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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.

Understand why

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It is valid to divide by erte^{rt} because the exponential function is never zero for any real values of rr and tt. Since the factor is strictly nonzero, dividing both sides of the equation by it does not lose any solutions for rr, allowing the differential equation to reduce to the algebraic characteristic equation.

Conditions: The equation is of the form 0=ert[ar2+br+c]0 = e^{rt}[ar^2 + br + c].; rr and tt are real numbers.