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Math Help - Differential Equations

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    Differential Equations

    Solve the following differential equation using the substitution given:

    2x \displaystyle{\frac{dy}{dx}} = 2y - y^2 , by substituting y = zx.

    I got to a stage where zx = 2 (not sure if I am right)...and then I don't know how to continue. Please help! Thank you!
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by Tangera View Post
    Solve the following differential equation using the substitution given:

    2x \displaystyle{\frac{dy}{dx}} = 2y - y^2 , by substituting y = zx.

    I got to a stage where zx = 2 (not sure if I am right)...and then I don't know how to continue. Please help! Thank you!
    Don't forget that z is a function of x.

    \frac{dy}{dx}=z+x \frac{dz}{dx}

    You must arrive to 2 \frac{dz}{dx}+z^2=0
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    Quote Originally Posted by Moo View Post
    Hello,

    Don't forget that z is a function of x.

    \frac{dy}{dx}=z+x \frac{dz}{dx}

    You must arrive to 2 \frac{dz}{dx}+z^2=0
    Yes I reached there and got xz = 2 after intergrating. And then I am stuck. X.x
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    Moo
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    Quote Originally Posted by Tangera View Post
    Yes I reached there and got xz = 2 after intergrating. And then I am stuck. X.x
    Oops sorry ^^

    Then, you know that y=xz... and you substitute
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    Quote Originally Posted by Moo View Post
    Oops sorry ^^

    Then, you know that y=xz... and you substitute
    Hmm...just to check...do I substitute y = xz = 2 into the given differential equation? But if so, then I get 2x\displaystyle{\frac{dy}{dx}} = 0...and I am lost! >.<
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    Moo
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    Quote Originally Posted by Tangera View Post
    Hmm...just to check...do I substitute y = xz = 2 into the given differential equation? But if so, then I get 2x\displaystyle{\frac{dy}{dx}} = 0...and I am lost! >.<
    Yes, and 2(2)-(2)^2=0 too
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    Quote Originally Posted by Moo View Post
    Yes, and 2(2)-(2)^2=0 too
    The answer I was given is 2x/(x+c)....but I don't know how to get there...Thanks for the help!
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  8. #8
    Moo
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    Haha I know why

    Yes I reached there and got xz = 2 after intergrating. And then I am stuck. X.x
    The exact solution shall be x+c=\frac 2z (c is the constant of integration !)

    \implies z=\frac{2}{x+c}

    Since xz=y ~,~ y=\frac{2x}{x+c}
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  9. #9
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    Ahhh I see!!

    Thank you very much, Moo!
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