Homogeneous Equations with Constant Coefficients Up until now, we have only worked on first order differential equations.  The next step is to investigate second order differential equations.  The general second order differential equation has the form         y''  =  f(t,y,y') The general solution to such an equation is very rough.  Instead, we will focus on special cases.  In particular, if the differential equation is linear, then it can be written in the form         P(t)y''  +  Q(t)y'  +  R(t)y  =  G(t) If P(t) is nonzero, then we can divide by P(t) to get         y''  +  p(t)y'  +  q(t)y  =  g(t) We call a second order linear differential equation homogeneous if g(t)  =  0. In this section we will be investigating homogeneous second order linear differential equations with constant coefficients.  They can be written in the form         ay'' + by' + cy  =  0   Example Solve         y'' + 3y' - 4y  =  0 Solution The strategy is to search for a solution of the form          y  =  ert  The reason for this is that long ago some geniuses figured this stuff out and it works. Now calculate derivatives         y'  =  rert     y''  =  r2ert Substituting into the differential equation gives         r2ert + 3(rert) - 4(ert)         =  (r2 + 3r - 4)ert  =  0 Now divide by ert to get         r2 + 3r - 4  =  0         (r - 1)(r + 4)  =  0         r  =  1        r  =  -4 We can conclude that two solutions are          y1  =  et    and    y2  =  e-4t  Now let          L(y)  =  y'' + 3y' - 4 It is easy to verify that if y1 and y2 are solutions to          L(y)  =  0 then          c1y1 + c2y2  is also a solution.  More specifically we can conclude that            y  =  c1et  + c2e-4t  Represents a two dimensional family (vector space) of solutions.   Later we will prove that this is the most general description of the solution space.   Example  Solve         y'' - y' - 6y  =  0    y(0)  =  1    y'(0)  =  2 As before we seek solutions of the form         y  =  ert  Now calculate derivatives         y'  =  rert     y''  =  r2ert Substituting into the differential equation gives         r2ert - (rert) - 6(ert)         =  (r2 - r - 6)ert  =  0 Now divide by ert to get         r2 - r - 6  =  0         (r - 3)(r + 2)  =  0 We can conclude that two solutions are          y1  =  e3t    and    y2  =  e-2t  We can conclude that            y  =  c1e3t  + c2e-2t  Represents a two dimensional family (vector space) of solutions.   Now use the initial conditions to find that         1  =  c1 + c2  We have that            y'  =  3c1e3t  - 2c2e-2t  Plugging in the initial condition with y', gives         2  =  3c1 - 2c2  This is a system of two equations and two unknowns.  We can use a matrix to arrive at          c1  =  4/5    and     c2  =  1/5 The final solution is                     y  =  4/5 e3t  + 1/5e-2t    In general for         ay'' + by' + cy  =  0 we call          ar2 + br + c  =  0 The characteristic equation for this differential equation.  Our examples demonstrated how to solve it if we have two distinct real roots.  For complex or repeated roots, a somewhat different strategy is needed.  We will discuss these other cases later on.  For real distinct roots we can use the quadratic formula an obtain a general solution                    y  =  c1er1t  + c2er2t  where                 and            Back to the Differential Equations Home Page