Question Number 83639 by niroj last updated on 04/Mar/20
$$ \\ $$$$\: \\ $$$$\:\boldsymbol{\mathrm{Find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{differential}}\:\boldsymbol{\mathrm{equations}}: \\ $$$$\:\:\:\left(\mathrm{i}\right)\:\mathrm{log}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\mathrm{ax}+\mathrm{by} \\ $$$$\:\:\:\left(\mathrm{ii}\right)\:\mathrm{x}\:\mathrm{cos}\:\mathrm{y}\:\mathrm{dy}=\left(\mathrm{x}\:\mathrm{e}^{\mathrm{x}} \mathrm{log}\:\mathrm{x}\:+\mathrm{e}^{\mathrm{x}} \right)\mathrm{dx} \\ $$$$ \\ $$
Answered by TANMAY PANACEA last updated on 05/Mar/20
$$\left.\mathrm{1}\right){ln}\left(\frac{{dy}}{{dx}}\right)={ax}+{by} \\ $$$$\frac{{dy}}{{dx}}={e}^{{ax}+{by}} \\ $$$$\frac{{dy}}{{e}^{{by}} }={e}^{{ax}} {dx} \\ $$$$\int{e}^{−{by}} {dy}=\int{e}^{{ax}} {dx} \\ $$$$\frac{{e}^{−{by}} }{−{b}}=\frac{{e}^{{ax}} }{{a}}+{c} \\ $$
Commented by niroj last updated on 05/Mar/20
$${its}\:\bigstar\:{great} \\ $$
Commented by TANMAY PANACEA last updated on 05/Mar/20
$${most}\:{welcome}\: \\ $$
Answered by TANMAY PANACEA last updated on 05/Mar/20
$$\left.\mathrm{2}\right){xcosydy}−\left({xe}^{{x}} {lnx}+{e}^{{x}} \right){dx} \\ $$$$\int{cosydy}−\int\left({e}^{{x}} {lnx}+\frac{{e}^{{x}} }{{x}}\right){dx}=\mathrm{0} \\ $$$$\int{cosydy}−\int\frac{{d}}{{dx}}\left({e}^{{x}} {lnx}\right){dx}=\mathrm{0} \\ $$$${siny}−{e}^{{x}} {lnx}={c} \\ $$
Commented by niroj last updated on 05/Mar/20
$$\:{thank}\:{you}\:{panacea}. \\ $$
Commented by TANMAY PANACEA last updated on 05/Mar/20
$${most}\:{welcome} \\ $$
Answered by M±th+et£s last updated on 05/Mar/20
$${log}\left(\frac{{dy}}{{dx}}\right)={ax}+{by}\rightarrow{log}\left({p}\right)={ax}+{by}\rightarrow{y}=\frac{\mathrm{1}}{{b}}\left({log}\left({p}\right)−{ax}\right) \\ $$$${Diff}.\:{w}.{r}.{t}.\:{x}\rightarrow{p}=\frac{\mathrm{1}}{{b}}\left(\frac{{d}\:{log}\left({p}\right)}{{dx}}\:−{a}\right) \\ $$$${p}=\frac{\mathrm{1}}{{b}}\left(\frac{\mathrm{1}}{{p}}\:\frac{{dp}}{{dx}}−{a}\right)\rightarrow\frac{{dx}}{{dp}}=\frac{\mathrm{1}}{{p}\left({a}+{bp}\right)}=\frac{\mathrm{1}}{{a}}\:\frac{\mathrm{1}}{{p}}−\frac{{b}}{{a}}\:\frac{\mathrm{1}}{{a}+{bp}} \\ $$$${x}=\frac{\mathrm{1}}{{a}}{ln}\mid{p}\mid−\frac{\mathrm{1}}{{a}}{ln}\mid{a}+{bp}\mid+\frac{\mathrm{1}}{{a}}{ln}\left({c}\right) \\ $$$${x}=\frac{\mathrm{1}}{{a}}{ln}\left(\frac{{cp}}{{a}+{bp}}\right)\rightarrow{e}^{{ax}} =\frac{{cp}}{{a}+{bp}}\rightarrow{p}=\frac{{ae}^{{ax}} }{{c}−{be}^{{ax}} } \\ $$$${y}=\:−\frac{{ln}\left({c}−{be}^{{ax}} \right)}{{b}} \\ $$$$ \\ $$$$ \\ $$
Answered by M±th+et£s last updated on 05/Mar/20
$${cos}\left({y}\right){dy}=\frac{{xe}^{{x}} \:{log}\left({x}\right)+{e}^{{x}} }{{x}}{dx} \\ $$$$\rightarrow{sin}\left({y}\right)=\int\left({e}^{{x}} {log}\left({x}\right)+\frac{{e}^{{x}} }{{x}}\right){dx}=\int{e}^{{x}} {log}\left({x}\right){dx}+{e}^{{x}} {log}\left({x}\right)−\int{e}^{{x}} \:{log}\left({x}\right)\:{dx} \\ $$$${sin}\left({y}\right)={e}^{{x}} {log}\left({x}\right)+{c}\rightarrow{y}={sin}^{−\mathrm{1}} \left({c}+{e}^{{x}} {log}\left({x}\right)\right) \\ $$$$ \\ $$