Mathematical Methods for Physics and Engineering : A Comprehensive Guide

15.2 LINEAR EQUATIONS WITH VARIABLE COEFFICIENTS

Using table 13.1,

q 1 (t)=^12 V 0 C(cosω 1 t−cosω 2 t),

whereω^21 (L+M)=Gandω 22 (L−M)=G. Thus the current is given by

i 1 (t)=^12 V 0 C(ω 2 sinω 2 t−ω 1 sinω 1 t).

Solution method.Perform a Laplace transform, as defined in (15.31), on the entire

equation, using (15.32) to calculate the transform of the derivatives. Then solve the

resulting algebraic equation for ̄y(s), the Laplace transform of the required solution

to the ODE. By using the method of partial fractions and consulting a table of

Laplace transforms of standard functions, calculate the inverse Laplace transform.

The resulting functiony(x)is the solution of the ODE that obeys the given boundary

conditions.

15.2 Linear equations with variable coefficients

There is no generally applicable method of solving equations with coefficients

that are functions ofx. Nevertheless, there are certain cases in which a solution is

possible. Some of the methods discussed in this section are also useful in finding

the general solution or particular integral for equations with constant coefficients

that have proved impenetrable by the techniques discussed above.

15.2.1 The Legendre and Euler linear equations

Legendre’s linear equation has the form

an(αx+β)n

dny

dxn

+···+a 1 (αx+β)

dy

dx

+a 0 y=f(x), (15.36)

whereα,βand theanare constants and may be solved by making the substitution

αx+β=et. We then have

dy

dx

=

dt

dx

dy

dt

=

α

αx+β

dy

dt

d^2 y

dx^2

=

d

dx

dy

dx

=

α^2

(αx+β)^2

(

d^2 y

dt^2

−

dy

dt

)

and so on for higher derivatives. Therefore we can write the terms of (15.36) as

(αx+β)

dy

dx

=α

dy

dt

,

(αx+β)^2

d^2 y

dx^2

=α^2

d

dt

(

d

dt

− 1

)

y,

..

.

(αx+β)n

dny

dxn

=αn

d

dt

(

d

dt

− 1

)

···

(

d

dt

−n+1

)

y.

(15.37)