Solution: We need to solve the differential equation \begin{equation}\label{eq:14June2024-1} (D-1)(D^2 - 2D +2)y = e^x . \end{equation} The corresponding auxiliary equation is \begin{align*} (m-1)(m^2 - 2m + 2) = 0. \end{align*} The roots of the auxiliary equation are $1, 1 + \iota $, and $1-\iota $. Thus, the complimentary function, that is, the solution corresponding to the homogeneous equation will be \begin{equation}\label{eq:14June2024-2} y_{\mathsf{cf} } = c_1 e^x + e^x \left( c_2 \cos x + c_3 \sin x \right) . \end{equation} Let us recall a working rule for finding the particular integral of $f(D)y = e^{ax}$.

• If $f(a)\neq 0$, then \[ y_{\mathsf{PI} }(x) = \frac{1}{f(D)} e^{ax} = \frac{1}{f(a)}e^{ax}. \]
• If $f(D) = 0$, then the following two cases arise:
  • If $f(D) = (D-a)^n$, then the particular integral will be \[ y_{\mathsf{PI} }(x) = \frac{1}{(D-a)^n}e^{ax} = \frac{x^n}{n!}e^{ax}, \quad n\in \mathbb{N} . \]
  • If $f(D) = (D-a)^r \phi (D)$, where $\phi (a)\neq 0$ and $r \in \mathbb{N} $, then \begin{align*} y_{\mathsf{PI} }(x) & = \frac{1}{(D-a)^r \phi (D)} e^{ax} \\ & = \frac{1}{(D-a)^r} \cdot \frac{1}{\phi (a)}e^{ax} \\ & = \frac{1}{\phi (a)} \cdot \frac{x^r}{r!} \cdot e^{ax}. \end{align*}

Now using the above method, if $f(D) = (D-1)(D^2 - 2D +2)$, then we have \begin{align*} y_{\mathsf{PI} } & = \frac{1}{ (D-1)(D^2 - 2D +2)} e^x \\ & = \frac{1}{D - 1} \cdot \frac{1}{D^2 - 2D + 2 } e^x \\ & = \frac{1}{D-1} \frac{e^x}{1-2+2} = \frac{1}{D-1} e^x \\ & = \frac{x}{1!} e^x = xe^x. \end{align*} Thus, the solution to the differential equation \eqref{eq:14June2024-1} will be \[ y(x) = c_1 e^x + e^x \left( c_2 \cos x + c_3 \sin x \right) + xe^x. \]