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Finding a homogeneous solution

  1. Apr 14, 2005 #1
    Q. Determine a homogeneous linear differential equation with constant coefficients having having the following solution:

    y = C1sin3x + C2cos3x

    My idea is to differntiate both sides with respect to x and come up with an equation in dy/dx

    what else? can be done......

    Is my idea correct.
     
  2. jcsd
  3. Apr 14, 2005 #2
    Hello,

    Since the solution is in the form y=ASin3x+BCos3x,
    The original equation must have complex roots 3i and -3i.
    Thus, a possible solution is d^2y/dx^2+9=0. =)
     
  4. Apr 14, 2005 #3

    Hurkyl

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    Not every differential equation is a first order equation!
     
  5. Apr 14, 2005 #4

    dextercioby

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    Pay attention.The characteristic equation is

    [tex] \lambda^{2}+9=0 [/tex]

    ,but the ODE is

    [tex] \frac{d^{2}y}{dx^2}+9y=0 [/tex]

    Okay?


    Daniel.
     
  6. Apr 16, 2005 #5
    Can somebody explain how they arrived at [tex] \lambda^{2}+9=0 [/tex]

    I know that the two roots are 3i and -3i. I had figured out this already.
     
  7. Apr 16, 2005 #6
    Suppose you were given [tex] \lambda^{2}+9=0 [/tex]

    How would factor it , in order to find the two values for [tex] \lambda
     
  8. Apr 16, 2005 #7

    dextercioby

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    By multiplying [itex] (\lambda-3i)(\lambda+3i) [/itex] and equating it to 0...?

    Daniel.
     
  9. Apr 16, 2005 #8
    Yup I got it thanks!!
     
  10. Apr 23, 2005 #9
    Thanks for pointing that out Daniel! I guess that I took the "y" there for granted every time I used the characteristic solution to get the [tex]y_h[/tex]
     
  11. Apr 23, 2005 #10

    HallsofIvy

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    By the way- this was clearly a simple problem because the given combination was clearly a solution to a linear equation with constant coefficients. It's not always that simple. In general, given a combination of functions with TWO "unknown constants", you form the simplest equation, involving differentials, the eliminates those constants.

    If you did NOT recognize y= C1cos(3x)+ C2sin(3x) as coming from λ= 3i and -3i, you could have done this:
    Since you are seeking a differential equation: DIFFERENTIATE-
    y'= -3 C1 sin(3x)+ 3 C2 cos(3x).
    Since there are two unknown constants, DIFFERENTIATE AGAIN-
    y"= -9 C1 cos(3x)- 9 C2 sin(3x).

    Now do whatever algebraic manipulations you need to eliminate the two constants.

    (In this example, of course, just add y" and 3y.)
     
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