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Differential EquationsHardSecond-Order Linear Equations

Second-Order Linear Equations · Hard problem

Solve y′′+3y′+2y=0y'' + 3y' + 2y = 0 with y(0)=1y(0) = 1, y′(0)=−2y'(0) = -2. The solution is y=c1e−2t+c2e−1ty = c_1e^{-2t} + c_2e^{-1t}. Find c1c_1.