“linear differential equation with constant coefficients”的概念、定义、翻译、参考文献-科学参考

linear differential equation with constant coefficients

Mathematics

For simplicity, consider such an equation of second‐order,

$a \frac{d^{2} y}{d x^{2}} + b \frac{d y}{d x} + c y = f (x), 1$

where a, b, and c are given constants and f is a given function. (Higher‐order equations can be treated similarly.) Suppose that f is not the zero function. Then the equation

$a \frac{d^{2} y}{d x^{2}} + b \frac{d y}{d x} + c y = 02$

is the homogeneous equation that corresponds to the inhomogeneous equation 1. The two are connected by the following result:
Theorem
If y = G(x) is the general solution of 2 and y = y₁(x) is a particular solution of 1, then y = G(x) + y₁(x) is the general solution of 1.

Thus the problem of solving 1 is reduced to the problem of finding the complementary function (C.F.) G(x), which is the general solution of 2, and a particular solution y₁(x) of 1, usually known in this context as a particular integral (P.I.).

The complementary function is found by looking for solutions of 2 of the form y = e^mx and obtaining the auxiliary equation am² + bm + c = 0. If this equation has distinct real roots m₁ and m₂, the C.F. is $y = A e^{m_{1} x} + B e^{m_{2} x};$ if it has one (repeated) root m, the C.F. is y = (A + Bx)e^mx; if it has non‐real roots α‎ ± β‎i, the C.F. is y = e^α‎x(Acosβ‎x + Bsinβ‎x).

The most elementary way of obtaining a particular integral is to try something similar in form to f(x). Thus, if f(x) = e^kx, try as the P.I. y₁(x) = pe^kx. If f(x) is a polynomial in x, try a polynomial of the same degree. If f(x) = coskx or sinkx, try y₁(x) = pcoskx + qsinkx. In each case, the values of the unknown coefficients are found by substituting the possible P.I. into the equation 1. If f(x) is the sum of two terms, a P.I. corresponding to each may be found and the two added together.

For example, the general solution of y″−3y′+2y = 4x + e^3x, is found to be $y = A e^{x} + B e^{2 x} + 2 x + 3 + \frac{1}{2} e^{3 x} .$

单词	linear differential equation with constant coefficients
释义	linear differential equation with constant coefficients Mathematics For simplicity, consider such an equation of second‐order, $a \frac{d^{2} y}{d x^{2}} + b \frac{d y}{d x} + c y = f (x), 1$ where a, b, and c are given constants and f is a given function. (Higher‐order equations can be treated similarly.) Suppose that f is not the zero function. Then the equation $a \frac{d^{2} y}{d x^{2}} + b \frac{d y}{d x} + c y = 02$ is the homogeneous equation that corresponds to the inhomogeneous equation 1. The two are connected by the following result: Theorem If y = G(x) is the general solution of 2 and y = y₁(x) is a particular solution of 1, then y = G(x) + y₁(x) is the general solution of 1. Thus the problem of solving 1 is reduced to the problem of finding the complementary function (C.F.) G(x), which is the general solution of 2, and a particular solution y₁(x) of 1, usually known in this context as a particular integral (P.I.). The complementary function is found by looking for solutions of 2 of the form y = e^mx and obtaining the auxiliary equation am² + bm + c = 0. If this equation has distinct real roots m₁ and m₂, the C.F. is $y = A e^{m_{1} x} + B e^{m_{2} x};$ if it has one (repeated) root m, the C.F. is y = (A + Bx)e^mx; if it has non‐real roots α‎ ± β‎i, the C.F. is y = e^α‎x(Acosβ‎x + Bsinβ‎x). The most elementary way of obtaining a particular integral is to try something similar in form to f(x). Thus, if f(x) = e^kx, try as the P.I. y₁(x) = pe^kx. If f(x) is a polynomial in x, try a polynomial of the same degree. If f(x) = coskx or sinkx, try y₁(x) = pcoskx + qsinkx. In each case, the values of the unknown coefficients are found by substituting the possible P.I. into the equation 1. If f(x) is the sum of two terms, a P.I. corresponding to each may be found and the two added together. For example, the general solution of y″−3y′+2y = 4x + e^3x, is found to be $y = A e^{x} + B e^{2 x} + 2 x + 3 + \frac{1}{2} e^{3 x} .$
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