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3xyy' = 3y2 + 4x sqrt(x2+y2) For​ x, y>​0, a general solution is ____?

Posted: Wed Jul 06, 2022 12:16 pm
by answerhappygod
3xyy' = 3y2 +4x sqrt(x2+y2)
For​ x, y>​0, a general solution is ____?