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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)=5y(0) = 5, y′(0)=−7y'(0) = -7. The solution is y=c1e−1t+c2e−3ty = c_1e^{-1t} + c_2e^{-3t}. Find c1c_1.