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Differential EquationsChallengeSecond-Order Linear Equations

Second-Order Linear Equations · Challenge problem

The equation y′′−3y′+2y=0y'' - 3y' + 2y = 0 has general solution y=Aex+Be2xy = Ae^{x} + Be^{2x}. With y(0)=3y(0) = 3 and y′(0)=5y'(0) = 5, compute y(ln⁡2)y(\ln 2).