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Easy Differential Equation Calc 1

  1. Oct 1, 2008 #1
    1. The problem statement, all variables and given/known data
    Is y = x^2 a solution to the DE [tex]y'' - 4xy' + 4y = 0[/tex]


    3. The attempt at a solution
    So:
    [tex]y = x^2[/tex]
    [tex]y' = 2x[/tex]
    [tex]y'' = 2[/tex]

    [tex] 2 - 4x(2x) + 4x^2 = 0[/tex]
    [tex] 2 - 8x^2 + 4x^2 = 0[/tex]
    [tex] 2 - 4x^2 = 0[/tex]

    But doing this I won't get an answer of whether it is a solution or not. I get a variable equal to a number. So what can I do differently? How do I solve this? Thanks in advance.
     
    Last edited: Oct 1, 2008
  2. jcsd
  3. Oct 1, 2008 #2

    HallsofIvy

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    Which means that this must be true for all x.

    Suppose you were asked whether x= 2 is a solution to x3- 3x2[/sup+ 10x+ 4= 0. Putting 2 in place of x gives you 23- 3(22)+ 10(2)+ 4= 8- 12+ 20+ 4= 20, not 0. Since the equation is not true, 2 is NOT a solution to the problem.


    Is that true for all x? If yes, then y= x2 is a solution to the differential equation. If not, then it is not a solution!

     
  4. Oct 1, 2008 #3
    So for any x, y = x^2 is a solution to that DE? And that is because I got a variable equal to a number? And just for a reference the answer in the back of the book said that it wasn't a solution to that DE.
     
  5. Oct 1, 2008 #4

    gabbagabbahey

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    Is [itex]2-4x^2[/itex] zero for all x?!
     
  6. Oct 1, 2008 #5
    Sorry the post I responded to before didn't load fully and I only saw a small part. Which I was equally confused about.
     
  7. Oct 1, 2008 #6
    So let me get this straight. Since [tex]2 - 4x^2 = 0[/tex] is not true for all x then, [tex] y= x^2[/tex] is not a solution to that DE. If however, that equation had been true for all x it would be a solution?
     
  8. Oct 1, 2008 #7

    gabbagabbahey

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    Yes, by plugging in [itex] y=x^2[/itex] to the left side of the DE, you showed that:

    [tex]y'' - 4xy' + 4y=2-4x^2 [/tex]

    which does not equal the right side of the DE (zero) for all [itex]x[/itex], so [itex] y=x^2[/itex] does not satisfy the DE.
     
  9. Oct 1, 2008 #8
    Furthermore, if they had said [tex]x = \sqrt{1/2}[/tex] a solution to the DE the answer would be yes, since that yields a result of 0 = 0?
     
  10. Oct 1, 2008 #9

    gabbagabbahey

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    Not really, for a solution to satisfy the DE it must satisfy it for all values of [itex]x[/itex].
     
  11. Oct 1, 2008 #10
    with x at that value, and y the same as above, then y would be a solution to that DE
     
  12. Oct 1, 2008 #11

    gabbagabbahey

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    It's true that [itex]2-4x^2=0[/itex] for [itex]x=1/\sqrt{2}[/itex], but [itex]y(1/\sqrt{2})=1/2[/itex] which is a constant, so dy/dx and d^2y/dx^2 are both zero and your left with 4*(1/2)=0 which is false. It is meaningless to say that y(a certain value) satisfies the DE, since y(a certain value) is a value, not a function. Only functions satisfy DE's.
     
  13. Oct 1, 2008 #12
    Forget that y part, I didn't mean to put that. If I said is [tex]y = x^2 [/tex] at [tex]x = 1/\sqrt{2}[/tex] a solution to that DE what would the answer be?
     
  14. Oct 1, 2008 #13

    gabbagabbahey

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    Still no; [itex]y=x^2[/itex] at [itex]x=1/\sqrt{2}[/itex] is just a value not a function. Functions satisfy DE's, values don't.
     
  15. Oct 1, 2008 #14
    Alright thanks
     
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