Solve differential equation \frac{dy}{dx}= (x+y+1)^2-(x+y-1)^2

The system described by $x^{\prime }=2x^2-8$ is linearized about the equilibrium point -2. What is the resulting linearized equation?
Answer is $x^{\prime }=-8x-16$.
How? I have no idea how it went from the first equation to the 2nd. Thanks.

a) Solve the differential equation:
${\displaystyle (x+1)\frac{dy}{dx}-3y={(x+1)}^4}$
given that ${\displaystyle y=16}$ and ${\displaystyle x=1}$, expressing the answer in the form of ${\displaystyle y=f\left(x\right)}$.
b) Hence find the area enclosed by the graphs ${\displaystyle y=f\left(x\right),y={(1-x)}^4}$ and the ${\displaystyle x-a\xi s}$.
I have found the answer to part a) using the first order linear differential equation method and the answer to part a) is: ${\displaystyle y={(1+x)}^4}$. However how would you calculate the area between the two graphs (${\displaystyle y={(1-x)}^4}$ and ${\displaystyle y={(1+x)}^4}$) and the x-axis.

If ${\displaystyle p\left(x\right)}$ is a solution to a first-order differential equation in the form of ${\displaystyle \frac{df}{dx}=g\left(f\right)}$, then ${\displaystyle p(x+c)}$, with c-constant, is a solution as well.
I know the idea of the proof, but I am having trouble expressing it in a rigorous proof. Namely, my idea is that if we shift ${\displaystyle p\left(x\right)}$ to the left by ${\displaystyle c}$, then both the LHS and RHS will shift by ${\displaystyle cP}$ (invoking the chain rule), so they are equal. Should I define ${\displaystyle u=x+c}$ and do a change of variable?

I have first-order differential equation
${\displaystyle y=xy\prime +\frac{1}{2}{\left(y\prime \right)}^2}$
Maybe, with this someone will find way to solve it
${\displaystyle \frac{1}{2}{y}^{\prime }(2x+y^{\prime })=y}$
I thought I can use ${\displaystyle x^2+y=t}$ for subtitution and when I derivate, I have ${\displaystyle t^{\prime }=2x+y^{\prime }}$ which is acctualy the
${\displaystyle (t^{\prime }-2x)t^{\prime }=2t=2x^2}$
same as previous. Who knows?

Solve the differential equation $(x-3y)dx-xdy=0$

Solve differential equation $dy/dx-12x^{3}y=x^3$