Question Number 91840 by M±th+et+s last updated on 03/May/20
$${solve}\: \\ $$$${y}\:{y}''=\left({y}'\right)^{\mathrm{2}} +{y}'\sqrt{{y}^{\mathrm{2}} +\left({y}'\right)^{\mathrm{2}} } \\ $$
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Question Number 91840 by M±th+et+s last updated on 03/May/20 | ||
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$${solve}\: \\ $$$${y}\:{y}''=\left({y}'\right)^{\mathrm{2}} +{y}'\sqrt{{y}^{\mathrm{2}} +\left({y}'\right)^{\mathrm{2}} } \\ $$ | ||