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January 31, 2015

January 31, 2015

Posted by **Claire** on Wednesday, April 6, 2011 at 1:17am.

dy/dx = 2/27(x-3)(x^2-6x+23) / (y) (y>0. For which y=2 when x =1, and then give this particular solution in explicit form.

regards Claire

- Maths -
**MathMate**, Wednesday, April 6, 2011 at 8:01amAssuming stands for √, and assuming parentheses are as follows:

dy/dx = (2/27)((x-3)√(x^2-6x+23)) / (y) (y>0. For which y=2 when x =1

Separate variables:

ydy = (2/27)(x-3)√(x^2-6x+23)

complete squares and use substitution

u=x-3

du=dx

ydy = (2/27)√(u^2+14) 2udu

use substitution v=u^2+14, dv=2udu, integrate and backsubstitute:

y^2/2 = (2(x^2-6x+23)^(3/2))/81 + C

Substitute x=1, y=2 to find C, and hence y.

I get C=2+2^(5/2).

Check me.

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