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
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.