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Finding a constant that makes the differential equation hold

  1. Feb 18, 2012 #1

    s3a

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    I get a = -49/4, 0, 49/4 by using the characteristic equation and quadratic formula and searching for values of a inside of the square root for which a is 0 or greater which isn't correct apparently. I've also wondered if the inside of the root should be less than 0 such that I can get a solution involving complex numbers using natural exponents which could lead me to the trigonometry involving reals but I don't have any concrete plans on getting this to work if this is what I am supposed to do.

    Any help would be greatly appreciated!
    Thanks in advance!

    Edit: Added the question.
     

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  3. Feb 18, 2012 #2

    HallsofIvy

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    I'm sorry but since you haven't said what the differential equation is, it's impossible to understand what you are doing.
     
  4. Feb 18, 2012 #3

    s3a

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    Oops! I just added the question!
     
  5. Feb 18, 2012 #4

    HallsofIvy

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    Solutions of such a second order, linear, differential equation can have three different forms:
    1) If the characteristic equation has a double real root, [itex]\alpha[/itex], then [itex]y= e^{\alpha x}(C_1+ C_2x)[/itex].

    2) If the characteristic equation has two real roots, [itex]\alpha[/itex] and [itex]\beta[/itex], [itex]y= C_1e^{\alpha x}+ C_2e^{\beta x}[/itex]

    3) If the characteristic equation has two complex roots, [itex]\alpha+ \beta i[/itex] and [itex]\alpha- \beta i[/itex], [itex]y= e^{\alpha x}(C_1 cos(\beta x)+ C_2 sin(\beta))[/itex]

    Now use the conditions y(0)= 0, y(7)= 0 to determine what those constants must be. What must a be in order that we are not forced to [itex]C_1= C_2= 0[/itex]?
     
    Last edited: Feb 18, 2012
  6. Feb 18, 2012 #5

    Ray Vickson

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    You need to show your work.

    RGV
     
  7. Feb 18, 2012 #6

    s3a

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    I attached my initial work.

    Also, HallsofIvy, I don't know how to proceed but I understand the constants should not have to be 0. Do the next steps involve trigonometry?
     

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  8. Feb 18, 2012 #7

    HallsofIvy

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    You don't give any reason for "choosing" a as you say.

    I pointed out that if the characteristic equation has a double root (49- 4a= 0) the solution must be of the form [itex]e^{\alpha x}(C_1+ C_2x)[/itex]. Now apply the boundary conditions: [itex]y(0)= e^0(C_1+ C_2(0))= C_1= 0[/itex] and [itex]y(7)= e^{7\alpha}(C_2(7))= 0[/itex] which gives [itex]C_2= 0[/itex]. So if a= 49/4, both [itex]C_1[/itex] and [itex]C_2[/itex] must be 0 which means y is never non-zero. a= 49/4 is NOT an answer.

    Now, try the other two cases.
     
  9. Feb 18, 2012 #8

    s3a

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    My current reasoning is to find a value to substitute for x which involves an a such that the whole thing is equal to 0 since x = 0 and x = 7 satisfy the boundary conditions but that has also failed. I attached my latest work.
     

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