Solve the following non-homogeneous second order linear differential equation. y''+py'-qy=-rt if p=4, q=8, r=7.

Solve the following non-homogeneous second order linear differential equation.
${y}^{″}+p{y}^{\prime }-qy=-rt$
if .
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Substituting in the given equation: ${y}^{″}+4{y}^{\prime }-8e=-7t$. Thus, auxiliary equation of ${y}^{″}+4{y}^{\prime }-8y=-7t$ is ${m}^{2}+4m-8=0$. Solving for m:
$m=\frac{4±\sqrt{{4}^{2}-4\left(-8\right)}}{2}$
$=-2±2\sqrt{3}$
Thus the complementary function is ${y}_{C}F={e}^{-2t}\left({c}_{1}{e}^{2\sqrt{3}t}+{c}_{2}{e}^{-2\sqrt{3}t}\right)$ where ${c}_{1}$ and ${c}_{2}$ are arbitrary constants.
Now to find Particular integral:
Using the operator $D=\frac{d}{dt}$ in ${y}^{″}+4{y}^{\prime }-8y=-7t$
$\left({D}^{2}+4D-8\right)y=-7t$. Thus
${y}_{PI}=\frac{1}{{D}^{2}+4D-8}7t$
$=\frac{7}{8}\cdot \frac{1}{1-\left(\frac{{D}^{2}+4D}{8}\right)}\left(t\right)$
Now using binomial expansion:${y}_{PI}=\frac{7}{8}\cdot \frac{1}{1-\left(\frac{{D}^{2}+4D}{8}\right)}\left(t\right)$
$=\frac{7}{8}\cdot \left(1+\frac{{D}^{2}+4D}{8}+\left(\frac{{D}^{2}+4D}{8}{\right)}^{2}+...\right)\left(t\right)$
$=\frac{7}{8}\left(t+\left(\frac{{D}^{2}+4D}{8}\right)t\right)$
$=\frac{7}{8}\left(t+\frac{1}{2}\right)$
$=\frac{7t}{8}+\frac{7}{16}$
Thus complete solution is ${y}_{CF}+{y}_{PI}$ or the complete solution is
${e}^{-2t}{c}_{1}{e}^{2\sqrt{3}t}+{c}_{2}{e}^{-2\sqrt{3}t}+\frac{7t}{8}+\frac{7}{16}$ where ${c}_{1}$ and ${c}_{2}$ are arbitrary constant