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While trying to solve the differential equation: $y'=\sqrt{|y|}$. I got confused how to deal with the absolute value. I want to draw a sketch for the direction field for that equation, and see for what initial values does this equation fulfill the conditions of the existence and uniqueness Theorem.

I tried to do the integral according to the sign, depends if $(y>0)$ or $(y<0)$. but I'm still not sure if my result is right.

result that I got:

  • while $y>0$ : $y=({\frac{x}{2}+c})^{2}$
  • while $y<0$ : $y=({-\frac{x}{2}+c})^{2}$

if that's the case, Its hard for me to imagine the Direction field by these two equations, because they are always above axis: $x$. How does it look? any hints?

F1sargyan
  • 727

2 Answers2

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For the $y<0$ case your DE is evaluated incorrectly:

$$\frac{dy}{dx}=\sqrt{-y}$$

$$(-y)^{-\frac{1}{2}}\frac{dy}{dx}=1$$

$$-2\sqrt{-y}=x+c$$

$$-4y=(x+c)^2$$

$$y=-\frac{(x+c)^2}{4}$$

This should hopefully help you to visualize the direction field now. Something like this:

enter image description here

Ian Miller
  • 12,140
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for $y>0$ $$(y)^{-\frac{1}{2}}y'=1$$ $$2y^{\frac{1}{2}}=x+c$$ $$y^{\frac{1}{2}}=\frac{1}{2}(x+c)$$ $$y=\frac{1}{4}(x+c)^2=(\frac{x}{2}+\frac{c}{2})^2=(\frac{x}{2}+k)^2$$ this answer same your answer

when the $y<0$ $$(-y)^{-\frac{1}{2}}y'=1$$ the solution is $$y=-(\frac{x}{2}+k)^2$$

E.H.E
  • 23,590