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  =  e^rt 

The reason for this is that long ago some geniuses figured this stuff out and it works.

Now calculate derivatives

        y'  =  re^rt     y''  =  r²e^rt

Substituting into the differential equation gives

        r²e^rt + 3(re^rt) - 4(e^rt)

        =  (r² + 3r - 4)e^rt  =  0

Now divide by e^rt to get

        r² + 3r - 4  =  0

        (r - 1)(r + 4)  =  0

        r  =  1        r  =  -4

We can conclude that two solutions are 

        y₁  =  e^t    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  =  c₁e^t  + 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  =  e^rt 

Now calculate derivatives

        y'  =  re^rt     y''  =  r²e^rt

Substituting into the differential equation gives

        r²e^rt - (re^rt) - 6(e^rt)

        =  (r² - r - 6)e^rt  =  0

Now divide by e^rt to get

        r² - r - 6  =  0

        (r - 3)(r + 2)  =  0

We can conclude that two solutions are 

        y₁  =  e^3t    and    y₂  =  e^-2t 

We can conclude that 

          y  =  c₁e^3t  + 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'  =  3c₁e^3t  - 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 e^3t  + 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  =  c₁e^r1t  + c₂e^r2t 

where

and

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