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Math Help - Second-Order Nonlinear DE

  1. #1
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    Second-Order Nonlinear DE

    Can someone please explain to me how to solve this equation:
    x*y''-y*y'=2...where I mus find the condition that the constants a and b must satisfy for y=ax+b to be a solution.Also how can I find the two solutions that satisfy y=3 when x=2.Thanks in advance
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  2. #2
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    Quote Originally Posted by AkilMAI View Post
    Can someone please explain to me how to solve this equation:
    x*y''-y*y'=2...where I mus find the condition that the constants a and b must satisfy for y=ax+b to be a solution.Also how can I find the two solutions that satisfy y=3 when x=2.Thanks in advance
    I think you have a typo

    y=ax+b

    y'=a

    y''=0

    If you put this into the ODE you get

    x\cdot (0)-(ax+b)\cdot (a)=2 \iff -a^2x-ab=2

    This implies that

    a=0

    but that makes the equation false!
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by AkilMAI View Post
    Can someone please explain to me how to solve this equation:
    x*y''-y*y'=2... how can I find the two solutions that satisfy y=3 when x=2.Thanks in advance
    Considering that is y(2)=3 and setting z=y' the DE is reduced to...

    z'= \frac{3}{2}\ z +1 (1)

    ... which is a linear first order DE and is solved in standard fashion if You know the 'initial value' of y'...

    Kind regards

    \chi \sigma
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  4. #4
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    Quote Originally Posted by chisigma View Post
    Considering that is y(2)=3 and setting z=y' the DE is reduced to...

    z'= \frac{3}{2}\ z +1 (1)

    ... which is a linear first order DE and is solved in standard fashion if You know the 'initial value' of y'...

    Kind regards

    \chi \sigma
    Please share - how did you arrive at this?
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  5. #5
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Danny View Post
    Please share - how did you arrive at this?
    Probably I entered a wrong way...

    ... my reasoning was that the in the DE...

    x\ y'' - y\ y'=2 (1)

    ... with 'initial condition'  y(2)=3 was possible the 'direct substitution' x=2, y=3 so that it becomes...

    2\ y'' - 3\ y' =2

    ... but of course it's a wrong way ...

    I had to remember that a second order non linear DE like (1), where all the variables x,y,y',y'' appear, is in general hard to attack...

    Sorry for the waste if time I caused...

    Kind regards

    \chi \sigma
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  6. #6
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    Not a waste of time - I just thought you had some cool magic!
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  7. #7
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    Quote Originally Posted by AkilMAI View Post
    Can someone please explain to me how to solve this equation:
    x*y''-y*y'=2...where I mus find the condition that the constants a and b must satisfy for y=ax+b to be a solution.Also how can I find the two solutions that satisfy y=3 when x=2.Thanks in advance
    This may help a bit but I can't see how to finish

    xy''-y\cdot y'=2

    Notice that

    \frac{d}{dx}\left( xy'\right)=xy''+y' \iff \frac{d}{dx}\left(xy'-y \right)=xy''

    and

    y\cdot y'=\frac{1}{2}\frac{d}{dx}y^2

    So we can rewrite the ODE as

    \frac{d}{dx}\left( xy'-y-\frac{1}{2}y^2\right)=2

    So we can integrate once to get

    xy'-y-\frac{1}{2}y^2=2x+C

    I can't see any obvious way to solve this.
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  8. #8
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    The ODE is now Ricatti. Note also that y = u-1 then we have

    2x u' = u^2 + 4x + 2c-1

    Further if we let

    u = -2x \frac{v'}{v}then the ODE becomes linear

    4x^2 v'' + 4xv' + (4x+2c-1)v = 0!
    Last edited by Ackbeet; May 19th 2011 at 11:45 AM.
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  9. #9
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    At the OP. Check on the ODE itself. Your question suggests that there is a solution of the form

    y = ax + b and as TheEmptySet points out - there's a contradiction.
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  10. #10
    MHF Contributor chisigma's Avatar
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    Another possible 'attack' to the DE...

    x\ y'' - y\ y'=2 (1)

    ... consists in supposing that y is analytic in x=2 so that is...

    y(x)= \sum_{n=0}^{\infty} \frac{y^{(n)} (2)}{n!}\ (x-2)^{n}= \sum_{n=0}^{\infty} a_{n}\ (x-2)^{n} (2)

    ... and that we know a_{0}= y(2) and a_{1}= y'(2). From (1) we derive first ...

    2\ y''(2)- y(2)\ y'(2) = 2 \implies y''(2)= 1+\frac{a_{0}\ a_{1}}{2} \implies a_{2}= \frac{1}{2}+ \frac{a_{0}\ a_{1}}{4} (2)

    ... and then...

    2\ y'''(2) + y''(2) -y(2)\ y''(2) - y'^{2}(2)=0 (3)

    ... that permits us to derive y'''(2) and then a_{3}= \frac{y'''(2)}{6}... and then all the derivatives of y in x=2 we want...

    Kind regards

    \chi \sigma
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