Math / CalculusQuestionH.w Solve: y′′+2y′+y=4e−xlnx\quad y^{\prime \prime}+2 y^{\prime}+y=4 e^{-x} \ln xy′′+2y′+y=4e−xlnxStuddy SolutionThe general solution to the differential equation y′′+2y′+y=4e−xlnxy^{\prime \prime}+2 y^{\prime}+y=4 e^{-x} \ln xy′′+2y′+y=4e−xlnx is y=c1e−x+c2xe−x+2x2e−xlnx−3x2e−xy = c_1e^{-x} + c_2xe^{-x} + 2x^2e^{-x}\ln x - 3x^2e^{-x}y=c1e−x+c2xe−x+2x2e−xlnx−3x2e−x.View Full Solution - FreeWas this helpful?
