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Math Help - how to determine separable & linear?

  1. #1
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    how to determine separable & linear?

    for separable, is it something that u can put all x to one side, and all y to another side ,then integrat both side?

    what about linear?
    and sometime it could be both linear and separable

    they keep confusing me...

    here i got a few examples, can someone explain?

    (1+x^2)dy/dx = 1 -2xy

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

    x/2 dy/dx = x^2 -y

    dy/dx = [(2x)(sqrt y)]/sqrt(1+x^2)

    thank you
    Last edited by mr fantastic; November 4th 2009 at 05:09 PM. Reason: Removed excessive ?'s in post title
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  2. #2
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    The way I distinct them is by trying to get them into the two basic forms for separable and linear differential equations; that is

    y' +f(x)y = g(x) for linear and
    f(y)y' = g(x)x' for separable

    For instance, in your examples;

    The first,
    (1+x^2)\frac{dy}{dx} = 1 -2xy [(1) say]
    can be rewritten as
    \frac{dy}{dx} +\frac{2x}{1+x^2}y =  \frac{1}{1+x^2} (assuming 1+x^2 can never be zero, in other words, no complex roots)
    thus is of form
    y' + f(x)y = g(x)
    where
    f(x)=\frac{2x}{1+x^2} and
    g(x)=\frac{1}{1+x^2}
    and so example (1) can be considered a linear differential equation.

    The second,
    (1+x^2)\frac{dy}{dx} = y - 2xy [(2) say]
    is much like (1) but, by writing y - 2xy as y(1-2x) you have a product of x terms and y terms (this is often - not always - a *hint* that the problem has a separable solution)
    So, lets consider than the problem is separable, then the form should be like
    f(y)y' = g(x)x'
    that is
    (1+x^2)\frac{dy}{dx} = y(1 - 2x)
    which then (by dividing by y and (1+x^2) - now watch since you're dividing by y there may be a singular solution at y=0 (or the general solution may include it) ) gives
    \frac{1}{y}\frac{dy}{dx} = \frac{1 - 2x}{1+x^2}
    then taking the integral of both sides with respect to x (you can think of this as multiplying each side by 'dx' and putting an integral sign on the front if its easier to remember - often it is) you get
    \int\!\frac{1}{y}\,dy = \int\!\frac{1 - 2x}{1+x^2}\,dx
    thus (2) can be considered a separable differential equation.

    Your third,
    \frac{x}{2} \frac{dy}{dx} = x^2 -y [3]
    looks like it'd be a separable (mainly because of that x/2 factor) however quickly thinking over it would give a y/x term which could add more work than the alternative - which would be to rewrite this in linear form)
    Now [3] can be written as
    \frac{x}{2} \frac{dy}{dx} +y = x^2 (taking y to the other side) and then,
    \frac{dy}{dx} +\frac{2}{x}y = \frac{2}{x}(x^2) (dividing by x/2, in other words multiplying by 2/x and not letting x be 0)
    Now the above is in the general form for a linear differential equation where our f(x) and g(x) are
    f(x)=\frac{2}{x} and
    g(x)=\frac{2}{x}x^2
    thus [3] can be solved as a linear differential equation

    Finally [4],
    \frac{dy}{dx} = \frac{2x\sqrt{y}}{\sqrt{1+x^2}}
    well, right away (if you can notice that the entire right hand side is a product of x and y terms and the left hand side has only y' ) theres a great this differential equation is separable. So, take sqrt(y) over to the left to get (meaning there could (and likely is) a singular solution at y = 0 not in the general solution)
    \frac{1}{\sqrt{y}}\frac{dy}{dx} = \frac{2x}{\sqrt{1+x^2}}
    then integrating with respect to x on both sides gives
    \int\!\frac{1}{\sqrt{y}}\,dy = \int\!\frac{2x}{\sqrt{1+x^2}}\,dx
    and so [4] can be solved by separating out the x-terms and y-terms.

    Generally, in practice, with a lot of math, the best method tends to come from thinking about the problem because more often than not you'll get a differential equation that isn't in standard form and you'll need to look for the best method to use. Sometimes - though unwise - it's still a good idea to plough through the equation assuming it can be linear (for example) and you hit a dead end or a really hard equation in which case, turn back and assume it is separable - and hopefully you'll find a nicer looking expression to solve.

    I hope some ideas of how make quicker deductions about an equation being separable or linear came from the above.

    Edit:
    Also, I though I might mention that there are occasions that even if you do get a nice looking, say separable, equation at the end - it may be impossible to solve in which case, the solution might lay with the linear equation - even if it looks incredibly 'horrible' and difficult to solve compared to the separable one at first.
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  3. #3
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    thx for your PERFECT reply :P

    just one more question, for the second equation (1+x^2)dy/dx = y - 2xy, the answer in my book says that its both linear and separable, what does it mean?
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  4. #4
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    [tex](1 + x^2)\frac{dy}{dx} = y - 2xy


    If you rearrange it like this:

    (1 + x^2)\frac{dy}{dx} = y(1 - 2x)

    \frac{1}{y}\frac{dy}{dx} = \frac{1 - 2x}{1 + x^2}

    You can see it is a separable DE (though the integral on the RHS will be difficult...)


    If you rearrange it like this:

    (1 + x^2)\frac{dy}{dx} = (1 - 2x)y

    (1 + x^2)\frac{dy}{dx} + (2x - 1)y = 0

    \frac{dy}{dx} + \frac{2x - 1}{1 + x^2}y = 0

    You can see that it is a first-order linear DE (though finding the integrating factor will be difficult...)


    So this DE is BOTH separable AND first order linear.
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