y1 dz+ ( 2 dy1Py1) z = R.
dx dx

From this equation z can be found by the rule given above for the linear equation of the first order, and will involve one arbitrary constant; thence y = y1 η = y1 ∫ zdx + Ay1, where A is another arbitrary constant, will be the general solution of the original equation, and, as was to be expected, involves two arbitrary constants.

The case of most frequent occurrence is that in which the coefficients P, Q are constants; we consider this case in some detail. If θ be a root of the quadratic equation θ² + θP + Q = 0, it can be at once seen that a particular integral of the differential equation with zero on the right side is y1 = eθx. Supposing first the roots of the quadratic equation to be different, and φ to be the other root, so that φ + θ = -P, the auxiliary differential equation for z, referred to above, becomes dz/dx + (θ − φ)z = Re-θx which leads to ze(θ-φ) = B + ∫ Re-θxdx, where B is an arbitrary constant, and hence to

y = Aeθ x + eθ x ∫ Be(φ-θ) x dx + eθ x ∫ e(φ-θ) x ∫ Re-θ x dxdx,

or say to y = Aeθ x + Ceθ x + U, where A, C are arbitrary constants and U is a function of x, not present at all when R = 0. If the quadratic equation θ² + Pθ + Q = 0 has equal roots, so that 2θ = -P, the auxiliary equation in z becomes dz/dx = Reθ x giving z = B + ∫ Reθ x dx, where B is an arbitrary constant, and hence

y = (A + Bx)eθ x + eθ x ∫ ∫ Re-θ x dxdx,

or, say, y = (A + Bx)eθ x + U, where A, B are arbitrary constants, and U is a function of x not present at all when R = 0. The portion Aeθ x + Beθ x or (A + Bx)eθ x of the solution, which is known as the complementary function, can clearly be written down at once by inspection of the given differential equation. The remaining portion U may, by taking the constants in the complementary function properly, be replaced by any particular solution whatever of the differential equation

d²v + P dy+ Qy = R;
dx² dx

for if u be any particular solution, this has a form

u = A0 eθ x + B0 eφ x + U,