Question Number 117396 by mnjuly1970 last updated on 11/Oct/20
$$\:\:\:\:\:\:\:\:…{differential}\:\:{equation}…\: \\ $$$$ \\ $$$$\:\:\:\:{solve}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{{xy}+\mathrm{2}{x}^{\mathrm{2}} {y}} \\ $$$$\:\:\:\:\:\:\:\:\:{general}\:\:{solution}\:=??? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{m}.{n}.\mathrm{1970} \\ $$$$\: \\ $$
Answered by mr W last updated on 11/Oct/20
$${ydy}=\frac{{dx}}{{x}\left(\mathrm{1}+\mathrm{2}{x}\right)} \\ $$$${ydy}=\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{2}}{\mathrm{2}{x}+\mathrm{1}}\right){dx} \\ $$$$\int{ydy}=\int\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{2}}{\mathrm{2}{x}+\mathrm{1}}\right){dx} \\ $$$$\frac{{y}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{ln}\:{x}−\mathrm{ln}\:\left(\mathrm{2}{x}+\mathrm{1}\right)+{C} \\ $$$$\Rightarrow{y}^{\mathrm{2}} =\mathrm{2}\:\mathrm{ln}\:\mid\frac{{Cx}}{\mathrm{2}{x}+\mathrm{1}}\mid \\ $$
Answered by Olaf last updated on 11/Oct/20
$$\:\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{1}}{{xy}\left(\mathrm{2}{x}+\mathrm{1}\right)} \\ $$$$\:\frac{{dy}}{{dx}}{y}\:=\:\frac{\mathrm{1}}{{x}\left(\mathrm{2}{x}+\mathrm{1}\right)}\:=\:\frac{\mathrm{1}}{{x}}−\frac{\mathrm{2}}{\mathrm{2}{x}+\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{y}^{\mathrm{2}} \:=\:\mathrm{ln}\mid{x}\mid−\mathrm{ln}\mid\mathrm{2}{x}+\mathrm{1}\mid+\mathrm{C} \\ $$$${y}\:=\:\pm\sqrt{\mathrm{2ln}\mid\frac{{x}}{\mathrm{2}{x}+\mathrm{1}}\mid+\mathrm{K}} \\ $$
Commented by mnjuly1970 last updated on 11/Oct/20
$${grateful}.. \\ $$
Answered by Dwaipayan Shikari last updated on 11/Oct/20
$$\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{{y}\left({x}+\mathrm{2}{x}^{\mathrm{2}} \right)} \\ $$$$\int{ydy}=\int\frac{{dx}}{{x}\left(\mathrm{1}+\mathrm{2}{x}\right)} \\ $$$$\frac{{y}^{\mathrm{2}} }{\mathrm{2}}=\int\frac{\mathrm{1}}{{x}}−\frac{\mathrm{2}}{\mathrm{1}+\mathrm{2}{x}}{dx} \\ $$$$\frac{{y}^{\mathrm{2}} }{\mathrm{2}}={log}\left({x}\right)−{log}\left(\mathrm{1}+\mathrm{2}{x}\right)+{C} \\ $$$$\frac{{y}^{\mathrm{2}} }{\mathrm{2}}={log}\left(\frac{{C}_{\mathrm{1}} {x}}{\mathrm{1}+\mathrm{2}{x}}\right) \\ $$$${y}=\pm\sqrt{{log}\left(\frac{{C}_{\mathrm{1}} ^{\mathrm{2}} {x}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{2}} }\right)} \\ $$
Commented by mnjuly1970 last updated on 11/Oct/20
$${thank}\:{you}\:{so}\:{much}.. \\ $$