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    Revising for a maths resit, simply cannot understand how you solve a non-separable first order differential equation by substitution. Have gone through forums, wikipedia, relevant chapters in my textbook and youtube tutorials and everyone seems to do something different, I can't translate it to any questions I attempt.
    If someone could explain how you're supposed to do this in teeny tiny baby steps I'll be very grateful, it's doing my nut in!
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    So maybe starting with a simple example? Say you have  \displaystyle x\frac{dy}{dx} =y , would you know where to start?
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    (Original post by B_9710)
    So maybe starting with a simple example? Say you have  \displaystyle x\frac{dy}{dx} =y , would you know where to start?
    I can do that one, because it's separable, could you take me through something like

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

    please? Very grateful for help!
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    Oh non separable. Apologies.
    So we have  \displaystyle 2xy \frac{dy}{dx}=-(x^2+y^2) .
    I'll be back in a few mins.
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    Are you given a substitution for that question? Usually you're given a substitution to use for a differential equation using substitution.
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    (Original post by dayflower2016)
    I can do that one, because it's separable, could you take me through something like

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

    please? Very grateful for help!
    Having a brief look, you can rearrange to make dy/dx the subject and then a substitution should "jump out", the following workings are maybe slightly tricky but should lead you to an answer. Some restrictions should probably be on x and y in this tbh but oh well
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    This is a Bernoulli type.
    If you rearrange it to give  \displaystyle \frac{dy}{dx} = -\frac{y}{x} - \frac{x}{y} so you should be able to see it is form the form  \displaystyle \frac{dy}{dx} = p(x)y+q(x)y^n .
    So we multiply through by  y and make the substitution  v=y^{2} . Then you can use an integrating factor.
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    (Original post by NeverLucky)
    Are you given a substitution for that question? Usually you're given a substitution to use for a differential equation using substitution.
    No substitution, I'm supposed to know how to work out what the substitution should be (which I don't). Also there are initial conditions given that y0=1 and x0=0
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    (Original post by B_9710)
    This is a Bernoulli type.
    If you rearrange it to give  \displaystyle \frac{dy}{dx} = -\frac{y}{x} - \frac{x}{y} so you should be able to see it is form the form  \displaystyle \frac{dy}{dx} = p(x)y+q(x)y^n .
    So we multiply through by  y and make the substitution  v=y^{2} . Then you can use an integrating factor.
    May I ask why you multiply through by  y before making the substitution? Won't it work anyway using the substitution without multiplying through?
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    (Original post by NeverLucky)
    May I ask why you multiply through by  y before making the substitution? Won't it work anyway using the substitution without multiplying through?
    Yeah but multiplying gets it into the standard form from which can easily see can be done via this sub
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    I get  \displaystyle y^2=\frac{A-x^3}{3x} for my GS.
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    (Original post by EnglishMuon)
    Yeah but multiplying gets it into the standard form from which can easily see can be done via this sub
    Hmm I'm a bit confused. If you multiply through by  y , then won't you get...

     y \dfrac {dy}{dx} = - \dfrac {y^2}{2x} - \dfrac {x}{2} ?

    And surely  \dfrac {dy}{dx} = - \dfrac {y}{2x} - \dfrac {x}{2y} is the standard form? Which can obviously be rearranged into:

     \dfrac {dy}{dx} + \dfrac {y}{2x} = - \dfrac {x}{2y} which is of the form:

     \dfrac {dy}{dx} + p(x)y = q(x)y^n
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    At a glance, you might be able to get away with making the sub u = \frac{y}{x} directly without any Bernoulli stuff.
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    (Original post by Zacken)
    At a glance, you might be able to get away with making the sub u = \frac{y}{x} directly without any Bernoulli stuff.
    That was my first thought, seems to work out. Get the same solution as above.
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    (Original post by 13 1 20 8 42)
    That was my first thought, seems to work out. Get the same solution as above.
    Nice, thought so; thanks!
 
 
 
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