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Math Help - separable DEQ

  1. #1
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    separable DEQ

    I'm getting a incorrect answer for Given 2xy(dy/dx) + y^2 - 1 = 0 find solution A) (1,0) B) (0,1) in x-y plane. Please tell me where i'm screw up and how to fix it.

    This is my workout:
    2xy(dy/dx) + y^2 - 1

    -seperate:
    y^2 +2y dy=-2x + 1 dx

    -intergrate:
    (y^3/3)+(y^2) = x-(x^2) +C
    This is where my problem is, I'm not sure from here how to do the following.

    -divide through by y
    y=square root (-x+C/y+3) I don't think that is right and if it is what is my next step in getting rid of y on the right side?
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  2. #2
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    Re: separable DEQ

    Quote Originally Posted by nivek0078 View Post
    I'm getting a incorrect answer for Given 2xy(dy/dx) + y^2 - 1 = 0 find solution A) (1,0) B) (0,1) in x-y plane. Please tell me where i'm screw up and how to fix it.

    This is my workout:
    2xy(dy/dx) + y^2 - 1

    -seperate:
    y^2 +2y dy=-2x + 1 dx

    -intergrate:
    (y^3/3)+(y^2) = x-(x^2) +C
    This is where my problem is, I'm not sure from here how to do the following.

    -divide through by y
    y=square root (-x+C/y+3) I don't think that is right and if it is what is my next step in getting rid of y on the right side?
    A DE is only separable if you can write it as \displaystyle \begin{align*} f(y)\,\frac{dy}{dx} = g(x) \end{align*}. To properly separate your variables in this case, you would need to do this...

    \displaystyle \begin{align*} 2x\,y\,\frac{dy}{dx} + y^2 - 1 &= 0 \\ 2x\,y\,\frac{dy}{dx} &= 1 - y^2 \\ \frac{2x\,y}{1 - y^2}\,\frac{dy}{dx} &= 1 \\ \frac{2y}{1 - y^2} \,\frac{dy}{dx} &= \frac{1}{x} \end{align*}

    And now you can integrate both sides...
    Thanks from nivek0078
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  3. #3
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    Re: separable DEQ

    Thank you for your help. I've been doing these problems way to long!
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  4. #4
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    Re: separable DEQ

    Still not getting the correct answer. Please tell me where i went wrong and how to fix it. So i took the intergal as given by Prove It:

    -intergal answer
    -ln(y^2-1) = ln(x) + C

    -solve for y have to remove the ln by using e
    e^(-ln(y^2-1)) = e^(ln(x) + C) => (1/(y^2)-1) = xe^C

    -general solution
    y=squart root x(e^C/2)
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  5. #5
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    Re: separable DEQ

    Quote Originally Posted by nivek0078 View Post
    Still not getting the correct answer. Please tell me where i went wrong and how to fix it. So i took the intergal as given by Prove It:

    -intergal answer
    -ln(y^2-1) = ln(x) + C

    -solve for y have to remove the ln by using e
    e^(-ln(y^2-1)) = e^(ln(x) + C) => (1/(y^2)-1) = xe^C

    -general solution
    y=squart root x(e^C/2)
    Here is a useful trick with arbitary constants

    rewrite C=\ln(C_2)

    This gives

    -\ln(y^2-1)=\ln(x)+\ln(C_2) \iff \ln(y^2-1)=\ln(C_2x)^{-1} \iff y^2-1=\frac{1}{C_2x} \iff   y^2-1=C_3\frac{1}{x}
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