Linear Differential Equations of Order n: I

Oppenheim is telling stories about linear differential equations; I review the material necessary to right myself.
Following  Tenenbaum and Pollard,

■ A linear differential equation of order n is an equation which can be written in the form
,
where f₀(x), f₁(x),  . . .  , f_n(x) and Q(x) are continuous functions of x defined on a common interval I, and f_n(x) is nonzero somewhere in I.   Note that y^(k) is the kth derivative of y.  For example,  y''' = y⁽³⁾.

Definition: Let the functions f₁(x), f₂(x),   . . .  , f_n(x), be defined on a common interval I. Then the functions are linearly dependent  if there exist  constants c₁, c₂, . . . , c_n, not all zero, such that

for every x in I.
     The functions are linearly independent if no such set of constants exists.


■ Theorem 19.3:  If f₀(x), f₁(x), . . . ,  f_n(x) and Q(x) are continuous functions of x on a common interval I and f_n(x) ≠ 0 when  x is in I, then

 1. The homogeneous linear differential equation

 
has n linearly independent solutions y₁(x), y₂(x), . . . , y_n(x)

2. The linear combination of these n solutions

,
c₁, c₂, . . . , c_n arbitrary constants,  is an n-parameter family of solutions of (a).

3. The function

where y_p is a particular solution of the nonhomogeneous equation (with Q(x)  ≠ 0),  is an n-parameter family of solutions of (18.11).

(Chapter 19, pg. 211:)
"It is extremely important that you prove the statements in Exercises 5 to 7 below."

■  5. If y_p is a solution of
     ,
then Ay_p is a solution of (19.5) with Q(x) replaced by AQ(x).

Proof:
If y_p is a solution of (19.5), we have
       
Mulitply through by A
       
And since
       
With the A's pulled into the derivatives, we have
       
And Ay_p is a solution of (19.5) with Q(x) replaced by AQ(x).


■ 6. Principle of Superposition.  If y_p1 is a solution of (19.5) with Q(x) replaced by Q₁(x) and y_p2 is a solution of (19.5) with Q(x) replaced by Q₂(x), then y_p = y_p1+ y_p2 is a solution of
          .
Proof:
Add the two equations in y_p1 and y_p2:

And since differentiation distributes:  (u' + v') = (u + v)',

After the substitution y_p = [y_p1+ y_p2],
       
Done.

  ■7. If y_p(x) =  u(x) + iv(x) is a solution of
      ,
where f₀(x), . . . , f_n(x) are real functions of x, then
     (a) the real part of y_p, i.e. u(x), is a solution of
               ,
     (b) the imaginary part of y_p, i.e. v(x), is a solution of
               .
Proof:
Writing y_p(x) as u + iv,

differentiate the terms separately,

and collect real and imaginary parts on the left-hand side:

Two complex quantities are equal iff their real parts are equal and their complex parts are equal.  That is,
          
AND THE RIGHTEOUSNESS IS COMPLETE.

                             Next:  The Homogeneous Case   →
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Notes: To make the material easier for me to find, I use the numbering from Tenenbaum and Pollard's Ordinary Differential Equations, Dover Press.  If you find my personal math blog useful, I am happy to change the numbering and layout for easier navigation.  Let me know.

Thanks as always to CodeCogs for the Latex Equation Editor.