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