Question

Use the method of variation of parameters to find a particular solution of the differential equation

y′′−2y′−15y=384e^−t.

Answer 1

$~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~y^{\prime \prime}-2 y^{\prime}-15 y=384 e^{-t}\\[1 em] \text{ Second order linear non-homogeneous differential equation . } \\[1 em] ~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~y=y_{h}+y_{p}\\[1 em] y_{h}~~~ \text{is the solution to the homogeneous .}\\[1 em] y^{\prime \prime}-2 y^{\prime}-15 y=0\\[1 em] r^{2}-2 r-15=0 \\[1 em] \begin{array}{l} (r-5)(r+3)=0 \\[1 em] \therefore r_{1}=5 \quad, \quad r_{2}=-3 \end{array}\\[1 em] ~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~\therefore y_{h}=C_{1} e^{5 t}+C_{2} e^{-3 t}\\[1 em] y_{p} \text{ is the particular solution is any function} \\[1 em] \text{ That satisfies the non-homogeneous equation .} \\[1 em] y_{p}=u_{1} v_{1}+u_{2} v_{2}\\[1 em] u_{1}=e^{5 t}, u_{2}=e^{-3 t} , f(t)=384 e^{-t}\\[2 em] D=\left|\begin{array}{ll}u_{1} & u_{2} \\ u_{1}^{\prime} & u_{2}^{\prime}\end{array}\right|=\left|\begin{array}{cc}e^{5 t} & e^{-3 t} \\ 5 e^{5 t} & -3 e^{-3 t}\end{array}\right|=-8 e^{2 t}\\[2 em] v_{1}=\int \frac{-u_{2} f(t)}{D} d x=\int \frac{-e^{-3 t} \cdot 384 e^{-t}}{-8 e^{2 t}} d t=48\int e^{-6 t} dt = -8 e^{-6 t}\\[2 em] v_{2}=\int \frac{u_{1} f(t)}{D} d t=\int \frac{e^{5 t} \cdot 384 e^{-t}}{-8 e^{2 t}} d t=-48\int e^{2 t} dt = -24 e^{2 t} \\[2 em] ~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~y_{p}=u_{1} v_{1}+u_{2} v_{2}\\[1 em] ~~~~~=e^{5 t}\times -8e^{-6 t}+e^{-3 t}\times -24e^{2 t}= -32e^{- t}\\[2 em] \therefore y=C_{1} e^{5 t}+C_{2} e^{-3 t} -32e^{- t}\\[2 em]$