Question Find a particular solution to \[ y^{\prime \prime}-8 y^{\prime}+16 y=\frac{-0.5 e^{4 t}}{t^{2}+1} . \] \( y_{p} \) Find a particular solution to y′′−8y′+16y=t2+1−0.5e4t​. yp​

y^{\prime \prime}-8 y^{\prime}+16 y=\frac{-0.5 e^{4 t}}{t^{2}+1}corresponding homogeneous differential equation is, y^{\prime \prime}-8 y^{\prime}+16 y=0 \ldots(2)\therefore The auxillary eqn of (2) is,\begin{array}{l}m^{2}-8 m+16=0 \\\therefore(m-4)^{2}=0 \ldots\left[(a-b)^{2}=a^{2}-2 a b+b^{2}\right] \\\therefore m=4,4 \text { (repeated goots.). } \\\end{array}\therefore General solution of (2) is,y_{h}(t)=c_{1} e^{a t}+c_{2} t e^{4 t}where C_{1} C_{2} are constants f y_{1}=e^{u t}+y_{2}=t e^{u t}.Now, To find the particulor solution of (T), for this we ase the variation of param eter method.Let y_{p}(t)=v_{1} y_{1}+v_{2} y_{2}where v_{1}=-\int \frac{y_{2} \cdot R(t)}{w\left(y_{1} y_{2}\right)} d t, v_{2}=\int \frac{y_{1}-R(t)}{w\left(y_{1} y_{2}\right)} d t.\begin{aligned}W\left(y_{1}, y_{2}\right)=\left|\begin{array}{ll}y_{1} & y_{2} \\y_{1}^{\prime \prime} & y_{2}^{\prime}\end{array}\right| & =\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| \\& =e^{4 t}\left(e^{4 t}+4 t e^{4 t}\right)-4 t e^{10 t}\end{aligned}\begin{array}{l}\because w\left(y_{1}, y_{2}\right)=e^{8 t}+4 t e^{8 t}-4 t e^{8 t} . \\ =e^{8 t} \\ \epsilon_{R(t)}=\frac{-0.5 e^{4 t}}{t^{2}+1} \\ \therefore v_{1}=-\int \frac{y_{2} \cdot R(t)}{w\left(y_{1}, y_{2}\right)} d t=-\int \frac{t e^{4 t} \cdot\left(-0.5 e^{(t)}\right.}{\left(t^{2}+1\right) e^{g t}} d t . \\ =\int \frac{0.5 t}{t^{2}+1} d t \\ =0.5 \int \frac{t}{t^{2}+1} d t \text {. } \\ =\frac{0.5}{2} \int \frac{2 t}{t^{2}+1} d t \quad\left(\begin{array}{l}\text { divide tmultply } \\ \text { by } 2\end{array}\right) \\ =\frac{0.5}{2} \ln \left(t^{2}+1\right) \quad \cdots\left[\because \int \frac{f^{\prime}(x)}{f(x)} d x=\ln f(x)\right] \\ \therefore v_{1}=\frac{0.5}{2} \ln \left(t^{2}+1\right) \text {. } \\ t v_{2}=\int \frac{y_{1} \cdot R(t)}{w\left(y_{1}, y_{2}\right)} d t=\int \frac{e^{4 t}\left(-0.5 e^{4 t)}\right.}{\left(t^{2}+1\right) e^{8 t}} d t . \\ =-0.5 \int \frac{1}{t^{2}+1} d t \\\end{array}\therefore V_{2}=-0.5 \tan ^{-1}(t)\therefore The parncular solution yp(t) is.\begin{aligned}y_{p}(t) & =y_{2} v_{1}+y_{2} v_{2} . \\& =e^{4 t}\left[\frac{0.5}{2} \ln \left(t^{2}+1\right)\right]+t e^{4 t}\left[-0.5 \tan ^{-1}(t)\right] \\y_{p}(t) & =e^{G t}\left[\frac{0.5}{2} \ln \left(t^{2}+1\right)-0.5 t \tan ^{-1}(t)\right]\end{aligned} ...