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    how do i solve
    dy/dx=(sqrt(x^2+y^2)-x)/y
    and
    dy/dx=(3x^4+y^3)/(3xy^2)

    so far i have:
    dy/dx=(sqrt(x^2+y^2)-x)/y
    y=vx dy/dx=xdv/dx+v
    xdv/dx+v=(xsqrt(1+v^2)-x)/vx=(sqrt(1+v^2)-1)/v
    xdv/dx=(sqrt(1+v^2)-1-v^2)/v
    d/dx(1/x)=d/dv(v/(sqrt(1+v^2)-1-v^2))
    lnx-ln(sqrt(v^2+1)-1)=0
    lnx-ln(sqrt((y/x)^2+1)-1)=0
    but this is wrong
    and
    dy/dx=(3x^4+y^3)/(3xy^2)
    dy/dx-(y/3x)=3x^3y^-2
    v=y^3
    dv/dx-(v/x)=9x^3
    d/dx(v/x)=9x^2
    v/x=y^3/x=3x^3
    y^3=3x^4
    but this too is wrong

    and i know i don't have any +c parts in here, for the question they don't need to be included

    and if anyone knows a good website that'd spit out the answer for me that would be equally helpful, i've tried wolfram and quickmath and the answers come out wrong (and by wrong i mean when i input them on the test i don't get the mark)
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     \frac{dy}{dx} = \dfrac{\sqrt{x^2+y^2-x}}{y}

     y dy = \sqrt{x^2+y^2-x}     dx

    hmm im gona do it on paper for a bit
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    Those equations are non-linear and scary and thus I'm sorry, I haven't a clue.

    You sure you're supposed to solve these?
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    (Original post by 3nTr0pY)
    Those equations are non-linear and scary and thus I'm sorry, I haven't a clue.

    You sure you're supposed to solve these?
    well thanks for trying anyhow

    and if it makes it easier the -x shouldn't be under the square root sign

    and i'm sure, question says find a general solution to the differential equation... , wish i didn't have to
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    2. I would rearrange to dy/dx - (y/3x) = x^3/y^2
    Then use the Integating factor which would be 1/x^(1/3)
    Then times the integrating factor by both sides
    Use the chain rule on the LHS and they integate both sides and rearrange to find y.
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    (Original post by abbii)
    well thanks for trying anyhow

    and if it makes it easier the -x shouldn't be under the square root sign

    and i'm sure, question says find a general solution to the differential equation... , wish i didn't have to
    To be honest, I didn't really try. I saw them, got scared and ran away. (after posting about it)

    Not very helpful of me, I'm afraid. Presumably there's some clever way that I can't think of for doing it. Other than the really clever way of just putting it into Wolframalpha, although whoever's marking it may not like that!
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    (Original post by murrayrj)
    2. I would rearrange to dy/dx - (y/3x) = x^3/y^2
    Then use the Integating factor which would be 1/x^(1/3)
    Then times the integrating factor by both sides
    Use the chain rule on the LHS and they integate both sides and rearrange to find y.
    you would have to rearrange to dy/dx - (y/3x) = 3x^3/y^2
    and integrating factor would surely be x^(-1/3)
    then d/dx(yx^(-1/3))=3x^3/y^2
    but you can't directly integrate 3x^3/y^2 because of the y term
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    (Original post by 3nTr0pY)
    To be honest, I didn't really try. I saw them, got scared and ran away. (after posting about it)

    Not very helpful of me, I'm afraid. Presumably there's some clever way that I can't think of for doing it. Other than the really clever way of just putting it into Wolframalpha, although whoever's marking it may not like that!
    presumably, and i used wolfram alpha for all the actual integration, but if you can solve the whole differential equation using it i don't know how, and it's an online test so i literally just need the answer
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    x^(-1/3) = 1/x^(1/3)
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    (Original post by abbii)
    presumably, and i used wolfram alpha for all the actual integration, but if you can solve the whole differential equation using it i don't know how, and it's an online test so i literally just need the answer
    http://www.wolframalpha.com/input/?i...2%29-x%29%2Fy+

    There is a God and yes, he does have all the answers.
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    I haven't had a proper look, but perhaps letting z=\dfrac{y}{x} and making the appropriate substitutions would work...
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    (Original post by murrayrj)
    x^(-1/3) = 1/x^(1/3)
    ok, that was me being thick
    you still can't use that method though
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    (Original post by olipal)
    I haven't had a proper look, but perhaps letting z=\dfrac{y}{x} and making the appropriate substitutions would work...
    i did with the first, and with the second it's bernoulli so you use a different substitution
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    (Original post by 3nTr0pY)
    http://www.wolframalpha.com/input/?i...2%29-x%29%2Fy+

    There is a God and yes, he does have all the answers.
    there is a god indeed
    you are brilliant, that link is going to be the most useful thing in the world
    (it only worked for the second by i only needed one so that's fine by me)
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    (Original post by abbii)
    ok, that was me being thick
    you still can't use that method though
    Yeah that method is only useful at A-Level: I didn't realise this was an "Undergraduate" question.
    If I may ask, what method of solving differential equations are you learning about/doing for these questions? As I may read up on them.
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    (Original post by murrayrj)
    Yeah that method is only useful at A-Level: I didn't realise this was an "Undergraduate" question.
    If I may ask, what method of solving differential equations are you learning about/doing for these questions? As I may read up on them.
    yes, id be interested in this too. I have only learned the integration factor method.
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    (Original post by murrayrj)
    Yeah that method is only useful at A-Level: I didn't realise this was an "Undergraduate" question.
    If I may ask, what method of solving differential equations are you learning about/doing for these questions? As I may read up on them.

    (Original post by meraphox)
    yes, id be interested in this too. I have only learned the integration factor method.
    for both integration by substitution, the first using y=vx (although i got it wrong so not sure if that's right) and the second v=y^3 since it is a bernoulli equation (that is to say of the form dy/dx+p(x)y=q(x)y^n) http://www.sosmath.com/diffeq/first/...ernouilli.html and you can probably find a derivation of that substitution if you're interested but i don't know it, each substitution removes the y variable to leave a simpler differential equation in terms of v and x that can be integrated by some known method

    if you want to read up i've found this site useful http://www.sosmath.com/diffeq/first/first.html it explains it all clearly (at least the pages i've looked at) and other than bifurcations i think we've covered everything there this term, in my first year (i'm in my second) we covered further maths methods and variation of parameters http://www.sosmath.com/diffeq/second...variation.html and partial differentiation so maybe look at them first, i'm assuming that you probably do further maths if you've done integrating factors so know about separable variables and y=yh+yp (yh= the homogenous solution and yp= some particular solution)
 
 
 
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