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Math Help - General solution.

  1. #1
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    General solution.

    What is the general solution of the differential equation: xy'= y+2 √x √y?

    i got:

    x(dy/dx)=y+2√x √y
    xdy = ydx + 2√x √ydx
    (xdy-ydx)/x^2 = 2√ydx/[(x)^(3/2)] ... dividing by x^2
    d(y/x) = 2√(y/x)(dx/x)
    d(y/x)/2√(y/x) = dx/x
    √(y/x) = logx +c ... integrating above expression
    e^√(y/x) = kx ... putting e^c=k

    but im not too sure if this is correct, could you clear this up for me?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by DasBill View Post
    What is the general solution of the differential equation: xy'= y+2 √x √y?

    i got:

    x(dy/dx)=y+2√x √y
    xdy = ydx + 2√x √ydx
    (xdy-ydx)/x^2 = 2√ydx/[(x)^(3/2)] ... dividing by x^2
    d(y/x) = 2√(y/x)(dx/x)
    d(y/x)/2√(y/x) = dx/x
    √(y/x) = logx +c ... integrating above expression
    e^√(y/x) = kx ... putting e^c=k

    but im not too sure if this is correct, could you clear this up for me?
    Hm...I would have manipulated it like this:

    y^{\prime}=\frac{y}{x}+2\sqrt{\frac{y}{x}}

    Take y=ux. So \frac{\,dy}{\,dx}=u+x\frac{\,du}{\,dx}.

    Therefore, the DE become u+x\frac{\,du}{\,dx}=u+2\sqrt{u}\implies\frac{\,du  }{2\sqrt{u}}=\frac{\,dx}{x}.

    Thus, \sqrt{u}=\ln x+C\implies u=\left(\ln x+C\right)^2.

    Therefore \frac{y}{x}=\left(\ln x + C\right)^2\implies y=x\left(\ln x+C\right)^2

    Does this make sense?
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