Solving Differential Equation with given conditions I have the following question:

Here is my attempt is this correct ?

 A: Your general solution should be $y=(C_1+C_2x)e^{-2x}$ instead of $y=C_1+C_2xe^{-2x}$!
Besides, don’t apply initial conditions until you finished everything!
You have 
$$y=(C_1+C_2x)e^{-2x}+x^2-2x-1$$
and $y(0)=0$ and $y’(0)=0$.
You will get $C_2=-1$ and $C_1=-2$.
Tidying up,
$$y=-(x+2)e^{-2x}+x^2-2x-1$$
A: Did you really need to use the quadratic formula to see that $\alpha^2+ 4\alpha+ 4= (\alpha+ 2)^2$? :).  Immediately after that you have $y= c_1+ c_2xe^{-2x}$.  I presume that was a typo and that you meant $y= (c_1+ c_2x)e^{-2x}= c_1e^{-2x}+ c_2xe^{-2x}$.
As Szeto said, the initial conditions apply to the entire equation, not just to the associated homogeneous equation- in fact, y(x) identically 0 is a solution too any linear homogeneous equation and satisfies y(0)= 0 and y'(0).   
Yes, to find a specific solution to the entire equation, look for a solution of the form $Ax^2+ Bx+ C$ and, just as you did, you get $x^2- 2x- 1$.  Now adding that to the solution to the associated homogeneous equation: $y(x)= (c_1+ c_2x)e^{-2x}+ x^2- 2x- 1$  Then $y'(x)= c_2e^{x}- 2(c_1+ c_2x)e^{-2x}+ 2x- 2= (c_2-c_1)e^{-2x}+ 2xe^{-2x}+ 2x- 2$. $y(0)= c_1- 1= 0$ and $y'(0)= c_2- c_1- 2= 0$.  Solve those equations for $c_1$ and $c_2$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\pars{\totald{}{x} + 2}^{2}\mrm{y}\pars{x} = 4x^{2} - 10 \implies
\left\{\begin{array}{rcl}
\ds{\pars{\totald{}{x} + 2}\mrm{y}\pars{x}} & \ds{=} & \ds{\mrm{z}\pars{x}\,,\qquad\mrm{y}\pars{0} = 0}
\\[1mm]
\ds{\pars{\totald{}{x} + 2}z\pars{x}} & \ds{=} & \ds{4x^{2} - 10\,,\qquad\mrm{z}\pars{0} = 0} 
\end{array}\right.
\\[5mm]
&\totald{\bracks{\expo{2x}\mrm{z}\pars{x}}}{x} = \expo{2x}\pars{4x^{2} - 10}
\\[5mm] &
\implies \mrm{z}\pars{x} =
\expo{-2x}\int_{0}^{x}\expo{2t}\pars{4t^{2} - 10}\dd t = 2x^{2} - 2x - 4
\end{align}

Repeat the same procedure with the other differential equation.

