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D operator - Differential equations

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Why bother with a post grad? Are they even worth it? Have your say! 26-10-2016

    You have already obtained the general solution.

    (Original post by Scary)
    the first part of the question was to use the auxiliary equation to find the yc and then finding the particular integral, the equation was d^2y/dx^2 -dy/dx -6y = e^-2x, the general solution i got was -1/5xe^(-2x) +Ae^(3x)+Be^(-2x)

    How did you get on?
    • Thread Starter

    thanks for all the help, one of the constants isnt the same though, its divided by 5 is it possible to let another constant say k=c/5?

    Hi, yes,  \frac{c}{5} is still a constant.

    This question overall demonstrates another way to solve second ODEs without having to try various functions to be the particular integral.

    Also, if the particular integral (PI) is the same as one of the solutions to the complementary function (in this case  PI(x) = Ae^{-2x} ), then when you substitute the PI back into the second order ODE, you end up with your PI equal to zero.
    • Thread Starter

    Hi yes to avoid the pi being a failing case you would change the trial function slightly? Or is there some other way around it?

    (Original post by Scary)
    Hi yes to avoid the pi being a failing case you would change the trial function slightly? Or is there some other way around it?
    One method is to multiply the particular integral (PI) with ascending powers of  x , staring from  x .

    In this case, instead of using the PI as  Ae^{-2x} we can try  Axe^{-2x} (where  A is a real value).

    Then substitute in  \frac{dy}{dx} \text{ and } \frac{d^{2}y}{dx^{2}} into the original ODE as normal.

    If this still provides a formula equal to zero, raise to the next power of  x for your new PI.

    Other methods to solve second order ODEs are Laplace transforms, Wronskians, variation of parameters, Frobenius method (to name a few).
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