Mathematical Methods for Physics and Engineering : A Comprehensive Guide

14.2 FIRST-DEGREE FIRST-ORDER EQUATIONS

Solve

dy

dx

=

− 1

2 yx

(

y^2 +

2

x

)

.

Rearranging we have
(
y^2 +

2

x

)

dx+2yx dy=0.

Givingyanddythe weightmandxanddxthe weight 1, the sums of the powers in each
term on the LHS are 2m+1, 0 and 2m+ 1 respectively. These are equal if 2m+1=0, i.e.
ifm=−^12. Substitutingy=vxm=vx−^1 /^2 , with the result thatdy=x−^1 /^2 dv−^12 vx−^3 /^2 dx,
we obtain

vdv+

dx

x

=0,

which is separable and may be integrated directly to give^12 v^2 +lnx=c.Replacingvby
y

√

xwe obtain the solution^12 y^2 x+lnx=c.

Solution method.Write the equation in the formAdx+Bdy=0. Givingyand

dyeach a weightmandxanddxeach a weight 1 , write down the sum of powers

in each term. Then, if a value ofmthat makes all these sums equal can be found,

substitutey=vxminto the original equation to make it separable. Integrate the

separated equation directly, and then replacevbyyx−mto obtain the solution.

14.2.7 Bernoulli’s equation

Bernoulli’s equation has the form

dy

dx

+P(x)y=Q(x)yn wheren=0or1. (14.21)

This equation is very similar in form to the linear equation (14.14), but is in fact

non-linear due to the extraynfactor on the RHS. However, the equation can be

made linear by substitutingv=y^1 −nand correspondingly

dy

dx

=

(

yn

1 −n

)

dv

dx

.

Substituting this into (14.21) and dividing through byyn, we find

dv

dx

+(1−n)P(x)v=(1−n)Q(x),

which is a linear equation and may be solved by the method described in

subsection 14.2.4.