Question

Find the particular solution of the differential equation from:
dy/dx = 2/27(x-3)ã(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

Answer 1

Assuming ã 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.