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Tricky second order differential equation help: Watch

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    I've been doing second order differential equations, finding an auxillary equation, then the complementary function, then the particular integral hence giving the general solution and finally using boundary curves.

    I now have this harder question:

    3x^2 d^2y/dx^2 -xdy/dx + y = 0 y(1) = 0, y'(1) = 1

    Now I don't think I should be following the same pattern to answer this. The auxillary equation would become 3x^2 k^2 - xk + 1 = 0


    What's the first step to solving this?

    I think it's an indicial equation
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    I've had a good attempt:

    Let y(x) = x^k

    y'(x) = kx^k-1

    y''(x) = k(k-1)x^k-2

    Plugging into equation:

    (3x^2.k(k-1)x^k-2) - (x.kx^k-1) + x^k = 0

    x^k[ 3k(k-1) - k + 1]

    Opening it up (3k-1)(k-1) = 0

    k=1/3 , 1


    I don't know how to finish the answer..!
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    (Original post by Cggh90)
    I've had a good attempt:

    Let y(x) = x^k

    y'(x) = kx^k-1

    y''(x) = k(k-1)x^k-2

    Plugging into equation:

    (3x^2.k(k-1)x^k-2) - (x.kx^k-1) + x^k = 0

    x^k[ 3k(k-1) - k + 1]

    Opening it up (3k-1)(k-1) = 0

    k=1/3 , 1


    I don't know how to finish the answer..!
    So you now have  y(x) = Ax^{\frac{1}{3}} + Bx

    Use your boundary conditions to find A and B.

    Just for the record, if you hadn't guessed that the complementary function was in the form x^k, you could have used a substitution of y =vx to get a new differential equation.
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    (Original post by Clarity Incognito)
    So you now have  y(x) = Ax^{\frac{1}{3}} + Bx

    Use your boundary conditions to find A and B.

    Just for the record, if you hadn't guessed that the complementary function was in the form x^k, you could have used a substitution of y =vx to get a new differential equation.

    I got (-3/2)x^1/3 + (3/2)x

    This ir correct, right?

    I'm still not particularly happy with how I did this apart from using lecture notes. Where does this y(x) = x^k actually come from?
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    But by the same token, where does e^k actually come from? Both are somewhat pulled out of a hat.

    I'm not sure there really is a "formal" justification here, but loosely speaking, everytime you differentiate x^k you "lose" a power of x. So when you see a DE where we multiply the 1st derivative by x, the 2nd derivative by x^2 etc., it's a reasonable guess that y=x^k might work.
 
 
 
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