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Math Help - transforming from a non-exact to an exact linear differential equation

  1. #1
    Senior Member MacstersUndead's Avatar
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    transforming from a non-exact to an exact linear differential equation

    The question asks me to solve the following equation.

    3(x^2 + y^2)dx+x(x^2 + 3y^2 + 6y)dy = 0

    I verified that this form is not exact by differentiating M with respect to y and N with respect to x and saw they were not equal. the problem now is to find a suitable u(x,y) such that when I multiply it by the equation, the equation becomes exact.

    one of the ways I tried to transform the equation was to see if it was linear. what I got was

    x' - x (3y^2 + 6y)/(x^2+y^2) = -x^3 / (x^2 + y^2) ...

    hints would be greatly appreciated. also any good site on typing with Latex. all I get are 'unknown errors'
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  2. #2
    A Plied Mathematician
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    Re: transforming from a non-exact to an exact linear differential equation

    Try u(x,y)=h(x^{m}y^{n}).

    It may be more algebra, calculus, and differential equations than other methods, but it takes out some of the guesswork. What do you get with this?
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  3. #3
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    Re: transforming from a non-exact to an exact linear differential equation

    To add to Ackbeet's suggestion, try u(x,y)=h(y) only!
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    Senior Member MacstersUndead's Avatar
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    Re: transforming from a non-exact to an exact linear differential equation

    Quote Originally Posted by Ackbeet View Post
    Try u(x,y)=h(x^{m}y^{n}).

    It may be more algebra, calculus, and differential equations than other methods, but it takes out some of the guesswork. What do you get with this?
    Well sir, I honestly don't know, but I decided to math dangerously. If I take your advice and multiplied both sides of the original equation by h(x^{m}y^{n})., then I get

    h(x^{m}y^{n})M dx + h(x^{m}y^{n})N dy = 0

    If the equation is exact, then
    (d/dy) h(x^{m}y^{n})M = (d/dx) h(x^{m}y^{n})N which implies

    (d/dy)M / (d/dx)N = h(nx/my) ?
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  5. #5
    A Plied Mathematician
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    Re: transforming from a non-exact to an exact linear differential equation

    Hmm. Not quite sure I would go there. This is what I do. I do start out with your initial equation there:

    \frac{\partial}{\partial y}\left[h(x^{m}y^{n})M(x,y)\right]=\frac{\partial}{\partial x}\left[h(x^{m}y^{n})N(x,y)\right].

    But then I use the chain and product rules thus:

    ny^{n-1}h'(x^{m}y^{n})M(x,y)+h(x^{m}y^{n})\frac{\partial M(x,y)}{\partial y}=mx^{m-1}h'(x^{m}y^{n})N(x,y)+h(x^{m}y^{n})\frac{\partial N(x,y)}{\partial x}.

    You plug in what \frac{\partial M(x,y)}{\partial y} and \frac{\partial N(x,y)}{\partial x} are for your DE. Then, you let u=x^{m}y^{n}, and try to get a differential equation for h(u) that you can solve. Because you have lots of freedom here, you can often choose n and/or m judiciously to greatly simplify your problem.
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  6. #6
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    Re: transforming from a non-exact to an exact linear differential equation

    As suggested by Danny, it is easy to obtain h(y) = exp(y)
    Attached Thumbnails Attached Thumbnails transforming from a non-exact to an exact linear differential equation-integrating-factor.jpg  
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