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Differential EquationsHardSecond-Order Linear Equations

Second-Order Linear Equations · Hard problem

Find the value of v0v_0 for which the solution of y′′+2y′+5y=0y''+2y'+5y=0 with y(0)=1,y′(0)=v0y(0)=1,y'(0)=v_0 has a horizontal tangent at t=π4t=\dfrac{\pi}4.