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Math Help - Differential Equations

  1. #1
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    Differential Equations

    I'm not sure if I've put this in the correct section, but here goes...
    Please may someone help me with this question. I understand how to do (a) but not (b).

    The gradient of a curve at the point (x,y) is given by the differential equation
    dy/dx = (2-x)/y
    a) Find the general solution of dy/dx = (2-x)/y
    b) Given that the curve passes through the point (4,2) show that the curve is a circle and find its radius and the coordinates of its centre.

    Thank you all
    Rachel
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  2. #2
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    mr fantastic's Avatar
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    Quote Originally Posted by racheltllong View Post
    I'm not sure if I've put this in the correct section, but here goes...
    Please may someone help me with this question. I understand how to do (a) but not (b).

    The gradient of a curve at the point (x,y) is given by the differential equation
    dy/dx = (2-x)/y
    a) Find the general solution of dy/dx = (2-x)/y
    b) Given that the curve passes through the point (4,2) show that the curve is a circle and find its radius and the coordinates of its centre.

    Thank you all
    Rachel
    You should have \frac{y^2}{2} = 2x - \frac{x^2}{2} + C

     \Rightarrow y^2 = 4x - x^2 + K \Rightarrow (x - 2)^2 + y^2 = A

    where A = K - 4 is just as arbitrary as K and K = 2C is just as arbitrary as C and C is an arbitrary constant.

    You can solve for A using the given condition, right?

    And you can recognise the equation of a circle. And get the radius and coordinates of centre from that equation, right?
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  3. #3
    Member Jonboy's Avatar
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    Is this really something taught in Pre-Calc ?
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  4. #4
    Math Engineering Student
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    He probably did not see the calculus' forum, but it's not taught at pre-calc. level.
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  5. #5
    Member Jonboy's Avatar
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    I was about to say, I'm missing out.
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