using you knowledge of derivatives (particularly setting y=e^kt) solve the following equation:

y'' + 7y' +3y=0

3 Answers

  • 1 decade ago
    Favorite Answer

    the general equation is y(t)= Ae^(-7+sqr root37/2) + Be(-7-sqr root37/2). Where A and B are some arbitrary constants.

  • 1 decade ago

    I would just add a little further explanation. If you start with

    y = constant*e^kt then as shown above you get two values of k. Call these k1 and k2. This means that y = constant*e^k1t and

    y = constant*e^k2t are both possible solutions of the differential equation and the constants don't have to be the same. It can further be shown that these two added together are also a solution to the differential equation so the most general answer

    is y = Ae^k1t + Be^k2t as given by the first person. (By the way, I think that helping a person find the answer themselves is far more important than just giving it to them.) Also this satisfies the general idea that a differential equation should contain the same number of arbitrary constants as the degree of the highest differential in it.

  • Thermo
    Lv 6
    1 decade ago

    Guess y = ae^bx

    Then y ' = abe^bx and y"= ab^2e^bx

    Poke this into y'' + 7y' +3y=0

    ab^2e^bx + 7abe^bx + 3ae^bx = 0

    This must be true for any x.

    e^bx never equals 0, so

    ab^2 + 7ab + 3a = 0

    b^2 + 7b + 3 = 0 or a=0

    If a=0 then b can be any

    if b^2 + 7b + 3 = 0 then b = ... or ... and b can be any real number


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