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Math Help - Solve the following ODE .. #8

  1. #1
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    Solve the following ODE .. #8

    Hello

    I have this problem

    Problem:
    Solve the following differential equation by using three different methods :

    (xy-1+y)dx+xdy=0

    Solution:


    One method is finding the integrating factor which is e^x which found by the formula \displaystyle e^{\displaystyle \int \dfrac{\dfrac{\partial M}{\partial y}-\dfrac{\partial N}{\partial x}}{N} \, dx}

    what are the other 2 methods??
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  2. #2
    MHF Contributor chisigma's Avatar
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    With some algebraic steps the DE becomes...

    \displaystyle y^{'} + y\ (1+\frac{1}{x^{2}}) = \frac{1}{x} (1)

    ... which is linear and can be solved with 'standard approach'...

    Kind regards

    \chi \sigma
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  3. #3
    A Plied Mathematician
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    You could always do a series solution method!
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  4. #4
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    If you had some I.C's then a numerical method like runge-kutta?
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  5. #5
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by chisigma View Post
    With some algebraic steps the DE becomes...

    \displaystyle y^{'} + y\ (1+\frac{1}{x^{2}}) = \frac{1}{x} (1)

    ... which is linear and can be solved with 'standard approach'...

    Kind regards

    \chi \sigma
    The 'standard approach' seems to lead to a solution like...

    \displaystyle y(x)= e^{-(x-\frac{1}{x})} \ \{\int \frac{e^{(x-\frac{1}{x})}}{x}\ dx + c \} (1)

    ... that contains a 'not very pleasant' integral ...

    Kind regards

    \chi \sigma
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  6. #6
    A Plied Mathematician
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    The substitution u=xy-1 renders the equation separable:

    du=x\,dy+y\,dx, and hence if we have

    (xy-1)\,dx+y\,dx+x\,dy=0, then we arrive at

    u\,dx+du=0.
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  7. #7
    Super Member General's Avatar
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    Another one is by using inspection.

    (xy-1)dx+ydx+xdy=0

    (xy-1)dx+d(xy)=0

    \dfrac{d(xy)}{xy-1}+dx=0 ... (1)

    Your equation is ready for integrating now.

    Actually, equation (1) is equivelant to the last equation in Ackbeet's replay.
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  8. #8
    Super Member General's Avatar
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    Quote Originally Posted by chisigma View Post
    With some algebraic steps the DE becomes...

    \displaystyle y^{'} + y\ (1+\frac{1}{x^{2}}) = \frac{1}{x} (1)

    ... which is linear and can be solved with 'standard approach'...

    Kind regards

    \chi \sigma
    Check your algebra.

    You should get : y'+\left(1+\dfrac{1}{x}\right) \cdot y = \dfrac{1}{x}
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  9. #9
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by General View Post
    Check your algebra.

    You should get : y'+\left(1+\dfrac{1}{x}\right) \cdot y = \dfrac{1}{x}
    Of course!... and the 'standard approach' leads to the 'confortable' solution...

    \displaystyle y(x)= c\ \frac{e^{-x}}{x} + \frac{1}{x} (1)

    Kind regards

    \chi \sigma
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  10. #10
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    Thanks All.
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  11. #11
    A Plied Mathematician
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    You're welcome for my (this time, real!) contribution.
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  12. #12
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    Thanks.
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