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Answer

(Refer to the method of undetermined coefficients) * We first check that the coefficient of \(\difftwo{y}\) is 1. * We solve the homogenous ODE first.

\[ \begin{aligned} \difftwo{y} + 3\diffone{y} + 2.25y &= 0\newline \implies m^{2} + 3m + 2.25 &= 0\newline m &= -\frac{3}{2} = -1.5 \end{aligned} \]
* The solution to the homogenous equation becomes
\[ \begin{aligned} \roundbr{c_{1} + c_{2}x}e^{-1.5x} \end{aligned} \]
* Based on $r(x) = -10e^{-1.5x}$, the solution to the homogenous equation will be of the same form. However, using the modification rule, we multiply an extra term of $x^{2}$ to it making $y_{p} = Cx^{2}e^{1.5x}$
* Substituing this solution in the differential equation we obtain
\[ \begin{aligned} \roundbr{2Ce^{-1.5x} - 6Cxe^{-1.5x} + 2.25Cx^{2}e^{-1.5x}} + 3\roundbr{2Cxe^{-1.5x} - 1.5Cx^{2}e^{-1.5x}} + 2.25Cx^{2}e^{-1.5x} &= -10e^{-1.5x} \end{aligned} \]
    Talking out the common factor of $e^{-1.5x}$ and simplifying,
\[ \begin{aligned} 2C = -10 \implies C = -5 \end{aligned} \]
    It is important to determine the constants relating to $y_{p}$ first, before using the initial conditions and getting the constants related to $y_{h}$
* Thus, the general solution is
\[ \begin{aligned} \roundbr{c_{1} + c_{2}x}e^{-1.5x} - 5x^{2}e^{-1.5x} \end{aligned} \]
* Using the initial conditions, we obtain
\[ \begin{aligned} \roundbr{1 + 1.5x}e^{-1.5x} - 5x^{2}e^{-1.5x} = \roundbr{1 + 1.5x - 5x^{2}}e^{-1.5x} \end{aligned} \]