![\begin{displaymath}
{dy \over dx} + p(x)y = q(x)y^n.
\end{displaymath}](b_450.gif) |
(1) |
Let
for
, then
![\begin{displaymath}
{dv\over dx} = (1-n)y^{-n}{dy\over dx}.
\end{displaymath}](b_453.gif) |
(2) |
Rewriting (1) gives
![\begin{displaymath}
y^{-n}{dy \over dx}= q(x)-p(x)y^{1-n} = q(x)-vp(x).
\end{displaymath}](b_454.gif) |
(3) |
Plugging (3) into (2),
![\begin{displaymath}
{dv\over dx} = (1-n)[q(x)-vp(x)].
\end{displaymath}](b_455.gif) |
(4) |
Now, this is a linear First-Order Ordinary Differential Equation
of the form
![\begin{displaymath}
{dv\over dx}+vP(x)=Q(x),
\end{displaymath}](b_456.gif) |
(5) |
where
and
. It can therefore be solved analytically using an
Integrating Factor
where
is a constant of integration. If
, then equation (1) becomes
![\begin{displaymath}
{dy \over dx}=y(q-p)
\end{displaymath}](b_462.gif) |
(7) |
![\begin{displaymath}
{dy\over y}=(q-p)\,dx
\end{displaymath}](b_463.gif) |
(8) |
![\begin{displaymath}
y=C_2 e^{\int [q(x)-p(x)]\,dx}.
\end{displaymath}](b_464.gif) |
(9) |
The general solution is then, with
and
constants,
![\begin{displaymath}
y = \cases{
\left[{{{(1-n)\int e^{(1-n)\int p(x)\,dx} q(x)\...
...or $n\not=1$\cr
C_2 e^{\int [q(x)-p(x)]\,dx} & for $n=1$.\cr}
\end{displaymath}](b_465.gif) |
(10) |
© 1996-9 Eric W. Weisstein
1999-05-26