# ODE5.4

Multiply both sides by :

Observe that the numerator of the term on the right side
is the derivative (with respect to ) of the
denominator of that term (indeed, that is why the factor
was chosen -- a sort of integrating factor). Integrate both sides with respect
to ; the result is

(Alternatively, since the differential equation does not explicitly involve
, the substitution
converts the
equation to a first-order equation in and
which is separable. Integrating and back-substituting
for yields the same result as given above, as of course it must.)

Exponentiating both sides, we get

where (we are free to write since
for all real ).

Solving the above for ,

Divide both sides by and integrate:

(where we have used
in the integral on the left)

Use the trig substitution , to get

Integrating,

or after back-substitution,

Exponentiating both sides and
multiplying both sides by ,

Solving this for ,

Either case may be written in the form
where and are arbitrary constants, as may be verified by substituting this solution into the original differential equation.