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Differential EquationsHardSecond-Order Linear Equations

Second-Order Linear Equations · Hard problem

Solve y′′+4y′+3y=0y'' + 4y' + 3y = 0 with y(0)=3y(0) = 3, y′(0)=−9y'(0) = -9. The solution is y=c1e−3t+c2e−1ty = c_1e^{-3t} + c_2e^{-1t}. Find c1c_1.