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Using 3(x-3)(x^2-6x+23)^1/2, as the chain rule differentiation of f(x)=(x^2-6x+23). Please explain how I find the general solution to the differential equation dy/dx= 2/27(x-3)SQUARE ROOT BEGINS (x^2-6x+23)/y SQUARE ROOT ENDS (y>0). In implicit form I have an answer of 18*y*square root of y =TOP LINE OF FRACTION is 2*(x^2-6x+23)^3/2 divided by 3, plus constant.Is this right ans what is c and using the answer what is the particular solution for which y=2 when x=1, written then in explicit form? Thank you for any response. Best wishes.

  • Calculus - ,

    This is right

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