Differentiation

Please use separation of variables method to find particular solution

(x/3 ) * (dy/dx) - (1/9)y^2 = 1 (x>0)

the initial condition y(1) = pi

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  • 1 十年前
    最愛解答

    (x/3 ) * (dy/dx) - (1/9)y^2 = 1 (x>0)

    (x/3 ) * (dy/dx) = 1+(1/9)y^2

    9x (dy/dx) = 27+3y^2

    3x (dy/dx) = 9+y^2

    dy/(9+y^2)=dx/3x

    ∫1/(9+y^2) dy=∫1/3x dx

    (1/3)arctan(y/3)=(1/3)lnx+C

    Sub x=1, y=π

    (1/3)√3=C

    C=1/√3

    So the particular solution is

    (1/3)arctan(y/3)=(1/3)lnx+1/√3

    arctan(y/3)=lnx+√3

    Note:

    For calculate ∫1/(9+y^2) dy

    Let y=3tanθ ; dy=3sec^2θ dθ

    ∫1/(9+y^2) dy

    =∫[1/(9+9tan^2θ)] (3sec^2θ)dθ

    =∫sec^2θ/3(sec^2θ) dθ

    =(1/3)θ+C

    =(1/3)arctan(y/3)+C

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