# Nonhomogenous LODE (Method of Variation of Parameters)

1. Apr 20, 2006

### 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....

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

Making it..

$$y_p = -\frac{1}{10}x^6logx + \frac{1}{50}x^6 + \frac{1}{42}x^6logx - \frac{1}{294}x^6 + \frac{1}{12}x^6logx - \frac{1}{48}x^6$$

So..

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

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

Putting the two together..

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

$$y' = a - bx^{-2} + 2cx - \frac{3}{70}x^5logx + x^5 - \frac{179}{9800}x^5$$

$$y'' = 2bx^{-3} + 2c - \frac{3}{14}x^4logx x^4 + 5x^4 - \frac{19}{392}x^4$$

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

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

$$B = y'(1) = a - b + 2c - \frac{179}{9800} = -\frac{24}{32}$$

$$C = y''(1) = 2b + 2c - \frac{19}{392} = -\frac{11}{16}$$

And solving for a, b, c...

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

$$B = y'(1) = a - b + 2c = -\frac{7171}{9800}$$

$$C = y''(1) = 2b + 2c = -\frac{501}{784}$$

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?

Last edited: Apr 20, 2006
2. Apr 22, 2006

### 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!