[1] Such a differential equation, with y as the dependent variable, superscript (n) denoting nth-derivative, and an, an − 1, ..., a1, a0 as constants, will have a characteristic equation of the form whose solutions r1, r2, ..., rn are the roots from which the general solution can be formed.
For a differential equation parameterized on time, the variable's evolution is stable if and only if the real part of each root is negative.
For both types of equation, persistent fluctuations occur if there is at least one pair of complex roots.
The method of integrating linear ordinary differential equations with constant coefficients was discovered by Leonhard Euler, who found that the solutions depended on an algebraic 'characteristic' equation.
[2] The qualities of the Euler's characteristic equation were later considered in greater detail by French mathematicians Augustin-Louis Cauchy and Gaspard Monge.
This results from the fact that the derivative of the exponential function e rx is a multiple of itself.
This suggests that certain values of r will allow multiples of e rx to sum to zero, thus solving the homogeneous differential equation.
are arbitrary constants which need to be determined by the boundary and/or initial conditions.
If a characteristic equation has parts with distinct real roots, h repeated roots, or k complex roots corresponding to general solutions of yD(x), yR1(x), ..., yRh(x), and yC1(x), ..., yCk(x), respectively, then the general solution to the differential equation is The linear homogeneous differential equation with constant coefficients has the characteristic equation By factoring the characteristic equation into[further explanation needed] one can see that the solutions for r are the distinct single root r1 = 3 and the double complex roots r2,3,4,5 = 1 ± i.
[1][7] Therefore, if the characteristic equation has distinct real roots r1, ..., rn, then a general solution will be of the form If the characteristic equation has a root r1 that is repeated k times, then it is clear that yp(x) = c1e r1x is at least one solution.
By applying this fact k times, it follows that By dividing out e r1x, it can be seen that Therefore, the general case for u(x) is a polynomial of degree k − 1, so that u(x) = c1 + c2x + c3x2 + ⋯ + ckxk −1.
Similarly, if c1 = 1/2i and c2 = −1/2i, then the independent solution formed is y2(x) = e ax sin bx.