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# Tricky second order differential equation help: Watch

1. 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
2. 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..!
3. (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

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.
4. (Original post by Clarity Incognito)
So you now have

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?
5. 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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