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then substituting (112) in (80) we have
f(a) fudæ + fa), + smo ..

fr-1(a) d"-24, , f"(a) d"-14, -,
"1.2.3...(n-1) dan-2 +1.2. n do-= Xe 4 (114)
Now as a is undetermined in (112) and (114), let us suppose
it to be a root of (113), say a = a,, so that f(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 ob-

serving that f"(a) = 1.2.3...(n-1)n, it is of the form da-lu. fn-'(a) dn-20, f"(a) du, f'(a) dco-l' 1.2.3...(n-1) don–2 f

19-1, +° a=Xe-d13. (115)
Now supposing all the roots of the characteristic to be un.
equal, there are n different equations of this form, corresponding
to these roots, a, az, ... an; also to solve (115) let
= 284 / uzdx ;

(116)
substituting which in (115) we have
one fn-'(a)

S(a)
3 on-l+ 3n-_

+...+ 1.2...(n-1)

+B, 4, + B, + ... + B -3 -3 = xe-oto/*; (117) and expressing the first term of the left-hand member in the following form, and adding f (ay), which is equal to zero, because a, is a root of (113), we have

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f"(a)

fn-'(a). let en 1.2. tpn-lin

+ f(a) = 0; (119) .. flar+B) = 0, by reason of the form of (113); and therefore a+ß is a root of (113): let this root be az, then a, + B = az, and ß = ay-ay; and as (119) is an algebraical equation of (n-1) dimensions, the other roots are az-az, ... Qn- a,; let these be represented by By, B2, ... Br-1; and aş Bn-s is evidently unity in (118),

(118) becomes PRICE, VOL. II.

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and pursuing the same process, y = B2-B= 2z - az; and as the equation for determining y will be of n—2 dimensions, the other roots will be a4-ag, az , ... an-Ag; and the differential equation for determining uz will be of the form dn–3u, ... dnu

du .

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and the equation for the determination of 8 will be of n-3 dimensions, and its roots will be at — Az, az – az, ... Q.,- az; and we shall continue the processes until we ultimately arrive at

Un-1 = elan-an-1)= /vndx ;

u = xe-d%; and thus, returning through the several steps, y=2018 / elas-ax)* dx elaz-az« dx |... lan-on-1}* dx (xe-adx; (120) 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 y = c'eq;* + c"@agde + ... + c(mpa net

+ee (eng 4 da ...e --+ da xe-da. (121) If x = 0, that is, if the given differential equation has no second member, then y = c'eaza + o" eaza + ... +cím) esinden

(122) 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.

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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 = 402 ["e-axx dx; and the several integrations will plainly introduce n arbitrary constants. 456.] The following are examples solved by this process.

d5y 03, • Ex. 1. "

+82 +104y = 0.

dx The characteristic is a – 13 a3 +26 a2 +82 a +104 = 0); of which the roots are –1+ ✓-1,-1+ 1, 3+21–1,3–2V-1,

-4: therefore by (122) y=e-*{q;ev=1e+ce-v=14}+eše{Cze?v=i*+62-21-12} + 0; e-4x

=k_e-* cos (x +.) + k, e34 cos (2x+y)+03e-48.

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The roots of the characteristic are - 3a, a, 2 a; therefore by (120),

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3 dr

172 4a - 13

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x sin ax = k cos (ax + y) + 2a

Ex. 7. It is required to determine the form of the function, when f (x+y)+ f (xy) = 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"(xy) = f"(x) f(y). Again, taking the y-differential twice, we have

F"(x+y)+$"(x - y) = f(x)f"(y). Consequently equating the right-hand members,

f"(x) = $"(9) = + ao, say,

f(x) f(y) where a’ is an arbitrary constant, since x and y are independent lables ; i. f(x) + a2 f (x) = 0.

(123) si f() = C, €4+ + C, e-ar;

(124) f(x) = A cos (ax + a);

(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 c, and c, are undetermined constants, and substituting in the given equation we have c = c2=1, and a is undetermined. Also taking (125) to be the solution A=2, a=0; a being undetermined; so that

f(x) = ax +e-ux;

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 dny dn-ly

"I HAY = A; dx +41 .mn-1 +...+ Anna then it may be expressed as follows:

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din + + ... + Anchetean (1-ą) = 0. (126) Now replace y by y+ then (126) becomes

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