Substituting \(p=4,\ q=8,\ r=7\) in the given equation: \(y''+4y'-8e=-7t\). Thus, auxiliary equation of \(y''+4y'-8y=-7t\) is \(m^2+4m-8=0\). Solving for m:

\(m=\frac{4\pm\sqrt{4^2-4(-8)}}{2}\)

\(=-2\pm2\sqrt3\)

Thus the complementary function is \(y_CF=e^{-2t}(c_1e^{2\sqrt3t}+c_2e^{-2\sqrt3t})\) where \(c_1\) and \(c_2\) are arbitrary constants.

Now to find Particular integral:

Using the operator \(D=\frac{d}{dt}\) in \(y''+4y'-8y=-7t\)

\((D^2+4D-8)y=-7t\). Thus

\(y_{PI}=\frac{1}{D^2+4D-8}7t\)

\(=\frac{7}{8}\cdot\frac{1}{1-(\frac{D^2+4D}{8})}(t)\)

Now using binomial expansion: \(y_{PI}=\frac{7}{8}\cdot\frac{1}{1-(\frac{D^2+4D}{8})}(t)\)

\(=\frac{7}{8}\cdot(1+\frac{D^2+4D}{8}+(\frac{D^2+4D}{8})^2+...)(t)\)

\(=\frac{7}{8}(t+(\frac{D^2+4D}{8})t)\)

\(=\frac{7}{8}(t+\frac{1}{2})\)

\(=\frac{7t}{8}+\frac{7}{16}\)

Thus complete solution is \(y_{CF}+y_{PI}\) or the complete solution is

\(e^{-2t}c_1e^{2\sqrt3t}+c_2e^{-2\sqrt3t}+\frac{7t}{8}+\frac{7}{16}\) where \(c_1\) and \(c_2\) are arbitrary constant

\(m=\frac{4\pm\sqrt{4^2-4(-8)}}{2}\)

\(=-2\pm2\sqrt3\)

Thus the complementary function is \(y_CF=e^{-2t}(c_1e^{2\sqrt3t}+c_2e^{-2\sqrt3t})\) where \(c_1\) and \(c_2\) are arbitrary constants.

Now to find Particular integral:

Using the operator \(D=\frac{d}{dt}\) in \(y''+4y'-8y=-7t\)

\((D^2+4D-8)y=-7t\). Thus

\(y_{PI}=\frac{1}{D^2+4D-8}7t\)

\(=\frac{7}{8}\cdot\frac{1}{1-(\frac{D^2+4D}{8})}(t)\)

Now using binomial expansion: \(y_{PI}=\frac{7}{8}\cdot\frac{1}{1-(\frac{D^2+4D}{8})}(t)\)

\(=\frac{7}{8}\cdot(1+\frac{D^2+4D}{8}+(\frac{D^2+4D}{8})^2+...)(t)\)

\(=\frac{7}{8}(t+(\frac{D^2+4D}{8})t)\)

\(=\frac{7}{8}(t+\frac{1}{2})\)

\(=\frac{7t}{8}+\frac{7}{16}\)

Thus complete solution is \(y_{CF}+y_{PI}\) or the complete solution is

\(e^{-2t}c_1e^{2\sqrt3t}+c_2e^{-2\sqrt3t}+\frac{7t}{8}+\frac{7}{16}\) where \(c_1\) and \(c_2\) are arbitrary constant