First Order Linear Differential Equations   In this section we will concentrate on first order linear differential equations.  Recall that this means that only a first derivative appears in the differential equation and that the equation is linear.  The general first order linear differential equation has the form          y' + p(x)y  =  g(x) Before we come up with the general solution we will work out the specific example         y' + 2/x y  =  ln x The strategy for solving this is to realize that the left hand side looks a little like the product rule for differentiation.  The product rule is          (my)'  =  my' + m'y This leads us to multiplying both sides of the equation by m which is called an integrating factor.                 m y' + m 2/x y  = m ln x We now search for a m with          m'  =  m 2/x or         dm/m  =  2/x dx Integrating both sides, produces         ln m  =  2ln x  =  ln(x2)         m  =  x2     exponentiating both sides Going back to the original differential equation and multiplying both sides by x2, we get        x2y' + 2x y  =  x2 ln x Using the product rule in reverse gives         (x2y)'  =  x2 ln x Now integrate both sides.  Note that the integral of the derivative is the original.  Integrate by parts to get                                            u  =  ln x           dv  =  x2 dx                                                              du  =  1/x dx       v  =  1/3 x3  Hence         x2y  =  1/3 x3 ln x  -  1/9  x3  + C         y  =  1/3 x ln x  -  1/9  x  + C/x2    Divide by x2 Notice that when C is nonzero, the solutions are undefined at x  =  0.  Also given an initial value with x positive, there will be no solution for negative x.   Now we will derive the general solution to first order linear differential equations.  Consider         y' + p(t)y  =  g(t) We multiply both sides by m to get         my' + mp(x)y  =  mg(x) We now search for a m with          m'  =  mp(x) or         dm/m  =  p(x) dx Integrating both sides, produces         ln m  =          m  =      exponentiating both sides Going back to the original differential equation and multiplying both sides by m, we get         my' + mp(x)y  =  mg(x)         (my)'  =  mg(x)         my  =  Solving for y gives           Back to the Differential Equations Home Page