Mathematical Methods for Physics and Engineering : A Comprehensive Guide

15.2 LINEAR EQUATIONS WITH VARIABLE COEFFICIENTS

Solve

d^2 y

dx^2

+y=cosecx. (15.50)

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, the
complementary function of (15.50) is

yc(x)=c 1 sinx+c 2 cosx,

and so let us choose the solutionu(x)=cosx(we could equally well choose sinx)and
make the substitutiony(x)=v(x)u(x)=v(x)cosxinto (15.50). This gives

cosx

d^2 v

dx^2

−2sinx

dv

dx

=cosecx, (15.51)

which is a first-order linear ODE indv/dxand may be solved by multiplying through by
a suitable integrating factor, as discussed in subsection 14.2.4. Writing (15.51) as

d^2 v

dx^2

−2tanx

dv

dx

=

cosecx

cosx

, (15.52)

we see that the required integrating factor is given by

exp

{

− 2

∫

tanxdx

}

=exp[2ln(cosx)]=cos^2 x.

Multiplying both sides of (15.52) by the integrating factor cos^2 xwe obtain

d

dx

(

cos^2 x

dv

dx

)

=cotx,

which integrates to give

cos^2 x

dv

dx

=ln(sinx)+c 1.

After rearranging and integrating again, this becomes

v=

∫

sec^2 xln(sinx)dx+c 1

∫

sec^2 xdx

=tanxln(sinx)−x+c 1 tanx+c 2.

Therefore the general solution to (15.50) is given byy=uv=vcosx,i.e.

y=c 1 sinx+c 2 cosx+sinxln(sinx)−xcosx,

which contains the full complementary function and the particular integral.

Solution method.Ifu(x)is a known solution of thenth-order equation (15.49) with

f(x)=0, then make the substitutiony(x)=u(x)v(x)in (15.49). This leads to an

equation of ordern− 1 indv/dx, which might be soluble.