Differentiating we have
2xy + x2y' + y2 + 2xyy' = 0 Collecting terms, we have
y' = -(y2 + 2xy)/(x2 + 2xy)
(I assume it is ok to leave y' in terms of both x & y)
Mr C.
asked 10/04/19Differentiating we have
2xy + x2y' + y2 + 2xyy' = 0 Collecting terms, we have
y' = -(y2 + 2xy)/(x2 + 2xy)
(I assume it is ok to leave y' in terms of both x & y)
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