Mathematical Methods for Physicists: A concise introduction - Site Map

ORDINARY DIFFERENTIAL EQUATIONS
The right hand side is already integrable; the condition that the left hand side of
Eq. (2.12) be exact gives
@
‰Rf …x†yŠ ˆ@R
@y @x ;
which yields
and integrating gives
dR=dx ˆ Rf …x†; or dR=R ˆ f …x†dx;
Z
ln R ˆ f …x†dx
from which we obtain the integrating factor R we were looking for
Z
R ˆ exp f …x†dx :
It is now possible to write the general solution of Eq. (2.12). On applying the
integrating factor, Eq. (2.12) becomes
d…ye F †
ˆ g…x†e F ;
dx
where F…x† ˆR f …x†dx. The solution is clearly given by
Z
y ˆ e F e F g…x†dx ‡ C :
Example 2.7
Show that the equation xdy=dx ‡ 2y ‡ x 2 ˆ 0 is not exact; then ®nd a suitable
integrating factor that makes the equation exact. What is the solution of this
equation?
Solution: We ®rst write the equation in the standard form
…2y ‡ x 2 †dx ‡ xdy ˆ 0;
then we notice that
@
@y …2y ‡ x2 †ˆ2
and
@
@x x ˆ 1;
which indicates that our equation is not exact. To ®nd the required integrating
factor that makes our equation exact, we rewrite our equation in the form of Eq.
(2.12):
dy
dx ‡ 2y
x ˆx
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