8.8.2. Case I: Real Distinct Roots

Example 1.

Find a basis of solutions for the differential equation

        ( D 3 7D+6 )y=0                  (8.12)

Solution

Eq(8.12) can be written as L( y )=0  with

        L= D 3 7D+6   =( D1 )( D2 )( D+3 )

Now,

        ( D1 ) u 1 =0   Þ  u 1 ( x )= e x

i.e., the null space of D-1 is N( D1 )={ e x } , where L denotes the linear span.  Similarly, N( D2 )={ e 2x }  and N( D+3 )={ e 3x } .  Since these functions are independent, the general solution for (8.12) is

        y= c 1 e x + c 2 e 2x + c 3 e 3x

Theorem 8.8.

Let L be a constant coefficient operator whose associated polynomial pL(r) has n distinct roots r 1 ,, r n .  Then the general solution of L( y )=0  on ( , )  is

        y= k=1 n c k e r k x                      (8.13)

Proof

By (8.10), we have

        L= a 0 k=1 n ( D r k )

Since

        ( D r k ) u k =0   Þ  u k ( x )= e r k x

so that the null space N( D r k )={ e r k x } .  Since the e r k x ’s are independent, we have

        N( L )=  { e r k x   ;  k=1,,n }                       QED.