Question Number 166764 by amin96 last updated on 27/Feb/22
$$\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}=\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{y}}−\boldsymbol{\mathrm{x}}\right)\:\: \\ $$
Answered by mr W last updated on 27/Feb/22
$${let}\:{u}={y}−{x} \\ $$$${y}={u}+{x} \\ $$$$\frac{{dy}}{{dx}}=\frac{{du}}{{dx}}+\mathrm{1} \\ $$$$\frac{{du}}{{dx}}+\mathrm{1}=\mathrm{cos}\:{u} \\ $$$$\frac{{du}}{\mathrm{cos}\:{u}−\mathrm{1}}={dx} \\ $$$$\int\frac{{du}}{\mathrm{cos}\:{u}−\mathrm{1}}=\int{dx} \\ $$$$\int\frac{{d}\left(\frac{{u}}{\mathrm{2}}\right)}{−\mathrm{sin}^{\mathrm{2}} \:\frac{{u}}{\mathrm{2}}}=\int{dx} \\ $$$$\int{d}\left(\mathrm{cot}\:\frac{{u}}{\mathrm{2}}\right)=\int{dx} \\ $$$$\mathrm{cot}\:\frac{{u}}{\mathrm{2}}={x}+{C} \\ $$$${u}=\mathrm{2}\:\mathrm{cot}^{−\mathrm{1}} \left({x}+{C}\right) \\ $$$${y}=\mathrm{2}\:\mathrm{cot}^{−\mathrm{1}} \left({x}+{C}\right)+{x} \\ $$
Commented by amin96 last updated on 27/Feb/22
$$\frac{\boldsymbol{{du}}}{\boldsymbol{{cosu}}−\mathrm{1}}=\boldsymbol{{dx}}\:\:\:\:\:\:\int\frac{\boldsymbol{{du}}}{\boldsymbol{{cosu}}−\mathrm{1}}=\int\boldsymbol{{dx}} \\ $$$$\frac{\mathrm{1}+\boldsymbol{{cosu}}}{\boldsymbol{{sinu}}}=\boldsymbol{{x}}+\boldsymbol{{c}}\:\:\:\:\:\:\:\boldsymbol{{cosec}}\left(\boldsymbol{{u}}\right)+\boldsymbol{{ctg}}\left(\boldsymbol{{u}}\right)=\boldsymbol{{x}}+\boldsymbol{{c}} \\ $$$$\boldsymbol{\mathrm{cot}}\left(\frac{\boldsymbol{\mathrm{u}}}{\mathrm{2}}\right)=\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{c}} \\ $$$$\frac{\boldsymbol{\mathrm{u}}}{\mathrm{2}}=\boldsymbol{\mathrm{arccot}}\left(\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{c}}\right)\:\:\:\:\:\:\:\boldsymbol{\mathrm{u}}=\mathrm{2}\boldsymbol{\mathrm{arccot}}\left(\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{c}}\right) \\ $$$$\boldsymbol{\mathrm{y}}=\mathrm{2}\boldsymbol{\mathrm{arccot}}\left(\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{c}}\right)+\boldsymbol{\mathrm{x}} \\ $$
Commented by amin96 last updated on 27/Feb/22
$$\boldsymbol{{y}}−\boldsymbol{{x}}=\mathrm{2}\pi\boldsymbol{{k}}\:\:\:\:\:\boldsymbol{{k}}\in\boldsymbol{{Z}}\:\:\:\:{its}\:{correct}? \\ $$