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Homework Help: Easy DEQ problem(s)

  1. Jan 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Verify that the indicated function is a solution to the differential equation. Assume an appropriate interval I of definition for each solution.

    [tex]y'' - 6y' +13y = 0 [/tex]
    [tex]y = e^{3x}cos(2x)[/tex]

    2. Relevant equations

    3. The attempt at a solution
    Well, I started taking derivatives to see if y was indeed a solution, but that got rather complicated fast for what is some of the first problem sets from day 1 of class. I believe that this is NOT a solution because of the cosine, which would indicate that y(x) is not a linear function clearly. Since the DEQ is linear of the second order, I assume that the solution will be linear as well?

    Frankly, I am confused on when to consider a function linear. Does the dependent variable have to be linear, or is it the independent variable that matters? What if I replaced cos(2x) in the above equation with cos(2y).

    Is this the correct manner of thinking on this problem? (ie: see cosine, must not be correct)
  2. jcsd
  3. Jan 9, 2012 #2
    it is the solution to the homogeneous diff equ. Linear means that none of the terms are of some power like squared.
  4. Jan 9, 2012 #3


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    I think you need to recalibrate your sense of what qualifies as complicated. :wink:

    No, that's not correct.

    You're not interested in whether a function is linear here. You want to know if an operator is linear. The equation you have can be written in the form ##\mathcal{L}y = f(x)##, where ##\mathcal{L}## is an operator acting on y. If the operator is linear, it has the properties
    \mathcal{L}(y_1 + y_2) &= \mathcal{L}y_1 + \mathcal{L}y_2 \\
    \mathcal{L}(cy) &= c\mathcal{L}y
    \end{align*}and the concepts from linear algebra can be applied to solve the equation.

    To determine if the operator ##\mathcal{L}## is linear, you just need to make sure that y and its derivatives, if they appear, are only taken to the first power, which is indeed the case here.
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