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anonymous
 4 years ago
dy/dx = y/(x+y^2cos y)
anonymous
 4 years ago
dy/dx = y/(x+y^2cos y)

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JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2This 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 firstorder ODE techniques.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2i.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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0invert the already inverted one?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2I mean you'd need to find the inverse function of x(y).

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2Your 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. \[ f^{1} = g \]
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