Question #231865

verify whether y(x)= 2e-x + xe-x is a solution of y" + 2y' + y=0.


1
Expert's answer
2021-09-02T00:07:59-0400

y=2ex+xexy=2e^{-x} +xe^{-x} find

y=2ex+exxex=exxexy=exex+xex=xexy'=-2e^{-x}+e^{-x}-xe^{-x}=-e^{-x}-xe^{-x}\\ y''=e^{-x}-e^{-x}+xe^{-x}=xe^{-x}

Substitute in equation

xex+2(ex+xex)+2ex+xex==xex2ex2xex+2ex+xex=0xe^{-x}+2\cdot (-e^{-x}+xe^{-x})+2e^{-x}+xe^{-x}=\\ =xe^{-x}-2e^{-x}-2xe^{-x}+2e^{-x}+xe^{-x}=0

so y=2ex+xexy=2e^{-x} +xe^{-x}is solution of y+2y+y=0y''+2y'+y=0.


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