let y'' +p(x)y' +q(x)y =0 (1) and y′′ +p(x)y′ +q(x)y =g(x)≠0 (2) True or false: If y 1 and y 2 solve(2), then y= y 1 -y 2 solves(1); If y 1 is a solution of (2), 2y 1 is also a solution of (2); If y 1...

let

              y'' +p(x)y' +q(x)y =0                        (1)

and       y′′ +p(x)y′ +q(x)y =g(x)≠0               (2)

1. 
   

   If y₁
   and y₂
   solve(2), then y= y₁-y₂ solves(1);

   
   


2. 
   

   If y₁
   is a solution of (2), 2y₁
   is also a solution of (2);

   
   


3. 
   

   If y₁
   and y₂
   are linearly dependent solutions of (1), then y = c₁y₁
   + c₂y₂ solves (1);

   
   


4. 
   

   Since y₁(x) ≡ 0 is a solution of (1), using reduction of order we can always find the general solution of (1);