verify whether y(x)= 2e-x + xe-x is a solution of y" + 2y' + y=0.
y=2e−x+xe−xy=2e^{-x} +xe^{-x}y=2e−x+xe−x find
y′=−2e−x+e−x−xe−x=−e−x−xe−xy′′=e−x−e−x+xe−x=xe−xy'=-2e^{-x}+e^{-x}-xe^{-x}=-e^{-x}-xe^{-x}\\ y''=e^{-x}-e^{-x}+xe^{-x}=xe^{-x}y′=−2e−x+e−x−xe−x=−e−x−xe−xy′′=e−x−e−x+xe−x=xe−x
Substitute in equation
xe−x+2⋅(−e−x+xe−x)+2e−x+xe−x==xe−x−2e−x−2xe−x+2e−x+xe−x=0xe^{-x}+2\cdot (-e^{-x}+xe^{-x})+2e^{-x}+xe^{-x}=\\ =xe^{-x}-2e^{-x}-2xe^{-x}+2e^{-x}+xe^{-x}=0xe−x+2⋅(−e−x+xe−x)+2e−x+xe−x==xe−x−2e−x−2xe−x+2e−x+xe−x=0
so y=2e−x+xe−xy=2e^{-x} +xe^{-x}y=2e−x+xe−xis solution of y′′+2y′+y=0y''+2y'+y=0y′′+2y′+y=0.
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