Question Find a particular solution to \[ y^{\prime \prime}-8 y^{\prime}+16 y=\frac{5 e^{4 t}}{t^{2}+1} \] \[ y_{p}= \]

Step1/2Solution:The given differential equation is \(y^''-8y^'+16y=(5e^(4t))/(t^2+1)\)\(rArr(D^2-8D+16)y=(5e^(4t))/(t^2+1)\), where \(D=d/dt.\)\(rArr (D-4)^2y=(5e^(4t))/(t^2+1)\)We have to find the particular solution \(y_p.\)Explanation:We know that if    be a differential equation then tha particular solution is   .Step2/2Here the particular solution of the given differential equation is \(y_p= 1/(D-4)^2(5e^(4t))/(t^2+1)\)\(rArr y_p= 5e^(4t)1/(D+4-4)^2 1/(t^2+1)\)\(rArr y_p= 5e^(4t)1/D^2 1/(t^2+1)\)\(rArr y_p= 5e^(4t)(1/D).(1/D 1/(t^2+1))\)\(rArr y_p= 5e^(4t)(1/D).(int 1/(t^2+1)dt)\)\(rArr y_p= 5e^(4t)(1/Dtan^-1(t))\)\(rArr y_p= 5e^(4t)(inttan^-1(t)dt)\)Integrating by parts we get, \( y_p= 5e^(4t)[tan^-1(t) int dt-int( d/dt(tan^-1(t)int dt)dt]\)\(rArr y_p= 5e^(4t)[t. tan^-1(t) -int(t/(t^2+1))dt]\)\(rArr y_p= 5e^(4t)[t. tan^-1(t) -1/2log(t^2+1)]\)Explanation:Final Answer:The final answer is:The particular integral is\( y_p= 5e^(4t)[t. tan^-1(t) -1/2log(t^2+1)]\) ...