Question Number 14288 by tawa tawa last updated on 30/May/17
$$\mathrm{Solve}:\:\:\mathrm{y}'\:+\:\frac{\mathrm{y}}{\mathrm{x}}\:=\:\mathrm{0} \\ $$
Answered by Tinkutara last updated on 30/May/17
$$\frac{{dy}}{{dx}}\:=\:−\:\frac{{y}}{{x}}\: \\ $$$$\frac{{dy}}{{y}}\:=\:−\:\frac{{dx}}{{x}} \\ $$$$\mathrm{ln}\:{y}\:=\:−\:\mathrm{ln}\:{x}\:+\:{C} \\ $$$$\mathrm{ln}\:{xy}\:=\:{C} \\ $$$${xy}\:=\:{e}^{{C}} \\ $$
Commented by Tinkutara last updated on 30/May/17
$$\mathrm{Thanks}! \\ $$
Commented by tawa tawa last updated on 30/May/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by prakash jain last updated on 30/May/17
$${xy}={e}^{{C}} \\ $$$${can}\:{also}\:{be}\:{written}\:{as} \\ $$$${xy}={c} \\ $$