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Hey I have a problem including absolute values without an initial condition. I looked all over my textbook and the internet and cannot solve it.

$$\frac{dy}{dt} = |y|^{1/2}$$

I did

positive:
$$2y^{1/2} = t$$

negative:
$$2(-y)^{1/2} = t $$

and I thought it was just $t|t|$, but I applied initial conditions from the previous section of the question and they don't work. Please give me a hint or help. Thanks

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  • $\begingroup$ $y=t^{2} sgn (t)/4$. $\endgroup$ – Kavi Rama Murthy 2 days ago
  • $\begingroup$ Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. $\endgroup$ – José Carlos Santos 2 days ago
  • $\begingroup$ @KaviRamaMurthy Do you need "plus a constant" so here $y=\pm\frac14(t+c)^2$ ? $\endgroup$ – Henry 2 days ago
  • $\begingroup$ @Henry I don't think that is a solution if $c \neq 0$. $\endgroup$ – Kavi Rama Murthy 2 days ago
  • $\begingroup$ @KaviRamaMurthy Let's try $y=\frac14(t+5)^2$ then $\frac{dy}{dt}=\frac12(t+5) = \sqrt{y}$. So that is OK. But I now have doubts what happens when $y$ is negative. So perhaps $y=\frac14(t-c)^2 \mathrm{sgn}(t-c)$ or $y=\frac14(t-c)|t-c|$ with the sign changing depending on whether $t < c$ or $t>c$ $\endgroup$ – Henry 2 days ago
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I tried to put t^2 = y , and y' = 2 t t' also I reduces by making two conditions one if y >0 and other if y < 0 ; so first case if y >0 : 2 t t' = t you should take a factor and solve for t and transform it into y second case for y <0 : exact same thing but the equation would be 2 t t' = - t Hope I helped you with my solution!

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Adib Akkari is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
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