Mathematical Methods for Physics and Engineering - Matematica.NET

15.2 LINEAR EQUATIONS WITH VARIABLE COEFFICIENTS◮Solved 2 y+ y =cosecx. (15.50)dx2 We see that the RHS does not fall into any of the categories listed in subsection 15.1.2,and so we are at an initial loss as to how to find the particular integral. However, thecomplementary function of (15.50) isy c (x) =c 1 sin x + c 2 cos x,and so let us choose the solution u(x) =cosx (we could equally well choose sin x) andmake the substitution y(x) =v(x)u(x) =v(x)cosx into (15.50). This givescos x d2 v dv− 2sinx =cosecx, (15.51)dx2 dxwhich is a first-order linear ODE in dv/dx and may be solved by multiplying through bya suitable integrating factor, as discussed in subsection 14.2.4. Writing (15.51) asd 2 v dv− 2tanxdx2 dx = cosec xcos x , (15.52)we see that the required integrating factor is given by{ ∫ }exp −2 tan xdx =exp[2ln(cosx)] =cos 2 x.Multiplying both sides of (15.52) by the integrating factor cos 2 x we obtain(dcos 2 x dv )=cotx,dx dxwhich integrates to givecos 2 x dvdx =ln(sinx)+c 1.After rearranging and integrating again, this becomes∫∫v = sec 2 x ln(sin x) dx + c 1 sec 2 xdx=tanx ln(sin x) − x + c 1 tan x + c 2 .Therefore the general solution to (15.50) is given by y = uv = v cos x, i.e.y = c 1 sin x + c 2 cos x +sinx ln(sin x) − x cos x,which contains the full complementary function and the particular integral. ◭Solution method. If u(x) is a known solution of the nth-order equation (15.49) withf(x) =0, then make the substitution y(x) =u(x)v(x) in (15.49). This leads to anequation of order n − 1 in dv/dx, which might be soluble.507