quently occur in differential equations, it is convenient to keep in mind that the complementary function due to it is of the form С cos nx + C′ sin n x. d Then y + x" + C cos x + C1 sin æ -2 1 y = x" − n (n − 1), x2 - 2 + n (n − 1) (n − 2) (n − 3) x2 - 4 — &c. +C cos x + C, sin a. When the differential expression is divided into factors, this may be put under the form 2) ( Whence we find (m-2)* (m+2)(m2+9) +ε2* (C+C1x) + C2 € ̄2*+ C3 Co$ (3x+a). λπ where == n a {cos o + (-)3 sin }, λ receiving all values from 0 to n; and the roots corresponding to λ = 0 and λ= n being + a and -a. If now we take a pair of the impossible roots, which we may call a and B, the corresponding terms in the general value of y are X. But by the theory of the decomposition of rational fractions we know that d (a (~~ - a)" X = e^zcos + { cos (a æ sin p) + (−)a sin (uæ sin p)} and da d xfd Xe -az cos {cos (ax sin p) − (−) sin (a x sin p)}, -1 X = er cos {cos (a x sin ) − (−)3 sin (a æ sin ()} xd Xe -ar cos {cos (ax sin () + (−) sin (ax sin p)}. Therefore substituting these expressions, the two terms in the value of y become after reduction, Ear cos cos (a x sin p+) fdæ {Xe-ar cos cos (ax sin )} cos 1 na2n-1 1 + na2n-1 arcos € cos sin (a x sin+p) fdx {Xe-arcos sin & ** sin (a æ sin p) } . Hence we put the expression for y under the form The symbol assigning to arcos sin (ax sin & + 9) × fd x { X € implies the sum of terms derived from in the preceding expression all values from The complementary functions are, for the sake of shortness, supposed to be included in the signs of integration; but if we wish to see their form, we have only to make X = 0 in the preceding expression when it becomes 2π n + п - {C, cos (ar sin 2+2)+ C, sin (ax sin+)} + &c. n + &c. 6 This is evidently the solution of the equation d2n y dx2n Euler, Calc. Integ. Vol. 11. Sect. 2, Cap. IV. -ax ax The term e ["da" * X may either be integrated by successive steps, or by the general formula for integration by parts; or what will generally be more convenient, the function If n were negative or fractional, the first term would retain the same form, but the form of the complementary function would be different from the difference between the roots of (x + a)" = 0, when n is integer and when it is fractional or negative. I cannot however here enter into a discussion of the difficulties of this subject, which is closely connected with that of General Differentiation. Euler, Calc. Integ. Ib. |