## anonymous 5 years ago Reduction of order method problem :( x^2y''+2xy'-12y=0 for y1=x^3

1. anonymous

2. anonymous

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

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

$y_2=v(x)x^3 \rightarrow y^'_2=v.3x^2+v'x^3 \rightarrow y''_2=3xv''+6x^2v'+6xv$

5. anonymous

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

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

thankyou so much your amazing mate

8. anonymous

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...