1000 Solved Problems in Modern Physics

20 1 Mathematical Physics

Procedure:

First step: Replace the RHS member of the given equation (I) by zero and solve
the complimentary function ofIto gety=u.
Second step: Differentiate successively the given equation (I) and obtain, either
directly or by elimination, a differential equation of a higher order of type I.
Third step: Solving this new equation by the previous rule we get its complete
solution

y=u+v

where the partuis the complimentary function of (I) already found in the first step,
andvis the sum of additional terms found
Fourth step: To find the values of the constants of integration in the particular
solutionv, substitute

y=u+v

and its derivatives in the equation (I). In the resulting identity equation equate the
coefficients of like terms, solve for constants of integration, substitute their values
back in

y=u+v

giving the complete solution of (I).

Type III

dny

dxn

=X

whereXis a function ofxalone, or constant
Integratentimes successively. Each integration will introduce one arbitrary
constant.

Type IV

d^2 y

dx^2

=Y

whereYis a function ofyalone
Multiply the LHS member by the factor 2ddyxdxand the RHS member bykequiv-
alent factor 2dy

2

dy

dx

d^2 y

dx^2

dx=d

(

dy

dx

) 2

= 2 Ydy

∫

d

(

dy

dx

) 2

=

(

dy

dx

) 2

=

∫

2 Ydy+C 1