How do you solve the differential #dy/dx=(x-4)/sqrt(x^2-8x+1)#?

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Answer 1 · 2017-01-10T14:12:28

# y = sqrt(x^2-8x+1) + C #

#d y/dx=(x-4)/sqrt(x^2-8x+1) #

Is a First Order separable DE which we can sole by integrating:

# :. y = int \ (x-4)/sqrt(x^2-8x+1) \ dx #

Let # u=x^2-8x+1 => (du)/dx=2x-8 = 2(x-4) #

Substituting into the RHS integral we get:

# y = int \ (1/2)/sqrt(u) \ du #
# \ \ = 1/2 int \ u^(-1/2) \ du #
# \ \ = 1/2 u^(1/2)/(1/2) + C#
# \ \ = sqrt(u) + C #
# \ \ = sqrt(x^2-8x+1) + C #

which is the General Solution