This expression added to the complementary function found above gives the complete integral; and no generality is lost by omitting from the particular integral the terms −15⁄32x cos x + 9⁄32 sin x, which are of the types of terms already occurring in the complementary function.
The symbolical method which has been explained has wider applications than that to which we have, for simplicity of explanation, restricted it. For example, if ψ(x) be any function of x, and a1, a2, ... an be different constants, and [(t + a1) (t + a2) ... (t + an)]−1 when expressed in partial fractions be written Σcm(t + am)−1, a particular integral of the differential equation (D + a1)(D + a2) ... (D + an)y = ψ(x) is given by
y = Σcm(D + am)−1 ψ(x) = Σcm (D + am)−1 e−amxeamx ψ(x) = Σcme−amxD−1 (eamxψ(x) ) = Σcme−amx ∫ eamxψ(x)dx.
The particular integral is thus expressed as a sum of n integrals. A linear differential equation of which the left side has the form
| xn | dny | + P1xn−1 | dn−1y | + ... + Pn−1x | dy | + Pny, |
| dxn | dxn−1 | dx |
where P1, ... Pn are constants, can be reduced to the case considered above. Writing x = et we have the identity
| xm | dmu | = θ(θ − 1)(θ − 2) ... (θ − m + 1)u, where θ = d/dt. |
| dxm |
When the linear differential equation, which we take to be of the second order, has variable coefficients, though there is no general rule for obtaining a solution in finite terms, there are some results which it is of advantage to have in mind. We have seen that if one solution of the equation obtained by putting the right side zero, say y1, be known, the equation can be solved. If y2 be another solution of
| d²y | + P | dy | + Qy = 0, |
| dx² | dx |
there being no relation of the form my1 + ny2 = k, where m, n, k are constants, it is easy to see that