MHB Differential Equations by separation of variables

LAK
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Can someone please help me to calculate the following using separation of variables:

dy/dx = x*(1 - y^2)^(1/2)

to that the solution is in the form:

y =
 
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I have moved both your threads here to the Differential Equations subforum as this is a better fit for them.

What do you get when you separate the variables, before integrating?
 
LAK said:
Can someone please help me to calculate the following using separation of variables:

dy/dx = x*(1 - y^2)^(1/2)

to that the solution is in the form:

y =

For starters:

$\displaystyle \begin{align*} \frac{1}{\sqrt{1 - y^2}} \, \frac{dy}{dx} = x \end{align*}$

and now you can integrate both side w.r.t. x :)
 
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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