## Quick Reference

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

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

is the homogeneous equation that corresponds to the non-homogeneous equation **1**. The two are connected by the following result:

Theorem

If *y*=*G*(*x*) is the general solution of **2** and *y*=*y*_{1}(*x*) is a particular solution of **1**, then *y*=*G*(*x*)+*y*_{1}(*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 *y*_{1}(*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 *am*^{2}+*bm*+*c*=0. If this equation has distinct real roots *m*_{1} and *m*_{2}, the C.F. is if it has one (repeated) root *m*, the C.F. is *y*=(*A*+*Bx*)*e*^{mx}; if it has non-real roots α ± β i, the C.F. is *y* =*e*^{αx}(*A* cos β*x*+*B* sin β*x*).

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. *y*_{1}(*x*)=*pe*^{kx}. If *f*(*x*) is a polynomial in *x*, try a polynomial of the same degree. If *f*(*x*)=cos *kx* or sin *kx*, try *y*_{1}(*x*)=*p* cos *kx*+*q* sin *kx*. 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.

Using these methods, the general solution of *y*″−3*y*′+2*y*=4*x*+*e*^{3x}, for example, is found to be

**From:**
linear differential equation with constant coefficients
in
The Concise Oxford Dictionary of Mathematics »

*Subjects:*
Mathematics.

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