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Im trying to find the critical points of this first order differential equation.

$$\large{{dy\over dx}=y\ln(y+2)}$$

I already got one critical point of $y=0$.

If I set $\ln(y+2) = 0$ Im unsure how to isolate y. Would I multiply each side by $e$?

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    $\ln 1=0$ and $y+2=1$ i.e $y=-1$,if this is what you're asking. – 2017-01-29
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    Do you want the critical points of $\frac{\mathrm{d}y}{\mathrm{d}x}$, $y(x)$ or $\ln(y+2)$? – 2017-01-29
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    @MichaelMcGovern, "critical point of a differential equation" typically means points where the derivative is zero. I think I've only seen this in the context of systems of first-order ODEs. But I guess one equation is technically a system. Eh... – 2017-01-29
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    The derivative of the unknown function or the derivative of the equation itself? – 2017-01-29
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    @MichaelMcGovern I wanted to know how to find the critical points of ln(y+2) – 2017-01-29
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    yeah! thats exactly what I was asking. thanks @kingW3 – 2017-01-29

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To find the critical points of the equation, set $dy/dx = 0$. So then $y \ln(y+2) = 0$.

You've already found $y = 0$.

The other one is from $\ln(y+2) = 0$. This is a logarithmic equation and is equivalent to the exponential equation $y+2 = e^0$. Since $e^0 = 1$, then we have $y = -1$.

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    Thanks , this is a great help! – 2017-01-29