14.3 HIGHER-DEGREE FIRST-ORDER EQUATIONS
whereFi=Fi(x, y). We are then left with solving thenfirst-degree equations
p=Fi(x, y). Writing the solutions to these first-degree equations asGi(x, y)=0,
the general solution to (14.28) is given by the product
G 1 (x, y)G 2 (x, y)···Gn(x, y)=0. (14.29)
Solve
(x^3 +x^2 +x+1)p^2 −(3x^2 +2x+1)yp+2xy^2 =0. (14.30)
This equation may be factorised to give
[(x+1)p−y][(x^2 +1)p− 2 xy]=0.
Taking each bracket in turn we have
(x+1)
dy
dx
−y=0,
(x^2 +1)
dy
dx
− 2 xy=0,
which have the solutionsy−c(x+1) = 0 andy−c(x^2 + 1) = 0 respectively (see
section 14.2 on first-degree first-order equations). Note that the arbitrary constants in
these two solutions can be taken to be the same, since only one is required for a first-order
equation. The general solution to (14.30) is then given by
[y−c(x+1)]
[
y−c(x^2 +1)
]
=0.
Solution method.If the equation can be factorised into the form (14.28) then solve
the first-order ODEp−Fi=0for each factor and write the solution in the form
Gi(x, y)=0. The solution to the original equation is then given by the product
(14.29).
14.3.2 Equations soluble forx
Equations that can be solved forx, i.e. such that they may be written in the form
x=F(y, p), (14.31)
can be reduced to first-degree first-order equations inpby differentiating both
sides with respect toy,sothat
dx
dy
=
1
p
=
∂F
∂y
+
∂F
∂p
dp
dy
.
This results in an equation of the formG(y, p) = 0, which can be used together
with (14.31) to eliminatepand give the general solution. Note that often a singular
solution to the equation will be found at the same time (see the introduction to
this chapter).