Question Find a particular solution to \[ y^{\prime \prime}-8 y^{\prime}+16 y=\frac{-8.5 e^{4 t}}{t^{2}+1} \] \[ y_{p}= \] Find a particular solution to y′′−8y′+16y=t2+1−8.5e4t​ yp​=

y^{\prime \prime}-8 y^{\prime}+16 y=\frac{-8 \cdot 5 e^{4 t}}{t^{2}+1}-(1)A.E of (1) is f(r)=0 ie, r^{2}-8 r+16=0\begin{aligned}(r-4)^{2} & =0 \\r & =4,4 .\end{aligned}\begin{aligned}\therefore \quad y_{c} & =\left(c_{1}+c_{2} t\right) e^{4 t} . \\y_{1} & =e^{4 t}, \quad y_{2}=t e^{4 t} \text { and } \quad R=\frac{-8 \cdot 5 e^{4 t}}{t^{2}+1} \\y_{1}^{\prime} & =4 e^{4 t}, \quad y_{2}^{\prime}=1 \cdot e^{4 t}+4 t e^{4 t}=(1+4 t) e^{4 t} \\\omega & =\left|\begin{array}{ll}y_{1} & y_{2} \\y_{1}^{\prime} & y_{2}^{\prime}\end{array}\right| \\& =\left|\begin{array}{ll}e^{4 t} & t e^{4 t} \\4 e^{4 t} & \left(1+e^{4 t}\right) e^{4 t}\end{array}\right| \\& =e^{8 t}[1+4 t-4 t] \\& =e^{8 t} \neq 0 .\end{aligned}\begin{aligned}f(t) & =-\int \frac{y_{2} R}{w} d t \\& =-\int \frac{t e^{4 t}\left(-\frac{8.5 e^{4 t t}}{t^{2}+1}\right)}{e^{8 t}} d t \\& =8.5 \int \frac{t}{t^{2}+1} d t \\& =\frac{8.5}{2} \ln \left|t^{2}+1\right| \\& =4.25 \ln \left|t^{2}+1\right| .\end{aligned}\begin{aligned}g(t) & =\int \frac{y_{1} R}{\omega} d t \\& =\int \frac{e^{4 t} \cdot\left(\frac{-8 \cdot 5 e^{4 t}}{\left(t^{2}+1\right)}\right.}{e^{8 t}} d t \\& =-8.5 \int \frac{1}{t^{2}+1} d t \\& =-8.5 \tan ^{-1}(t)\end{aligned}\begin{aligned}y_{p} & =y_{1}(t) f(t)+y_{2}(t) g(t) \\& =e^{4 t} \cdot\left[4.25 \ln \left|t^{2}+1\right|\right]+t e^{4 t}\left(-8.5 \tan ^{-1}(t)\right) \\y_{p} & =\left[4.25 \ln \left|t^{2}+1\right|-8.5 t \cdot \tan ^{-1}(t)\right] e^{4 t}\end{aligned}Thus, y=y c+y p=\left(c_{1}+c_{2} t\right) e^{4 t}+4.25 e^{4 t} \ln \left|t^{2}+1\right|-8.5 t e^{4 t} \cdot \tan ^{-1}(t) ...