Now as a is undetermined in (112) and (114), let us suppose it to be a root of (113), say a = a1, so that ƒ(a,) = 0; then the first term of the left-hand member of (114) vanishes, and there remains a differential equation of the (n-1)th order and observing that f" (a) = 1.2.3...(n-1)n, it is of the form dn-lu1 fn-1 (a1) f" (a) du, f'(a1) dan-1 1.2.3...(n-1) dan-2 +...+ 1.2 dx + + u1 = xe¬a1. (115) 1 Now supposing all the roots of the characteristic to be unequal, there are n different equations of this form, corresponding to these roots, a, a, a; also to solve (115) let ... and expressing the first term of the left-hand member in the by reason of the form of (113); and therefore a1+ẞ is a root of ... the other roots are a3-a1, a-a; let these be represented PRICE, VOL. II. 4 N P. and pursuing the same process, y = B2-B1 = а3-a; and as the equation for determining y will be of n-2 dimensions, the other roots will be a4-a2, a5-a2,... a-a; and the differential equation for determining u, will be of the form and the equation for the determination of 8 will be of n-3 dimensions, and its roots will be a-a, a -a,... a-a; and we shall continue the processes until we ultimately arrive at and as a constant is to be introduced at each successive integration, it is manifest that in the course of the process n such will be introduced, and therefore that the integral is general. And the general form of it is +ear.x +c(n)ea nx aq_a1)x dx +efel de f...dxfxe-dr. (121) If x = O, that is, if the given differential equation has no second member, then An examination of the form of the constant which will be introduced at the several integrations of the multiple integral in (120) shews that the result is of a form precisely the same as that indicated in equation (106). I may observe that this method of solution is the same as that investigated in Art. 443; but the general form of that Article is too complicated to be of useful employment, and therefore I have chosen to give a special inquiry. Should there be a pair of imaginary and conjugate roots in the characteristic, the corresponding result may be reduced to a circular function. This process is also applicable when two or more of the roots of the characteristic are equal; also the general result in equation (120) holds good. Thus suppose all the roots to be equal, then y = eax e-axx dx"; and the several integrations will plainly introduce n arbitrary constants. 456.] The following are examples solved by this process. The characteristic is a-13 a3 +26a2 +82a+104 = 0; of which the roots are −1+√ −1, −1 + √−1, 3+2√−1, 3−2√−1, -4 therefore by (122) y=e ̃ ̄2 {C,e√−12+Cye ̄√—Ix} + e3× {C2e2 √=1 x + c2e ̄2 √ = 1 × } + ©2 e ̄12 = k1e-* cos(x+Y1) + k2 e3* cos (2 x + Y1⁄2) + Cze−4× ̧ The characteristic is a3-7 a a2+16a2 a-12a3 =0; of which the roots are 2a, 2a, 3a; and therefore by (121) of which then roots are equal, and each is equal to -a; so The roots of the characteristic are -3a, a, 2a; therefore by (120), The roots of the characteristic are equal, and each is equal to 1. Therefore (120) becomes effxe de dr dx = x + 2 + (c2x + C12) e*. y = ex + a2y = x. The characteristic is a2+a2 = 0, and therefore the roots of the characteristic are a√−1, −a√−1. whereby when x is given the general integral can be found. Ex. 7. It is required to determine the form of the function, when f(x+y)+ƒ(x−y) = f (x)f(y); x and y being two variables independent of each other. Taking the x-differential twice, we have f(x +y)+f”(−y) = f”()f(y). Again, taking the y-differential twice, we have ƒ” (x+y)+ƒ” (x−y) = f (x)ƒ”(y). Consequently equating the right-hand members, where a2 is an arbitrary constant, since x and y are independent variables; C1 ·· ƒ"(x) ± a2ƒ (x) = 0. (123) (124) (125) (124) or (125) being the solution according as the lower or the upper sign is taken in (123). Taking (124) to be the solution, where c1 and c2 are undetermined constants, and substituting in the given equation we have c1 = c2 = 1, and a is undetermined. Also taking (125) to be the solution A=2, a=0; a being undetermined; so that f(x) = ex+e-ax; f(x) = 2 cos ax; either of which equations satisfies the proposed functional equation. 457.] If the right-hand member of the linear equation contains a constant only, so that the equation is of the form Now replace y by y+; then (126) becomes dy +An-1 dx (126) |