- #1

- 25

- 0

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

[tex] x^3y''' + x^2y'' - 2xy' + 2y = x^3log(x) [/tex]

[tex] y(1) = \frac{10}{32} [/tex]

[tex] y'(1) = -\frac{24}{32} [/tex]

[tex] y''(1) = -\frac{11}{16} [/tex]

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

----------------

[tex] y_h [/tex]

Substituting [tex] \inline m^n [/tex] for y and dividing by [tex] \inline x^m [/tex] I got:

[tex] m(m-1)(m-2) + m(m-1) - 2m + 2 = 0 [/tex]

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

[tex] (m-1)(m-2)(m+1) = 0 [/tex]

[tex] m = \pm1, 2 [/tex]

[tex] y_h = ax + bx^{-1} + cx^2 [/tex]

----------------

[tex] y_p [/tex]

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

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

[tex] W = -\frac{6}{x} [/tex]

[tex] 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} [/tex]

[tex] W_1 = 3 [/tex]

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

[tex] W_2 = -x^2 [/tex]

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

[tex] W_3 = -\frac{2}{x} [/tex]

Therefore:

[tex] 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 [/tex]

Which, er, is....

[tex] 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) [/tex]

Making it..

[tex] 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

[/tex]

So..

[tex] y_p = -\frac{1}{140}x^6logx - \frac{83}{19600}x^6 [/tex]

--------------------------

[tex] y = y_h + y_p [/tex]

Putting the two together..

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

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

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

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

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

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

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

And solving for a, b, c...

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

[tex] B = y'(1) = a - b + 2c = -\frac{7171}{9800} [/tex]

[tex] C = y''(1) = 2b + 2c = -\frac{501}{784} [/tex]

A - B = D

[tex] 2b - c = 1\frac{387}{9800} [/tex]

C - D

[tex] 3c = -\frac{501}{784} - 1\frac{387}{9800} [/tex]

[tex] c = -\frac{7849}{58800} [/tex]

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

[tex] 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 [/tex]

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

-----------

Question 1:

What went wrong?!

-----------

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?

[tex] x^3y''' + x^2y'' - 2xy' + 2y = x^3log(x) [/tex]

[tex] y(1) = \frac{10}{32} [/tex]

[tex] y'(1) = -\frac{24}{32} [/tex]

[tex] y''(1) = -\frac{11}{16} [/tex]

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

----------------

[tex] y_h [/tex]

Substituting [tex] \inline m^n [/tex] for y and dividing by [tex] \inline x^m [/tex] I got:

[tex] m(m-1)(m-2) + m(m-1) - 2m + 2 = 0 [/tex]

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

[tex] (m-1)(m-2)(m+1) = 0 [/tex]

[tex] m = \pm1, 2 [/tex]

[tex] y_h = ax + bx^{-1} + cx^2 [/tex]

----------------

[tex] y_p [/tex]

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

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

[tex] W = -\frac{6}{x} [/tex]

[tex] 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} [/tex]

[tex] W_1 = 3 [/tex]

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

[tex] W_2 = -x^2 [/tex]

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

[tex] W_3 = -\frac{2}{x} [/tex]

Therefore:

[tex] 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 [/tex]

Which, er, is....

[tex] 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) [/tex]

Making it..

[tex] 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

[/tex]

So..

[tex] y_p = -\frac{1}{140}x^6logx - \frac{83}{19600}x^6 [/tex]

--------------------------

[tex] y = y_h + y_p [/tex]

Putting the two together..

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

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

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

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

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

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

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

And solving for a, b, c...

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

[tex] B = y'(1) = a - b + 2c = -\frac{7171}{9800} [/tex]

[tex] C = y''(1) = 2b + 2c = -\frac{501}{784} [/tex]

A - B = D

[tex] 2b - c = 1\frac{387}{9800} [/tex]

C - D

[tex] 3c = -\frac{501}{784} - 1\frac{387}{9800} [/tex]

[tex] c = -\frac{7849}{58800} [/tex]

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

[tex] 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 [/tex]

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

-----------

Question 1:

What went wrong?!

-----------

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: