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Differential EquationsHardSecond-Order Linear Equations

Second-Order Linear Equations · Hard problem

Solve y′′+1y′−6y=0y'' + 1y' - 6y = 0 with y(0)=3y(0) = 3, y′(0)=1y'(0) = 1. The solution is y=c1e−3t+c2e2ty = c_1e^{-3t} + c_2e^{2t}. Find c1c_1.