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Finding the equation of a curve

  1. Mar 13, 2013 #1
    1. The problem statement, all variables and given/known data

    The tangent of any point that belongs to a curve, cuts Y axis in such a way, that the cut off segment in Y axis is twice as big as the X value of the point. Find the equation of the curve, if point (1,4) is part of it.


    2. Relevant equations

    The ultimate solution is [itex]y = x(4-lnx^2)[/itex]

    3. The attempt at a solution

    This might be a little confusing, so I'll try to clarify the situation: assuming for example, that point (1,4) is part of the curve, if we try to find the tangent of the curve at that point, we know that it will cut Y axis where y=2x=2. So in that case the tangent is a line going through (1,4) and (0,2).


    I guess we get [itex]y = x(4-lnx^2)[/itex] when we put x=1 and y=4 into the general solution, which gives us the specific constant value, then insert the constant value into the general solution to get the final solution. But I don't know how to obtain the general solution. Could you please help me solve this problem?
     
  2. jcsd
  3. Mar 13, 2013 #2

    Simon Bridge

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    The curve is y=f(x)
    In general: the slope of the tangent at point x=q is f'(q)
    The equation of the line is y=f'(q)x+c(q) so that c(q) is the y intercept.
    The problem statement is saying that c(q)=2q... which, in general, will be a function of q.
     
  4. Mar 13, 2013 #3
    So we have [itex]y=y'x+2x[/itex] which is the same as [itex]y'-y/x=-2[/itex] which is a linear differential equation, solving it by using the Bernoulli method y=uv, we get the general solution [itex]y=x(C-lnx^2)[/itex]. Inserting x=1, y=4 gives C=4, thus the final answer is [itex]y=x(4-lnx^2)[/itex].
    You're awesome, thanks!
     
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