The general solution of the nonhomogeneous equation y''-6y'-7=8e(-t)-7

1.	The general solution of the nonhomogeneous equation y''-6y'-7=8e(-t)-7t-6 is
Answer» CORRECT QUESTION: Solve the following differential equation $\quad D^{2}y-6Dy-7y=8e^{-x}-7x-6$ SOLUTION: The given differential equation is $\quad D^{2}y-6Dy-7y=8e^{-x}-7x-6$ To find C. F. The AUXILIARY equation is $\quad m^{2}-6m-7=0$ $\Rightarrow (m+1)(m-7)=0$ $\Rightarrow m=-1,\:7$ C. F. is therefore, $C_{1}e^{-x}+C_{2}e^{7x}$ . To find P. I. Let, $y_{p}=Axe^{-x}+Bx+C$ Then, $Dy_{p}=-Axe^{-x}+Ae^{-x}+B$ and $D^{2}y_{p}=Axe^{-x}-2Ae^{-x}$ Since $y_{p}$ is a root of the given differential equation, $\quad D^{2}y_{p}-6Dy_{p}-7y_{p}=8e^{-x}-7x-6$ $\Rightarrow Axe^{-x}-2Ae^{-x}+6Axe^{-x}-6Ae^{-x}-6B-7Axe^{-x}-7Bx-7C=8e^{-x}-7x-6$ $\Rightarrow -8Ae^{-x}-7Bx-6B-7c=8e^{-x}-7x-6$ Equating like terms in both sides, we GET $\quad\quad -8A=8$ $\quad\quad -7B=-7$ $\quad\quad -6B-7C=-6$ $\implies$ $\quad\quad A=-1,\:B=1,\:C=0$ $\therefore y_{p}=-xe^{-x}+x$ $\therefore$ complete solution is $\quad y=C_{1}e^{-x}+C_{2}e^{7x}-xe^{-x}+x$