Suppose y ( x ) = x 2 is a solution of y " + P ( x ) y ′ + Q ( x ) y = 0 on (0,1) where P and Q are continuous functions on (0,1). Can both P and Q be bounded functions? Justify your answer.I have tried differentiating y = x 2 , y ′ = 2 x and y ′ / x = 2. Differentiating this once again yielded y " − y ′ / x = 0. So I don't think P ( x ) can be bounded in (0,1). But Iam not so sure about whether this is correct or not. So it would be great if anyone has some ideas to share.

Question

Suppose   y  (  x  )  =      x    2   is a solution of   y  &quot;  +  P  (  x  )      y    ′    +  Q  (  x  )  y  =  0 on (0,1) where   P and   Q are continuous functions on (0,1). Can both   P and   Q be bounded functions? Justify your answer.I have tried differentiating   y  =      x    2  ,       y    ′    =  2  x and       y    ′        /    x  =  2. Differentiating this once again yielded   y  &quot;  −      y    ′        /    x  =  0. So I don&#039;t think   P  (  x  ) can be bounded in (0,1). But Iam not so sure about whether this is correct or not. So it would be great if anyone has some ideas to share.

celatw0kej · Accepted Answer

The question you ask can be rewritten as : can one find   P  ,  Q bounded continuous functions such that :  2  +  2  x  P  (  x  )  +      x    2    Q  (  x  )  =  0for   x  ∈  (  0  ,  1  ).Letting   x  →  0 and assuming by contradiction that   P and   Q are bounded, you find   2  =  0, so it cannot hold.The reasoning you were doing had the right idea but in this kind of problem (proving something cannot exist) a useful method is to prove it by contradiction (indeed assuming something exists usually gives a lot of possibilities).

Suppose y ( x ) = x 2 </msup> is a solution of y "

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