Question Find a particular solution to \[ y^{\prime \prime}+8 y^{\prime}+16 y=\frac{1.5 e^{-4 t}}{t^{2}+1} \] \( y_{p}= \) Find a particular solution to y′′+8y′+16y=t2+11.5e−4t​ yp​=

Solution: y^{\prime \prime}+8 y^{\prime}+16 y=\frac{1.5 e^{-4 t}}{t^{2}+1}Auxiliary equation is\begin{array}{l}m^{2}+8 m+16=0 \\(m+4)^{2}=0 \Rightarrow m=-4,-4\end{array}\therefore Complementary function is\begin{array}{c}y_{c}(t)=\left(c_{1}+c_{2} t\right) e^{-4 t} \\\therefore y_{1}(t)=e^{-4 t}, \quad y_{2}(t)=t e^{-4 t} \text { are linearly }\end{array}independent solutions\text { wronskian } w=\left|\begin{array}{ll}y_{1} & y_{2} \\y_{1}^{\prime} & y_{2}^{\prime}\end{array}\right|\begin{array}{l}W=\left|\begin{array}{cc}e^{-4 t} & t e^{-4 t} \\-4 e^{-4 t} & e^{-4 t}-4 t e^{-4 t}\end{array}\right| \\W=e^{-8 t}+\varphi(t)=\frac{1.5 e^{-4 t}}{t^{2}+1}\end{array}\therefore by variation of parameter\begin{aligned}y_{p} & =y_{2} \int \frac{y_{1} \varphi}{w} d t-y_{1} \int \frac{y_{2} \varphi}{W} d t \\& =t e^{-4 t} \int \frac{e^{-4 t}}{e^{-8 t}} 1.5 e^{-4 t} d t-e^{-4 t} \int \frac{t e^{-4 t}}{e^{-8 t}} \frac{1.5 e^{-4 t}}{t^{2}+1} d t \\& =t e^{-4 t} \cdot 1.5 \int \frac{1}{t^{2}+1} d t-1.5 e^{-4 t} \int \frac{t}{t^{2}+1} d t \\y_{p} & =1.5 t e^{-4 t} \tan ^{-1}(t)-\frac{1.5}{2} e^{-4 t} \ln \left|1+t^{2}\right|\end{aligned}y_{p}=1.5 t e^{-4 t} \tan ^{-1}(t)-0.75 e^{-4 t} \ln \left(1+t^{2}\right) ...