## 8.10. The Relation Between The Homogeneous And Nonhomogeneous Equations

### Theorem 8.10.

Let L: C ( n ) ( J )C( J )  be a linear differential operator of order n.  Let u 1 ,, u n  be n independent solution of the homogeneous equation L( y )=0 .  Let y1 be a particular solution of the nonhomogeneous equation L( y )=R , where RC( J ) .  Then the general solution for L( y )=R  is

y=f( x )   = y 1 ( x )+ k=1 n c k u k ( x )                       (8.16)

where c k  are constants.

#### Proof

Let f be any solution of L( y )=R .  Since L is linear, we have

L( f y 1 )=L( f )L( y 1 )   =RR=0

Hence, ( f y 1 )N( L )  so that f y 1 = k=1 n c k u k .  QED.

### Comment

Thus, Theorem 8.10 breaks down the problem of solving a nonhomogeneous equation into 2 parts.  One is to find a particular solution, the other is to solve the homogeneous equation.  This has a simple geometric analogy: to find all points on a plane, we find a particular point on the plane, then add to it all points on the parallel plane that goes through the origin.