# general solution: absolute value [closed]

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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• $y=t^{2} sgn (t)/4$. – Kavi Rama Murthy 2 days ago
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• @KaviRamaMurthy Do you need "plus a constant" so here $y=\pm\frac14(t+c)^2$ ? – Henry 2 days ago
• @Henry I don't think that is a solution if $c \neq 0$. – Kavi Rama Murthy 2 days ago
• @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 $t>c$ – Henry 2 days ago