \ge x_0$.

Then there is a unique solution $y(x) = (\frac{1}{2}(x-x_0)+ \sqrt{y(x_0)})^2$.

If $y(x_0) = 0$, two solutions are $y(x) = 0$ and $y(x) = (\frac{1}{2}(x-x_0))^2$.

If $y(x_0) <0$, then the solution is $y(x) = -(\frac{1}{2}(x_0-x)+ \sqrt{-y(x_0)})^2$, and this is valid for $x \in [x_0, x_0+2 \sqrt{-y(x_0)})$. For $x \ge x_0+2 \sqrt{-y(x_0)}$, the two solutions described above apply (with appropriate changes for $x_0$, of course). The solutions is unique on $[x_0, x_0+2 \sqrt{-y(x_0)})$.

share|cite|improve this answer
sir please tell me about the editted problem and thanks for this answer – abc Oct 4 '13 at 6:36
I have given all possible answers for all $x_0$ and all $y(x_0)$. If $x_0 = 0$ and $y(x_0) = 0$, then the two solutions are $y(x) = 0$ and $y(x) = \frac{1}{4} x^2$, for $x \ge 0$. – copper.hat Oct 4 '13 at 6:40
thanks sir is then y(x)=-$\frac{x^2}\4$. – abc Oct 4 '13 at 6:45
Was that a rhetorical question? – copper.hat Oct 4 '13 at 6:47

\begin{align} {\rm d}\sqrt{\left\vert y\right\vert\,} &= {1 \over 2\sqrt{\left\vert y\right\vert\,}}\,\,{\rm sgn}\left(y\right){\rm d}y \\[3mm] {{\rm d}y \over \sqrt{\left\vert y\right\vert\,}} &= 2{\rm sgn}\left(y\right){\rm d}\sqrt{\left\vert y\right\vert\,} = 2{\rm d}\left[{\rm sgn}\left(y\right)\sqrt{\left\vert y\right\vert\,}\right] - 2\sqrt{\left\vert y\right\vert\,}\,\left[2\delta\left(y\right)\right] = 2{\rm d}\left[{\rm sgn}\left(y\right)\sqrt{\left\vert y\right\vert\,}\right] \end{align}
$$ 2{\rm sgn}\left(y\right)\sqrt{\left\vert y\right\vert\,} - 2{\rm sgn}\left(y_{0}\right)\sqrt{\left\vert y_{0}\right\vert\,} = x - x_{0} $$ $$ \sqrt{\left\vert y\right\vert\,} = {\rm sgn}\left(y\right)\left[% {\rm sgn}\left(y_{0}\right)\,\sqrt{\left\vert y_{0}\right\vert\,} + {1 \over 2}\,\left(x - x_{0}\right) \right] $$

$$ \left\vert y\right\vert = \left[% {\sqrt{\left\vert y_{0}\right\vert\,} + {1 \over 2}\,\rm sgn}\left(y_{0}\right)\left(x - x_{0}\right) \right]^{2} $$

$$ \begin{array}{|c|}\hline\\ \color{#ff0000}{\large\quad% y \color{#000000}{\ =\ } {\rm sgn}\left(y\right)\left[% {\sqrt{\left\vert y_{0}\right\vert\,} + {1 \over 2}\,\rm sgn}\left(y_{0}\right)\left(x - x_{0}\right) \right]^{2} \quad} \\ \\ \hline \end{array} $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.