Question Find a particular soluton to \[ y^{\prime}+8 y+16 y=\frac{-13 e^{4 t}}{t^{2}+1} \] \( y_{p}= \) Find a particular soluton to y′+8y+16y=t2+1−13e4t yp=
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Given DE: y^{\prime \prime}+8 y^{\prime}+16 y=\frac{-13 e^{-4 t}}{t^{2}+1}Auxitiary equn of the reduced equn (1) is\begin{aligned}& m^{2}+8 m+16=0 \\\Rightarrow & m^{2}+2 \cdot m \cdot 4+4^{2}=0 \\\Rightarrow & (m+4)^{2}=0 \\\Rightarrow & m=-4,-4\end{aligned}\therefore The c \cdot F is y_{c}=\left(c_{1}+c_{2} t\right) e^{-4 t} ; c_{1}, c_{2} are arbitraryTo find P.I Here f(t)=\frac{-13 e^{-4 t}}{t^{2}+1} constantswe have y_{1}=e^{-4 t} and y_{2}=t e^{-4 t}Now,\begin{aligned}W\left(y_{1}, y_{2}\right) & =\left|\begin{array}{cc}y_{1} & y_{2} \\y_{1}^{\prime} & y_{2}^{\prime}\end{array}\right|=y_{1} y_{2}-y_{2} y_{1}^{\prime} \\& =e^{-4 t} \cdot(-4 t+1) e^{-4 t}-t e^{-4 t}-4 e^{-4 t} \\& =-4 t e^{-8 t}+e^{-8 t}+4 t e^{-8 t} \\& =e^{-8 t} \neq 0\end{aligned}So, y_{1}, y_{2} are linearly independent sorution\begin{aligned}\therefore y_{p} & =-y_{1} \int \frac{f(t) \cdot y_{2}}{w} d t+y_{2} \int \frac{f(t) \cdot y_{1}}{w} d t \\& =-e^{-4 t} \int \frac{-13 e^{-4 t}}{t^{2}+1} \cdot \frac{t e^{-4 t}}{e^{-8 t}} d t+t e^{-4 t} \int \frac{-13 e^{-4 t}}{t^{2}+1} \cdot \frac{e^{-4 t}}{e^{-8 t}} d t \\& =13 e^{-4 t} \int \frac{t}{t^{2}+1} d t-13 t e^{-4 t} \int \frac{d t}{t^{2}+1} \\& =\frac{13}{2} e^{-4 t} \int \frac{2 t}{t^{2}+1} d t-13 t e^{-4 t} \tan ^{-1}(t) \\& =\frac{13}{2} e^{-4 t} \ln \left(t^{2}+1\right)-13 t e^{-4 t} \tan ^{-1}(t)\end{aligned}\therefore Which is the particular sorution of DE(1). ...