Differential Equation Questions

Question Number 100594 by Coronavirus last updated on 27/Jun/20

$${solve}\:\:{the}\:{differential}\:\:{equations} \\$$$$\mathrm{1}-\:\:{x}\mathrm{cos}\:\left(\mathrm{ln}\:\frac{{x}}{{y}}\right){dy}−{ydx}=\mathrm{0} \\$$$$\mathrm{2}-\:\:{ydx}+\mathrm{2}{xdy}\:=\mathrm{2}{y}\frac{\sqrt{{x}}}{{cos}^{\mathrm{2}} \left({y}\right)}{dy}\:\:\:\:\:{y}\left(\mathrm{0}\right)=\pi \\$$

Answered by smridha last updated on 27/Jun/20

$$\mathrm{2}.\left[\frac{\boldsymbol{{dx}}}{\mathrm{2}\sqrt{\boldsymbol{{x}}}}+\frac{\sqrt{\boldsymbol{{x}}}}{\boldsymbol{{y}}}\boldsymbol{{dy}}=\boldsymbol{{sec}}^{\mathrm{2}} \left(\boldsymbol{{y}}\right)\boldsymbol{{dy}}\right] \\$$$$\boldsymbol{{mult}}:\:\boldsymbol{{by}}\:\boldsymbol{{y}}\:\boldsymbol{{both}}\:\boldsymbol{{sides}}\:\boldsymbol{{and}}\:\boldsymbol{{integrating}} \\$$$$\int\boldsymbol{{d}}\left(\boldsymbol{{y}}.\sqrt{\boldsymbol{{x}}}\right)=\int\boldsymbol{{y}}.\boldsymbol{{sec}}^{\mathrm{2}} \left(\boldsymbol{{y}}\right)\boldsymbol{{dy}} \\$$$$\Rightarrow\boldsymbol{{y}}\sqrt{\boldsymbol{{x}}}=\boldsymbol{{ytan}}\left(\boldsymbol{{y}}\right)+\boldsymbol{{ln}}\left[\boldsymbol{{cos}}\left(\boldsymbol{{y}}\right)\right]+\boldsymbol{{c}} \\$$$$\boldsymbol{{put}}\:\boldsymbol{{the}}\:\boldsymbol{{condition}}\:\boldsymbol{{y}}\left(\mathrm{0}\right)=\boldsymbol{\pi} \\$$$$\Rightarrow\mathrm{0}=\mathrm{0}+\boldsymbol{{ln}}\left(−\mathrm{1}\right)+\boldsymbol{{c}} \\$$$$\boldsymbol{{so}}\:\boldsymbol{{c}}=\underset{−} {+}\boldsymbol{{i}\pi} \\$$$$\boldsymbol{{so}}\:\boldsymbol{{the}}\:\boldsymbol{{solution}}\: \\$$$$\boldsymbol{{y}}\sqrt{\boldsymbol{{x}}}=\boldsymbol{{ytan}}\left(\boldsymbol{{y}}\right)+\boldsymbol{{ln}}\left[\boldsymbol{{cos}}\left(\boldsymbol{{y}}\right)\right]\underset{−} {+}\boldsymbol{{i}\pi} \\$$

Answered by smridha last updated on 28/Jun/20

$$\left(\mathrm{1}\right).\frac{\boldsymbol{{dx}}}{\boldsymbol{{dy}}}=\frac{\boldsymbol{{x}}}{\mathrm{2}\boldsymbol{{y}}}\left[\left(\frac{\boldsymbol{{x}}}{\boldsymbol{{y}}}\right)^{\boldsymbol{{i}}} +\left(\frac{\boldsymbol{{x}}}{\boldsymbol{{y}}}\right)^{−\boldsymbol{{i}}} \right] \\$$$$\boldsymbol{{let}}\:\:\:\frac{\boldsymbol{{x}}}{\boldsymbol{{y}}}=\boldsymbol{{v}}\:\boldsymbol{{so}}\:\frac{\boldsymbol{{dx}}}{\boldsymbol{{dy}}}=\boldsymbol{{v}}+\boldsymbol{{y}}.\frac{\boldsymbol{{dv}}}{\boldsymbol{{dy}}} \\$$$$\boldsymbol{{now}}\: \\$$$$\:\:\:\boldsymbol{{v}}+\boldsymbol{{y}}.\frac{\boldsymbol{{dv}}}{\boldsymbol{{dy}}}=\frac{\boldsymbol{{v}}}{\mathrm{2}}\left[\frac{\boldsymbol{{v}}^{\mathrm{2}\boldsymbol{{i}}} +\mathrm{1}}{\boldsymbol{{v}}^{\boldsymbol{{i}}} }\right] \\$$$$\boldsymbol{{y}}.\frac{\boldsymbol{{dv}}}{\boldsymbol{{dy}}}=\boldsymbol{{v}}\left[\frac{\left(\boldsymbol{{v}}^{\boldsymbol{{i}}} −\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}\boldsymbol{{v}}^{\boldsymbol{{i}}} }\right] \\$$$$\frac{\boldsymbol{{dy}}}{\boldsymbol{{y}}}=\frac{\mathrm{2}\boldsymbol{{v}}^{\boldsymbol{{i}}−\mathrm{1}} }{\left(\boldsymbol{{v}}^{\boldsymbol{{i}}} −\mathrm{1}\right)^{\mathrm{2}} }\boldsymbol{{dv}} \\$$$$\boldsymbol{{integrating}}\:\boldsymbol{{both}}\:\boldsymbol{{sides}}\:\boldsymbol{{we}}\:\boldsymbol{{get}}.. \\$$$$\boldsymbol{{ln}}\left(\boldsymbol{{y}}\right)=\mathrm{2}\boldsymbol{{i}}.\frac{\mathrm{1}}{\left(\boldsymbol{{v}}^{\boldsymbol{{i}}} −\mathrm{1}\right)}+\boldsymbol{{ln}}\left(\boldsymbol{{c}}\right) \\$$$$\boldsymbol{{so}}\:\:\:\boldsymbol{{y}}=\boldsymbol{{c}}.\boldsymbol{{e}}^{\frac{\mathrm{2}\boldsymbol{{i}}}{\left[\left(\frac{\boldsymbol{{x}}}{\boldsymbol{{y}}}\right)^{\boldsymbol{{i}}} −\mathrm{1}\right]}} \\$$