calculus

consider the differential equation dy/dx= (y - 1)/ x squared where x not = 0

a) find the particular solution y= f(x) to the differential equation with the initial condition f(2)=0

(b)for the particular solution y = F(x) described in part (a) find lim F(x) X goes to infinity

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asked by sammy
  1. dy/dx = (y - 1)/ x^2
    Use the integration method of separation of variables.

    dy/(y-1) = dx/x^2

    ln (y-1) = -1/x + C

    You say that the initial condition is
    f(x) = y = 0 at x =2

    Are you sure the initial condition is not f(0) = 2 ? You cannot have a logarithm of a negative number.

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    posted by drwls
  2. Start like above, but once you gget to the logarithm part, it is in absolute values, so it is ln(1) which is 0. Solve from there

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    posted by alame
  3. The two statements above is wrong.

    you first move around the
    dy/dx = (y-1)/x^2 into

    dy/(y-1) = dx/x^2

    Then you integral where it turns to

    ln( (y-1)/C ) = -1/x

    Next you move the e from the ln to the other side

    (y-1)/C = e^(-1/x)

    After that you times C to both sides and move the -1 afterwords

    y-1 = Ce^(-1/x) --> y = Ce^(-1/x)+1

    I not sure how you do part (b) though.
    Sorry

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    posted by Andy

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