MHB -Identify the equilibrium values y'=5\sqrt{5},y>0

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$\textrm{ Given $y'=5\sqrt{5},y>0$ answer the following questions.}\\$
$\textrm{a. Identify the equilibrium values.}\\ $
$\textrm{Which are stable and which are unstable?}\\$
$\textrm{b. Construct a phase line. Identify the signs of $y′$ and $y′′$.}\\$
$\textrm{c. Sketch several solution curves.}\\$
ok just posting this now
have deal with it at school
basically clueless
 
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karush said:
$\textrm{ Given $y'=5\sqrt{5},y>0$ answer the following questions.}\\$
$\textrm{a. Identify the equilibrium values.}\\ $
$\textrm{Which are stable and which are unstable?}\\$
$\textrm{b. Construct a phase line. Identify the signs of $y′$ and $y′′$.}\\$
$\textrm{c. Sketch several solution curves.}\\$
ok just posting this now
have deal with it at school
basically clueless

Are you sure you copied the question correctly? In the given ODE, there are no equilibrium solutions, since $y'$ is a constant...(Wondering)
 
ok sorry, I will just skip this one for now..

I had to go on to another topic anyway.;););)
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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