Mathematical Methods for Physics and Engineering : A Comprehensive Guide

14.2 FIRST-DEGREE FIRST-ORDER EQUATIONS

Solve

dy

dx

=(x+y+1)^2.

Making the substitutionv=x+y+ 1, we obtain, as in (14.23),

dv

dx

=v^2 +1,

which is separable and integrates directly to give
∫
dv
1+v^2

=

∫

dx ⇒ tan−^1 v=x+c 1.

So the solution to the original ODE is tan−^1 (x+y+1)=x+c 1 ,wherec 1 is a constant of
integration.

Solution method.In an equation such as (14.22), substitutev=ax+by+cto obtain

a separable equation that can be integrated directly. Then replacevbyax+by+c

to obtain the solution.

Secondly, we discuss

dy

dx

=

ax+by+c

ex+fy+g

, (14.24)

wherea,b,c,e,fandgare all constants. This equation may be solved by letting

x=X+αandy=Y+β,whereαandβare constants found from

aα+bβ+c= 0 (14.25)

eα+fβ+g=0. (14.26)

Then (14.24) can be written as

dY

dX

=

aX+bY

eX+fY

,

which is homogeneous and can be solved by the method of subsection 14.2.5.

Note, however, that ifa/e=b/fthen (14.25) and (14.26) are not independent

and so cannot be solved uniquely forαandβ. However, in this case, (14.24)

reduces to an equation of the form (14.22), which was discussed above.

Solve

dy

dx

=

2 x− 5 y+3

2 x+4y− 6

.

Letx=X+αandy=Y+β,whereαandβobey the relations

2 α− 5 β+3=0

2 α+4β−6=0,

which solve to giveα=β= 1. Making these substitutions we find

dY

dX

=

2 X− 5 Y

2 X+4Y

,