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Differential EquationsHardSecond-Order Linear Equations

Second-Order Linear Equations · Hard problem

Solve y′′+0y′−4y=0y'' + 0y' - 4y = 0 with y(0)=3y(0) = 3, y′(0)=−2y'(0) = -2. The solution is y=c1e2t+c2e−2ty = c_1e^{2t} + c_2e^{-2t}. Find c1c_1.