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A linear ordinary differential equation has constant coefficients when the multipliers $a_j$ of the unknown function and its derivatives do not depend on the independent variable. The forcing $f$ may still vary. For the homogeneous equation $f=0$, its characteristic equation of a constant-coefficient differential equation determines exponential solutions.

The product rule gives $y^{\prime }=x^n{z}^{\prime }+n{x}^{n-1}z$ and $y^{\prime \prime }=x^n{z}^{\prime \prime }+2n{x}^{n-1}{z}^{\prime }+n(n-1)x^{n-2}z$. On an interval avoiding $x=0$, substitution and division by $x^{n+2}$ giveFor a constant-coefficient differential equation, the coefficient of $z^{\prime }$ forces $2n+4=0$. At $n=-2$, the remaining coefficient $n^2+3n+2$ also vanishes. ThusThe characteristic equation of a constant-coefficient differential equation is $r^2+2r+2=0$, with complex conjugate roots of a polynomial $-1\pm i$. Its real solutions $e^{-x}\cos x$ and $e^{-x}\sin x$ are linearly independent. Therefore the general solution isThe constants can be chosen independently on the two sides of the singular point. The power substitution reducing a differential equation to constant coefficients here is invertible wherever $x\ne 0$.

The characteristic equation of a constant-coefficient differential equation here is $r^2-3r+2=(r-1)(r-2)=0$. Its homogeneous solution is $C_1{e}^x+C_2{e}^{2x}$. Because the forcing $e^x$ is already a homogeneous mode, the method of undetermined coefficients requires $y_p=Ax{e}^x$ rather than $A{e}^x$.

Here $y_p^{\prime }=A(x+1)e^x$ and $y_p^{\prime \prime }=A(x+2)e^x$, so $y_p^{\prime \prime }-3y_p^{\prime }+2y_p=-A{e}^x$. Thus $A=-1$, and the general solution is