Question Number 116614 by bemath last updated on 05/Oct/20
$$\:\left(\mathrm{x}^{\mathrm{2}} \:\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{y}}\right)−\mathrm{y}\right)\mathrm{dx}\:+\:\mathrm{x}\:\mathrm{dy}\:=\:\mathrm{0} \\ $$$$\mathrm{where}\:\mathrm{y}\left(\mathrm{1}\right)=\:\frac{\pi}{\mathrm{4}} \\ $$
Commented by TANMAY PANACEA last updated on 05/Oct/20
$${i}\:{think}\:{it}\:{should}\:{be}\:{cos}^{\mathrm{2}} \left(\frac{{y}}{{x}}\right) \\ $$
Commented by bobhans last updated on 05/Oct/20
$$\mathrm{let}\:\frac{\mathrm{x}}{\mathrm{y}}\:=\:\mathrm{v}\:\Rightarrow\:\mathrm{y}\:=\:\mathrm{xv}^{−\mathrm{1}} \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{v}^{−\mathrm{1}} \:−\mathrm{xv}^{−\mathrm{2}} \:\frac{\mathrm{dv}}{\mathrm{dx}} \\ $$$$\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{y}−\mathrm{x}^{\mathrm{2}} \:\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{y}}\right)}{\mathrm{x}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{v}}−\frac{\mathrm{x}}{\mathrm{v}^{\mathrm{2}} }\:\frac{\mathrm{dv}}{\mathrm{dx}}\:=\:\frac{\frac{\mathrm{x}}{\mathrm{v}}−\mathrm{x}^{\mathrm{2}} \:\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{v}\right)}{\mathrm{x}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{v}}−\frac{\mathrm{x}\:\mathrm{dv}}{\mathrm{v}^{\mathrm{2}} \:\mathrm{dx}}\:=\:\frac{\mathrm{1}}{\mathrm{v}}\:−\mathrm{x}\:\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{v}\right) \\ $$$$\Rightarrow\frac{\mathrm{x}\:\mathrm{dv}}{\mathrm{v}^{\mathrm{2}} \:\mathrm{dx}}\:=\:\mathrm{x}\:\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{v}\right) \\ $$$$\Rightarrow\frac{\mathrm{dv}}{\mathrm{cos}\:^{\mathrm{2}} \:\left(\mathrm{v}\right)}\:=\:\mathrm{dx}\:;\:\int\:\mathrm{sec}\:^{\mathrm{2}} \left(\mathrm{v}\right)\:\mathrm{dv}\:=\:\int\:\mathrm{dx} \\ $$$$\Rightarrow\:\mathrm{tan}\:\left(\mathrm{v}\right)\:=\:\mathrm{x}+\mathrm{c} \\ $$$$\Rightarrow\:\mathrm{tan}\:\left(\frac{\mathrm{x}}{\mathrm{y}}\right)\:=\:\mathrm{x}+\mathrm{c}\:,\:\mathrm{from}\:\mathrm{given}\:\mathrm{condition} \\ $$$$\mathrm{y}\left(\mathrm{1}\right)=\frac{\pi}{\mathrm{4}}\:\Rightarrow\mathrm{tan}\:\left(\frac{\mathrm{4}}{\pi}\right)\:=\:\mathrm{1}+\:\mathrm{c}\: \\ $$$$\Rightarrow\mathrm{0}.\mathrm{022}−\mathrm{1}\:=\:\mathrm{c}\:,\:\mathrm{c}\:=−\mathrm{0}.\mathrm{978} \\ $$$$\mathrm{therefore}\:\mathrm{solution}\:\mathrm{is}\:\frac{\mathrm{x}}{\mathrm{y}}\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}−\mathrm{0}.\mathrm{978}\right) \\ $$$$\mathrm{or}\:\mathrm{y}\:=\:\frac{\mathrm{x}}{\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}−\mathrm{0}.\mathrm{978}\right)} \\ $$
Commented by bemath last updated on 05/Oct/20
$$\mathrm{sir}\:\mathrm{Tanmay}.\:\mathrm{in}\:\mathrm{my}\:\mathrm{book}\:\mathrm{the}\:\mathrm{question} \\ $$$$\mathrm{correct}\:\mathrm{sir} \\ $$
Answered by TANMAY PANACEA last updated on 05/Oct/20
$${recheck}\:{the}\:{problem}\:{pls} \\ $$
Answered by TANMAY PANACEA last updated on 05/Oct/20
$${assume}\:\boldsymbol{{cos}}^{\mathrm{2}} \left(\frac{\boldsymbol{{y}}}{\boldsymbol{{x}}}\right)\:\boldsymbol{{given}} \\ $$$$\boldsymbol{{xdy}}−\boldsymbol{{ydx}}+\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{cos}}^{\mathrm{2}} \left(\frac{\boldsymbol{{y}}}{\boldsymbol{{x}}}\right)\boldsymbol{{dx}}=\mathrm{0} \\ $$$$\frac{\boldsymbol{{xdy}}−\boldsymbol{{ydx}}}{\boldsymbol{{x}}^{\mathrm{2}} }×\boldsymbol{{sec}}^{\mathrm{2}} \left(\frac{\boldsymbol{{y}}}{\boldsymbol{{x}}}\right)+\boldsymbol{{dx}}=\boldsymbol{{dc}} \\ $$$$\boldsymbol{{sec}}^{\mathrm{2}} \left(\frac{\boldsymbol{{y}}}{\boldsymbol{{x}}}\right)\boldsymbol{{d}}\left(\frac{\boldsymbol{{y}}}{\boldsymbol{{x}}}\right)+\boldsymbol{{dx}}=\boldsymbol{{dc}} \\ $$$$\boldsymbol{{intregating}} \\ $$$$\boldsymbol{{tan}}\left(\frac{\boldsymbol{{y}}}{\boldsymbol{{x}}}\right)+\boldsymbol{{x}}=\boldsymbol{{c}} \\ $$