Question Number 85111 by niroj last updated on 19/Mar/20
$$\:\boldsymbol{\mathrm{Solve}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{differential}}\:\boldsymbol{\mathrm{equation}}: \\ $$$$\:\bigstar.\left(\mathrm{1}+\mathrm{x}+\mathrm{xy}^{\mathrm{2}} \right)\mathrm{dy}+\left(\mathrm{y}+\mathrm{y}^{\mathrm{3}} \right)\mathrm{dx} \\ $$$$\: \\ $$
Commented by jagoll last updated on 19/Mar/20
$$\left(\mathrm{1}+\mathrm{x}+\mathrm{xy}^{\mathrm{2}} \right)\mathrm{dy}+\left(\mathrm{y}+\mathrm{y}^{\mathrm{3}} \right)\mathrm{dx} \\ $$$$\mathrm{nothing}\:−\mathrm{1}\:\mathrm{sir} \\ $$
Commented by jagoll last updated on 19/Mar/20
$$\left(\mathrm{1}+\mathrm{x}+\mathrm{xy}^{\mathrm{2}} \right)\mathrm{dy}+\left(\mathrm{y}+\mathrm{y}^{\mathrm{3}} \right)\mathrm{dx}\:=\mathrm{0}?? \\ $$
Commented by mr W last updated on 19/Mar/20
$$\left(\mathrm{1}+\mathrm{x}+\mathrm{xy}^{\mathrm{2}} \right)\mathrm{dy}+\left(\mathrm{y}+\mathrm{y}^{\mathrm{3}} \right)\mathrm{dx}\:{is}\:{no}\:{eqn}.\:! \\ $$$${it}\:{should}\:{be}: \\ $$$$\left(\mathrm{1}+\mathrm{x}+\mathrm{xy}^{\mathrm{2}} \right)\mathrm{dy}+\left(\mathrm{y}+\mathrm{y}^{\mathrm{3}} \right)\mathrm{dx}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{1}+\mathrm{x}\left(\mathrm{1}+\mathrm{y}^{\mathrm{2}} \right)+{y}\left(\mathrm{1}+\mathrm{y}^{\mathrm{2}} \right)\frac{{dx}}{{dy}}=\mathrm{0} \\ $$$$\mathrm{x}\left(\mathrm{1}+\mathrm{y}^{\mathrm{2}} \right)+{y}\left(\mathrm{1}+\mathrm{y}^{\mathrm{2}} \right)\frac{{dx}}{{dy}}=−\mathrm{1} \\ $$$$\frac{{dx}}{{dy}}+\frac{{x}}{{y}}=−\frac{\mathrm{1}}{{y}\left(\mathrm{1}+{y}^{\mathrm{2}} \right)} \\ $$
Commented by jagoll last updated on 19/Mar/20
$$\mathrm{okay}\:\mathrm{sir}. \\ $$
Answered by mr W last updated on 19/Mar/20
$$\left(\mathrm{1}+{x}+{xy}^{\mathrm{2}} \right){dy}+\left({y}+{y}^{\mathrm{3}} \right){dx}=\mathrm{0} \\ $$$${y}\left(\mathrm{1}+{y}^{\mathrm{2}} \right)\frac{{dx}}{{dy}}+{x}\left(\mathrm{1}+{y}^{\mathrm{2}} \right)=−\mathrm{1} \\ $$$$\Rightarrow\frac{{dx}}{{dy}}+\frac{\mathrm{1}}{{y}}{x}=−\frac{\mathrm{1}}{{y}\left(\mathrm{1}+{y}^{\mathrm{2}} \right)} \\ $$$$\int\frac{{dy}}{{y}}=\mathrm{ln}\:{y}\:\Rightarrow{IF}={u}\left({y}\right)={e}^{\mathrm{ln}\:{y}} ={y} \\ $$$$\Rightarrow{x}=\frac{−\int\frac{{u}\left({y}\right){dy}}{{y}\left(\mathrm{1}+{y}^{\mathrm{2}} \right)}+{C}}{{u}\left({y}\right)}=−\frac{\mathrm{1}}{{y}}\left(\int\frac{{dy}}{\mathrm{1}+{y}^{\mathrm{2}} }−{C}\right) \\ $$$$\Rightarrow{x}=−\frac{\mathrm{1}}{{y}}\left(\mathrm{tan}^{−\mathrm{1}} {y}−{C}\right) \\ $$$$\Rightarrow{xy}+\mathrm{tan}^{−\mathrm{1}} {y}={C} \\ $$
Commented by jagoll last updated on 19/Mar/20
$$\mathrm{how}\:\mathrm{get}\:−\mathrm{1}\:\mathrm{sir}?\:\mathrm{the}\:\mathrm{original}\:\mathrm{question} \\ $$$$\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{see}\:−\mathrm{1} \\ $$
Commented by mr W last updated on 19/Mar/20
$${please}\:{check}\:{again},\:{you}\:{can}\:{see}\:{why}\:−\mathrm{1}. \\ $$
Commented by niroj last updated on 19/Mar/20
$${thanks}\:{both}\:{of}\:{mr}.\:{w}\:\&\:{jagoll} \\ $$$${for}\:{your}\:{effort}..\:{answer}\:{is}\:{done}. \\ $$$$\:\:{equation}\:{always}\:{something}\:{to}\:{equal} \\ $$$${otherwise}\:{put}\:\mathrm{0}\:{if}\:{not}\:{mention}\:{any}\:{orbitary}\:{values}. \\ $$$$\: \\ $$
Commented by mr W last updated on 19/Mar/20
$${you}\:{have}\:{had}\:{saved}\:{us}\:{alot}\:{of}\:{time}\:{if} \\ $$$${you}\:{had}\:{given}\:{the}\:{question}\:{complett} \\ $$$${right}\:{away},\:{sir}… \\ $$