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anonymous

  • 5 years ago

(2x^2 + y^2)dx + 2xy=0 Solve DE

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  1. anonymous
    • 5 years ago
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    I think you might be missing a dy somewhere. Are you?

  2. anonymous
    • 5 years ago
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    If your problem is (2x^2 + y^2)dx + 2xy dy = 0, then this is equivalent to (2x^2 + y^2) + 2xy dy/dx = 0, or 2xy y' + y^2 = -2x^2. If you know the method of integrating factors, this problem is now easy. The idea is that you might recognize the Left Hand Side as the derivative of the product [x*y^2] with respect to x. Why? Because if you take the derivative of x*y^2 with respect to x, you use the product rule and you would get y^2 + x*(2y)*y' (the y' comes from the chain rule since y is a function of x). So now in our equation, if the Left Hand Side is the derivative of the product of x*y^2, then so is the Right Hand Side. So integrate both sides to get that x*y^2 = -2(x^3)/3 + C. Then solve for y if you want an explicit solution, or you can move things around so you get C on the Right Hand Side and everything else on the Left to get an implicit solution.

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