an equation whose complete solution is y = (A+Bx) cos nx + (C+ Dx) sin nx. Substituting this in the given equation we find B = 0, 1 D= whence 2n' which agrees with the previous solution. The latter method, which is general, consists in forming a new equation of a higher order, but with its second member free from that term which is the cause of failure. As by the elevation of the order of the equation superfluous constants are introduced, the relations which connect them must be found by substitution of the result in the given equation. 12. To the class of linear equations with constant coefficients all equations of the form (a+bx)nd" y d+A (a+bx)*-10"-1+B(a+bæ)n-2 dn~??! dx" +Ly=X, A, B,... L being constant and X a function of x, may be reduced. It suffices to change the independent variable by assuming a + bx = e*. Ex. Given (a+b)2+b (a+bx) dy+n2y = 0. Assuming a + bx = e', we find Hence, by substitution in the given equation, we have in which it only remains to substitute for tits value log (a+bx). 13. Beside the properties upon which the above methods are founded, linear equations possess many others, of which we shall notice the most important. We suppose, as before, y to be the dependent, a the independent variable. 1st. The complete value of y when the linear equation has a second member X will be found by adding to any particular value of y that complementary function which would express its complete value were the second member 0. 6.4.192 be Representing the linear equation in the form (7), Tet the particular value of y which satisfies it, Y the complete value which would satisfy it were the second member 0; and assume y = y1+ Y. The equation then becomes and this becomes an identity, the first line of its left-hand member being by hypothesis equal to X, and the second line equal to 0. Ex. Thus a particular integral of the equation The above property, which relates to the generalizing of a particular solution, is important, because, as we shall hereafter see, a particular solution of a linear equation may often be obtained by a symbolical process which does not involve even the labour of an integration. 2ndly. The order of a linear differential equation may always be depressed by unity if we know a particular value of y which would satisfy the equation were its second member equal to 0. It will suffice to demonstrate this property for the equation of the second order Let y, be a particular value of y when X=0, and assume y=y,v. Substituting, we have the first line of which is by hypothesis 0. In the reduced a linear equation of the first order for determining u. And this being found, we have In the particular case in which X=0, we find from (19) 3rdly. Linear equations are connected by remarkable analogies with ordinary algebraic equations. This subject has been investigated chiefly by Libri and Liouville, who have shewn that most of the characteristic properties of algebraic equations have their analogues in linear differential equations. Thus an algebraic equation can be deprived of its 2nd, 3rd,...th term by the solution of an algebraic equation of the 1st, 2nd,...(r-1)th degree. A linear differential equation can be deprived of its 2nd, 3rd,... th term by the solution of another linear differential equation of the 1st, 2nd,....... (r — 1)th order. This may be proved by assuming y=vy,, and properly determining v so as to make in the resulting equation y, assume the required form. Again, as from two simultaneous algebraic equations, we can by the process for greatest common measure obtain a depressed equation satisfied only by their common roots, so from two simultaneous linear differential equations we can by a formally equivalent process deduce a new equation of a depressed order satisfied only by their common integrals. This is best illustrated by example. Ex. Required the common integrals, if any, of the equations Differentiating the second equation and then eliminating da, we find the depressed equation and dy If we differentiate this we shall find that the result is merely an algebraic consequence of the two equations last written, not an algebraically new equation. Thus the process of reduction cannot be repeated. We have therefore y= Ce as the only common integral. [See the Supplementary Volume, Chapter XXII.] 5. dic dy 3 da2 + y = 0. V day dx3 d+4y=0, it being given that one of the roots of the auxiliary equation, m3 — 3m2 + 4 = 0, is — 1. 8. What form does the solution of the above equation assume when k = 1? dy ✓ =3y. dx t |