Differential equation finding from general solution

Question

Could you please find the differential equation by using the following method:

Delete Arbitrary constant (C) $$ Y = x^2+C_1e^x+C_2e^{−2x} $$

Answer 1

Yes eliminate $C_1, C_2)$ by differentiating. $$y=x^2+C_1 e^x+C_2 e^{-2x}$$ Multiply by $e^{-x}$
$$ye^{-x}=x^2e^{-x}+C_1+C_2 e^{-3x}$$ Differntiate w. r. $x$ to get $$(y'-y)e^{-x}=2xe^{-x}-x^2e^{-x}-3C_2 e^{-3x}$$ Next nultiply by $e^{3x}$ $$(y'-y)e^{2x}=2xe^{2x}-x^2e^{2x}-3C_2$$ D. w.r.t. $x$ $$(y''-y')e^{2x}+2(y'-y)e^{2x}=2e^{2x}+4xe^{2x}-2xe^{2x}-4x^2e^{2x}$$ $$\implies (y''+y'-2y)=2+4x-2x+4x^2e^{2x}$$ $$\implies y''-y'-2y=2+2x+4x^2$$

Answer 2

$$Y=x^2+C_1e^x+C_2e^{−2x}$$ You can conclude that $r=-2 $ and $r=1$ Where the characteristic polynomial is $$(r-1)(r+2)=0 \implies r^2+r-2=0 $$ The homgeneous equation is: $$y''+y'-2y=0$$ For the inhomogeneous equation, we have : $$y_p''+y_p'-2y_p=f(x)$$ Where $y_p=x^2$ $$\implies f(x)=2+2x-2x^2$$ Therefore the equation is: $$y''+y'-2y=2+2x-2x^2$$