Nonhomogenous LODE (Method of Variation of Parameters)

[kape]

Nonhomogenous LODE (Higher Order) - Method of Variation of Parameters

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

y(1) = \frac{10}{32}

y'(1) = -\frac{24}{32}

y''(1) = -\frac{11}{16}


I know that \inline y = y_h + y_p and that I probably should use the method of variation of parameters, so:


----------------
y_h

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

m(m-1)(m-2) + m(m-1) - 2m + 2 = 0

(m-1)[m^2-2m+m-2] = 0

(m-1)(m-2)(m+1) = 0

m = \pm1, 2

y_h = ax + bx^{-1} + cx^2


----------------
y_p

So, taking \inline y_1 = x, y_2 = \frac{1}{x}, y_3 = x^2

W = \begin{array}{|ccc|} x & \frac{1}{x} & x^2 \\ 1 & -\frac{1}{x^2} & 2x \\ 0 & \frac{2}{x^3} & 2 \end{array}

W = -\frac{6}{x}

W_1 = \begin{array}{|ccc|} 0 & \frac{1}{x} & x^2 \\ 0 & -\frac{1}{x^2} & 2x \\ 1 & \frac{2}{x^3} & 2 \end{array}

W_1 = 3

W_2 = \begin{array}{|ccc|} x & 0 & x^2 \\ 1 & 0 & 2x \\ 0 & 1 & 2 \end{array}

W_2 = -x^2

W_3 = \begin{array}{|ccc|} x & \frac{1}{x} & 0 \\ 1 & -\frac{1}{x^2} & 0 \\ 0 & \frac{2}{x^3} & 1 \end{array}

W_3 = -\frac{2}{x}

Therefore:

y_p = x\int -\frac{x}{2} x^3 \lnx dx + \frac{1}{x}\int \frac{x^3}{6} x^3 \lnx dx + x^2\int -\frac{x^3}{3} \lnx dx

Which, er, is...

y_p = x(-\frac{1}{10}x^5logx + \frac{1}{50}x^5) + \frac{1}{x}(\frac{1}{42}x^7logx - \frac{1}{294}x^7) + x^2(\frac{1}{12}x^4logx <br /> <br /> - \frac{1}{48}x^4)

Making it..

y_p = -\frac{1}{10}x^6logx + \frac{1}{50}x^6 + \frac{1}{42}x^6logx - \frac{1}{294}x^6 + \frac{1}{12}x^6logx - \frac{1}{48}x^6 <br /> <br />

So..

y_p = -\frac{1}{140}x^6logx - \frac{83}{19600}x^6


--------------------------
y = y_h + y_p

Putting the two together..

y = ax + bx^{-1} + cx^2 - \frac{1}{140}x^6logx - \frac{83}{19600}x^6

y&#039; = a - bx^{-2} + 2cx - \frac{3}{70}x^5logx + x^5 - \frac{179}{9800}x^5

y&#039;&#039; = 2bx^{-3} + 2c - \frac{3}{14}x^4logx x^4 + 5x^4 - \frac{19}{392}x^4

Putting in the initial values at y(1)..

A = y(1) = a + b + c - \frac{83}{19600} = \frac{10}{32}

B = y&#039;(1) = a - b + 2c - \frac{179}{9800} = -\frac{24}{32}

C = y&#039;&#039;(1) = 2b + 2c - \frac{19}{392} = -\frac{11}{16}

And solving for a, b, c...

A = y(1) = a + b + c = \frac{388}{1225}

B = y&#039;(1) = a - b + 2c = -\frac{7171}{9800}

C = y&#039;&#039;(1) = 2b + 2c = -\frac{501}{784}

A - B = D

2b - c = 1\frac{387}{9800}

C - D

3c = -\frac{501}{784} - 1\frac{387}{9800}

c = -\frac{7849}{58800}

And it is starting to look quite silly.. but carrying on to find b, a and putting them into the original equation gets:

y = -\frac{109}{39200}x + \frac{53273}{117600}x^{-1} - \frac{7849}{58800}x^2 - \frac{1}{140}x^6logx - \frac{83}{19600}x^6

Which is frankly, quite ridiculous. And unsurprisingly, it turns out to be wrong. So.



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Question 1:

What went wrong?!


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Question 2:

I am having a lot of trouble finding roots (factorizing).. right now I basically do it by trial and error and for me it takes an

inordinate amount of time.. Is there a way to do this quickly?

 

[kape]

I've also tried checking some of the calculations using Matlab and they seem to be correct.. I get the feeling that it's not the calculation but the method?

Also, I've managed to find info on factorizing.. I found out there was this thing called the Factor Theorem and Long Division!