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Answers for “这里展示的二阶情况的特征方程是什么?”

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The characteristic equation is the quadratic algebraic equation 0=ar2+br+c0 = ar^2 + br + c. It is obtained by substituting the trial solution y=erty = e^{rt} into the second-order linear homogeneous differential equation ay′′+by′+cy=0ay'' + by' + cy = 0, factoring out the nonzero exponential term erte^{rt}, and setting the remaining polynomial in rr to zero.

Conditions: The differential equation is ay′′+by′+cy=0ay'' + by' + cy = 0.; The trial solution is y=erty = e^{rt}.; a,b,ca, b, c are constants.