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anonymous

  • 5 years ago

dy/dx = y/(x+y^2cos y)

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  1. JamesJ
    • 5 years ago
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    This problem really belongs in mathematics, but it's so good I'll try and answer it here. It's easiest to think of this as dx/dy = (x + y^2 cos y)/y = x/y + y cos y or x' - (1/y)x = y cos y Given that form, you can solve using standard first-order ODE techniques.

  2. JamesJ
    • 5 years ago
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    i.e., x(y) = Ay + By.sin y Now you need to invert that equation. As you can see, the inverse in general won't be pretty.

  3. anonymous
    • 5 years ago
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    thanks dude

  4. anonymous
    • 5 years ago
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    invert the already inverted one?

  5. JamesJ
    • 5 years ago
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    I mean you'd need to find the inverse function of x(y).

  6. anonymous
    • 5 years ago
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    please explain

  7. JamesJ
    • 5 years ago
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    Your original equation implicitly asks for a solution y = f(x), y as a function of x. The solution I've provided gives x as a function of y. Hence if you actually need y = f(x), then you'll need to find the inverse function of x = g(y) = Ay + By.sin y The inverse function of g will be the function f. \[ f^{-1} = g \]

  8. anonymous
    • 5 years ago
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    Thanks again homie

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