14.2 FIRST-DEGREE FIRST-ORDER EQUATIONS
which may then be integrated directly to give
μ(x)y=
∫
μ(x)Q(x)dx. (14.16)
The required integrating factorμ(x) is determined by the first equality in (14.15),
i.e.
d
dx
(μy)=μ
dy
dx
+
dμ
dx
y=μ
dy
dx
+μP y,
which immediately gives the simple relation
dμ
dx
=μ(x)P(x) ⇒ μ(x)=exp
{∫
P(x)dx
}
. (14.17)
Solve
dy
dx
+2xy=4x.
The integrating factor is given immediately by
μ(x)=exp
{∫
2 xdx
}
=expx^2.
Multiplying through the ODE byμ(x)=expx^2 and integrating, we have
yexpx^2 =4
∫
xexpx^2 dx=2expx^2 +c.
The solution to the ODE is therefore given byy=2+cexp(−x^2 ).
Solution method.Rearrange the equation into the form (14.14) and multiply by the
integrating factorμ(x)given by (14.17). The left- and right-hand sides can then be
integrated directly, giving y from (14.16).
14.2.5 Homogeneous equations
Homogeneous equation are ODEs that may be written in the form
dy
dx
=
A(x, y)
B(x, y)
=F
(y
x
)
, (14.18)
whereA(x, y)andB(x, y) are homogeneous functions of the same degree. A
functionf(x, y) is homogeneous of degreenif, for anyλ,itobeys
f(λx, λy)=λnf(x, y).
For example, ifA=x^2 y−xy^2 andB=x^3 +y^3 then we see thatAandBare
both homogeneous functions of degree 3. In general, for functions of the form of
AandB, we see that for both to be homogeneous, and of the same degree, we
require the sum of the powers inxandyin each term ofAandBto be the same