LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
                                                      JAMESKEESLING
                   In this post we give solution of the most general first-order ordinary differential equation.
                 This equation has the form:
                 (1)                                  dx +p(t)x = g(t).
                                                      dt
                 Wesolve the equation by finding an integrating factor in much the same way that we did
                 to produce exact differential equations from equations that were not exact.
                   Consider the following expression.
                                                                                       
                 (2)            d   exp Z p(t)dt ·x =exp Z p(t)dt · dx +p(t)·x
                                dt                                               dt
                   Wecan put (1) and (2) together in the following way.
                                                                                           
                    d   exp Z p(t)dt ·x =exp Z p(t)dt · dx +p(t)x =exp Z p(t)dt ·g(t)
                    dt                                               dt
                   Wecan now solve.
                                      Z                Z       Z               
                                  exp      p(t)dt  · x =      exp     p(t)dt  · g(t)  dt +C
                                         Z           Z         Z                       
                               x=exp − p(t)dt ·                exp     p(t)dt   · g(t) dt +C
                                                                    R       

                   The basic trick is to note that the function exp    p(t)dt is an integrating factor for the
                 left-hand side of (1).
                                                         1. Example
                   Consider the differential equation.
                 (3)                                   dx + 1 x = sin(t)
                                                   R dt 
 t
                   Our integrating factor is exp      1 dt  = exp(ln(t)) = t. So, our differential equation
                                                      t
                 becomes
                                                               1
              2                      JAMESKEESLING
                                    d (t · x) = t · sin(t)
                                    dt
              and the solutiion is
                           x(t) = 1 Z t sin(t)dt + C = sin(t) −cos(t)+ C.
                               t         t    t        t