Higher Order Non-Homogenous ODE

[kape]

Higher Order Homogenous ODE (Euler-Cauchy)

*sigh* I am (yet again) stuck on a problem.. I would greatly appreciate any help!

x^3y''' - 3x^2y'' + (6-x^2)xy' - (6-x^2)y = 0

\inline y_1 = x is a solution to the equation above

y'(0) = 3
y''(0) = 9
y'''(0) = 18

I'm not quite sure what it means by "\inline y_1 = x is a solution to the equation above" so I decided to ignore it. I probably shouldn't have because it is probably something I need to know and it is probably the reason why I haven't been able to solve this problem.

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What I have done so far

Substituting \inline m^n for y and dividing by \inline x^m I got:

m(m-1)(m-2) - 3m(m-1) + (6-x^2)m - (6-x^2) = 0

m(m-1)(m-2) - 3m(m-1) + (6-x^2)(m-1) = 0

m(m-2) - 3m + (6-x^2) = 0

m^2 - 5m + (6-x^2) = 0

It is at this point that I am confused. It looks like a quadratic equation but what do I do with the (6-x^2)?!?


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Question 1:

What DO I do with (6-x^2) ?!


Someone told me that he got the roots 1, 2, 3... which I then tried, but got the answer wrong:

y = c_1x + c_2x^2 + c_3x^3

y' = c + 2c_2x + 3c_3x^2

y'' = 2c_2 + 6c_3x

y''' = 6c_3

And putting in the initial values I get:

c_1 = 3

c_2 = 9/2

c_3 = 3

Which doesn't seem right.. (can there be two different constants that have the same value?) But I put in the found constants into the original equation to get:

y = 3x + \frac{9}{2}x^2 + 3x^3

Which is.. wrong.


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Question 2:

Are the roots not 1, 2, 3 or did I do something wrong?

 

[benorin]

Use the substitution y=xu where the x is from the known solution and u is a function of x [i.e. u=u(x)], we also have

y=xu
y^{\prime}=xu^{\prime}+u
y^{\prime\prime}=xu^{\prime\prime}+2u^{\prime}
y^{\prime\prime\prime}=xu^{\prime\prime\prime}+3u^{\prime\prime}

substitute these into the given 3^rd order DE to get (after a good deal of cancelation)

x^3u^{\prime\prime\prime}-x^4u^{\prime}=0