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Differential EquationsHardSecond-Order Linear Equations

Second-Order Linear Equations · Hard problem

Solve y′′+2y′−3y=0y'' + 2y' - 3y = 0 with y(0)=−5y(0) = -5, y′(0)=3y'(0) = 3. The solution is y=c1e−3t+c2e1ty = c_1e^{-3t} + c_2e^{1t}. Find c1c_1.