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Math Help - Family of hyperbolas?

  1. #1
    Super Member fardeen_gen's Avatar
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    Family of hyperbolas?

    Show that the following differential equation (4x + 2y + 1)dx + (3x + 2y + 1)dy = 0 represents a family of hyperbolas having lines x + y = 0 and 2x + y + 1 = 0 as asymptotes.

    EDIT:
    As Danny rightly pointed out, the differential equation is actually:
    (4x + 3y + 1)dx + (3x + 2y + 1)dy = 0
    Last edited by fardeen_gen; May 5th 2009 at 09:30 AM.
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  2. #2
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    Quote Originally Posted by fardeen_gen View Post
    Show that the following differential equation (4x + 2y + 1)dx + (3x + 2y + 1)dy = 0 represents a family of hyperbolas having lines x + y = 0 and 2x + y + 1 = 0 as asymptotes.
    I believe there's a typo here. I think the ODE is (4x + 3y + 1)dx + (3x + 2y + 1)dy = 0.
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  3. #3
    Super Member fardeen_gen's Avatar
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    I copied the text correctly.
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  4. #4
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    I'll let you be the judge. The solution to your problem

     <br />
\frac{2}{\sqrt{7}} \tan^{-1} \frac{5x+4y+2}{\sqrt{7}x} + \ln \left| 8x^2+5(2y+1)x +(2y+1)^2\right| = c<br />

    and of mine

     <br />
(x+y)(2x+y+1)= c<br />
.

    I'm not sure how you see the asymptotes x + y = 0 and 2x + y + 1 = 0 coming out in your solution.
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  5. #5
    Super Member fardeen_gen's Avatar
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    Can you fill in the details?
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  6. #6
    Behold, the power of SARDINES!
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    Quote Originally Posted by fardeen_gen View Post
    Can you fill in the details?
    The equation is seperable

    so dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy

    So

    \frac{\partial z}{\partial x}=4x+3y+1

    Take the partial integral w.r.t x and we get

    z(x,y)=2x^2+3xy+x+g(y)

    now take the derivative w.r.t y

    \frac{\partial z}{\partial y}=3x+g'(y)=3x+2y+1

    So we get that

    g'(y)=2y+1 \implies g(y) =y^2+y

    So we get

    2x^2+3xy+x+y^2+y=c \iff 2x^2+xy+x+y^2+2xy+y=c

    x(2x+y+1)+y(y+2x+1)=c \iff (x+y)(2x+y+1)=c
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  7. #7
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    @ TheEmptySet - How about his
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