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This appears to be the solution of the equation, but it does not satisfy it unless C1 = 0, when it becomes

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which is only a particular integral, and therefore incomplete.

This arises from our implying in the use of equation (4) that nb, qb,- = 0 is generally true, whereas the equation

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shews that b, is not necessarily connected with b, since it may be satisfied by n = 1.

To complete the solution, we have from (2) which is always true

ao =

- bo,

and from (4) which is true for n = 1, we have

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These quantities are independent of a, a,, &c., therefore writing C, for a, as it is an arbitrary constant,

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is a particular integral of the proposed equation, and the complete solution is

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n (n − 1) ... (n − m + 2) { n − m (p + 1) + 1 } a„ + kTM ɑn-m = 0, from which

n (n − 1)... (n − m + 2) ... (n − m + 1) b, + kTM bn-m = 0.

But this is the equation which would result from substituting (b,x") in

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or

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y = C {(3 − k2x2) cos (ka + a) + 3kx sin (kx + a)}.

Ellis, Cam. Math. Jour. Vol. 11. p. 202.

CHAPTER VI.

PARTIAL DIFFERENTIAL EQUATIONS.

SECT. 1. Linear Equations with Constant Coefficients.

By the method of the separation of symbols the integration of Linear Partial Differential Equations is reduced to the same processes as those for the integration of ordinary differential equations of the same class. Hence the theory which is given in the beginning of Chap. IV. is equally applicable to the present subject, and it is unnecessary to repeat it here; I shall therefore content myself with referring to what has been previously said in the Chapter alluded to, adding that every differential equation of this class between two variables has an exact analogue among partial differential equations of the same class, and that the form of the solution of the latter is the same as that of the former. On this point one remark may be made which is of considerable importance in the interpretation of our results. As in the solution of ordinary differential equations we continually meet with expressions of the form

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so in partial differential equations we shall find expressions of the form

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in which the arbitrary function takes the place of the arbitrary constant. Now as the preceding formula is the symbolical expression for Taylor's Theorem, we know that

$(y) = $(y + ax).

Hence, in the solution of partial differential equations, arbitrary functions of binomials play the same parts as arbitrary constants multiplied by exponentials do in equations between two variables.

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Now supposing a to be the independent variable, and

d

dy

a constant, with respect to it, by the Theorem given in Ex. (11), Chap. xv. of the Diff. Calc. this is equivalent to

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or, effecting the integration, and adding an arbitrary function of y, instead of an arbitrary constant,

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or, as the form of is arbitrary, we may for

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It is obvious that if we had taken y for our independent

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The same method is applicable to any number of independent variables.

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If we expand the operating factor in ascending powers of

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the other terms being neglected, because when the operations

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