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anonymous

  • 5 years ago

Reduction of order method problem :( x^2y''+2xy'-12y=0 for y1=x^3

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  1. anonymous
    • 5 years ago
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    please help me

  2. anonymous
    • 5 years ago
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    If memory serves, reduction of order uses your first solution to find a second. You take your first solution and multiply it with some function, v(x), say, so that\[y_2(x)=v(x)y_1(x)=v(x)x^3 \]here.

  3. anonymous
    • 5 years ago
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    You then go through the motions: 1. Find first derivative of y_2 2. Find second derivative of y_2 Substitute into your original equation and set to zero (since the aim is to find v(x) for y_2=v*y_1 a solution.

  4. anonymous
    • 5 years ago
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    \[y_2=v(x)x^3 \rightarrow y^'_2=v.3x^2+v'x^3 \rightarrow y''_2=3xv''+6x^2v'+6xv\]

  5. anonymous
    • 5 years ago
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    Substitute into your original equation,\[x^2(3xv''+6x^2v'+6xv)+2x(3vx^2+x^2v')-12(x^3v)\]\[=3x^4v''+8x^4v'\]\[=0\]since we're trying to find v(x) that allows us to use y_2=v.y_1 as solution.

  6. anonymous
    • 5 years ago
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    You can then use a substitution of \[u=\frac{dv}{dx}\rightarrow{\frac{du}{dx}=\frac{d^2v}{dx^2}}\]and our equation becomes\[x^4(3u'+8u)=0\]x^4 is, in general, non-zero, so we have to find the solution for the second factor; i.e. find u(x)\[3\frac{du}{dx}+8u=0\]This is separable,\[\frac{du}{u}=-\frac{8}{3}dx{\rightarrow}\int\limits_{}{}\frac{du}{u}=\int\limits_{}{}-\frac{8}{3}dx\]

  7. anonymous
    • 5 years ago
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    thankyou so much your amazing mate

  8. anonymous
    • 5 years ago
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    Cheers :) Please check everything, though, as in, check that the final solution does work when you plug it into the DE. I had to rush...

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