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calculus

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I posted this below, and no one answered. Please help with this very complicated question! 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), giving answer in implicit form. I need details! Many thanks.

  • calculus -

    Separate the variables:
    27sqrt(y)dy=2(x-3)sqrt(x^2-6x+23)dx
    Let z=x^2-6x+23, then dz=(x^2-6x+23)'dx=
    (2x-6)dx=2(x-3)dx
    27sqrt(y)dy=sqrt(z)dz
    Integrating both sides gives
    18y*sqrt(y)=(2/3)*z*sqrt(z)+C
    18y*sqrt(y)=(2/3)*(x^2-6x+23)*sqrt(x^2-6x+23)+C

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