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$$y'=-2-y+y^2$$ and $y_{0}=2$ is a solution

So it is a riccati ODE:

$y=2+z\Rightarrow z=y-2$

$z'=y'$

substation into the original ODE gives:

$z'=-2-(2+z)+(z+2)^2=-4-z+z^2+4z+4=z^2+3z$

So we have:

$z'=z^2+3z$ which is bernoulli $z'=3z+z^{1+1}$ when $k=1$

$w=\frac{1}{z}$

$w'=-z'\cdot z^{-2}=-(3z+z^2)z^{-2}=\frac{-3}{z}-1=-3w-1$

$w'+3w=-1$

$w_{h}=e^{-3\int dx}=e^{-3x+c}=e^{-3x}k$

$w_{p}=e^{-3x}k(x)$

$w'_{p}=-3e^{-3x}k(x)+k'(x)e^{-3x}$

Substation into $w'+3w=-1$:

$-3e^{-3x}k(x)+k'(x)e^{-3x}+3e^{-3x}k(x)=-1\Rightarrow k'(x)e^{-3x}=-1\Rightarrow k'(x)=-e^{3x}$

integrating gives:

$k(x)=\frac{-e^{3x}}{3}+c$

So

$w_{p}=e^{-3x}(\frac{-e^{3x}}{3}+c)=-\frac{1}{3}+ce^{3x}$

$w=w_{h}+w_{p}=e^{-3x}k-\frac{1}{3}+ce^{3x}$

is it correct?

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    An easy way to check if your solution is correct is to plug it into the original equation and see if it works. Always, ALWAYS, do this with complicated ODEs (or hell even IEs). – 2017-01-23
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    no need for ricatti here, this DE is simply seperable – 2017-01-23
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    Hint: $$ x+C=\int\frac{dy}{y^2-y-2}=\int\frac{dy}{(y-1/2)^2-5/4}=....... $$ – 2017-01-23
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    @CameronWilliams I get $-3e^{-3x}k+3ce^{3x}=-2+e^{-3x}k-\frac{1}{3}+xe^{3x}+(+e^{-3x}k-\frac{1}{3}+xe^{3x})^2$ – 2017-01-23
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    @tired so I cant turn it to $\frac{1}{x^2+1}$ to get $arctan$ shouldn't I use partial fractions? – 2017-01-23
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    $\text{arctanh}$ is simply to invert so go for it! =) – 2017-01-23
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    What is the meaning of "and $y_0=2$ is a solution"? Is it that you are looking for the solution $y$ such that $y(0)=2$? – 2017-01-23
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    @Did it is given that 2 is a solution – 2017-01-23
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    And this fact is useful how? – 2017-01-23
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    @Did to use riccati we must have a least one solution – 2017-01-23
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    OK, so they really want you to solve it with this approach. Pure sadism, I guess... – 2017-01-23
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    @Did it was given as an example of how to use riccati – 2017-01-23
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    I believe you, but there are better examples... – 2017-01-23

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Clearly $y=-1$ and $y=2$ are trivial solutions. Note that $$ y^2-y-2=(y-2)(y+1)$$ and hence $$ y'=(y-2)(y+1)$$ or $$ \frac{dy}{(y-2)(y+1)}=dx.$$ Integrating both sides gives $$ \int\frac{dy}{(y-2)(y+1)}=\int dx+C.$$ So $$ \frac13\ln\left|\frac{y-2}{y+1}\right|=x+C$$ or $$ \frac{y-2}{y+1}=Ce^{3x}$$ So the solution is $$ y=\frac{2+Ce^{3x}}{1-Ce^{3x}}.$$

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    These, yes, and also the solution $y=-1$. – 2017-01-23
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    @Did, I almost forget $y=-1$ and $y=2$ are trivial solutions. – 2017-01-23