I'm given a problem with the ODE
y'[x]=y[x]*Sqrt[4-2y[x]]
I've tried
DSolve[{w'[x] == w[x]*Sqrt[4 - 2*w[x]]}, w[x], x]
which gives me
{{w[x] -> -2 (-1 + Tanh[1/2 (-2 x - Sqrt[2] C[1])]^2)}} but that doesn't seem correct for some reason.
Can someone please point me in the direction of how to do this problem correctly? I don't really understand Mathematica at all.
here is a soultion by Maple
$$ x+\frac{1}{2}\,\ln \left( \sqrt {4-2\,y \left( x \right) }+2 \right)- \frac{1}{2}\, \ln \left( -2+\sqrt {4-2\,y \left( x \right) } \right) +{\it \_C1}=0\,.$$
solving the above equation for $y$, gives
$$ y(x) = -{\frac {{{8\rm e}^{2\,x+2\,{\it C1}}}}{ \left( {{\rm e}^{2\,x+2\, {\it C1}}}-1 \right) ^{2}}} \,.$$
The differential equation can be solved by the method of separation of variables
$$ \frac{dy}{dx}=y \sqrt{4-2y}\implies \int \frac{dy}{y \sqrt{4-2y}}=\int dx + C \,.$$
You need to work out the above integrals and then solve the resulting equation for y to get y as a function of $x$.