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I am trying to evaluate the equation:

$$y'=y\left(y^2-\frac12\right)$$

I multiplied the y over and tried to solve it in seperable form (M and N). The partial deritives did not work out to be equal to eachother so I am now stuck finding an integrating factor. Is this the right approach?

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  • $\begingroup$ Finding and applying the integrating factor is in this case equivalent to separation of variables. $\endgroup$ – Dr. Lutz Lehmann Oct 27 '18 at 9:51
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Separation will work, just that partial fractions have to be dealt with: $$\int\frac1{y(y^2-1/2)}\,dy=\int1\,dx$$ $$\int\left(\frac{4y}{2y^2-1}-\frac2y\right)\,dy=\int1\,dx$$ $$\ln(2y^2-1)-2\ln y=\ln\frac{2y^2-1}{y^2}=x+K$$ $$\frac{2y^2-1}{y^2}=2-\frac1{y^2}=Ae^x$$ $$y=\pm\frac1{\sqrt{2-Ae^x}}$$

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This is also a Bernoulli equation, set $u=y^{-2}$ to get $$ u'=-2y^{-3}y'=-2+u\implies u=2+Ce^x(u_0-2)\implies y(x)=\frac{y_0}{\sqrt{2y_0^2+(1-2y_0^2)e^x}} $$

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