Question Number 50140 by peter frank last updated on 14/Dec/18
$${Solve}\:{the}\:{differential} \\ $$$${equation} \\ $$$$\left.{a}\right){x}\left({x}+{y}\right)\frac{{dy}}{{dx}}={x}^{\mathrm{2}} +{xy}−\mathrm{3}{y}^{\mathrm{2}} \\ $$$$\left.{b}\right){y}+{xy}^{\mathrm{2}} −{x}\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$${c}\:\left[\:\:{x}^{\mathrm{2}} \frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}\:} }−\mathrm{2}{x}\frac{{dy}}{{dx}}+\mathrm{2}\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}\right){y}=\mathrm{24}{x}^{\mathrm{3}} \right. \\ $$$${given}\:{that}\:\frac{{dy}}{{dx}}=\mathrm{6}\:\:\:,\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{0} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 15/Dec/18
$$\left.{a}\right)\frac{{dy}}{{dx}}=\frac{{x}^{\mathrm{2}} +{xy}−\mathrm{3}{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{xy}}=\frac{\mathrm{1}+\frac{{y}}{{x}}−\frac{\mathrm{3}{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} }}{\mathrm{1}+\frac{{y}}{{x}}}\:\:\left[{deviding}\:{N}_{{r}} {and}\:{D}_{{r}} \:{by}\:{x}^{\mathrm{2}} \right] \\ $$$$\frac{{y}}{{x}}={t}\:\:\:\:{y}={tx}\:\:{so}\:\:\frac{{dy}}{{dx}}={t}+{x}\frac{{dt}}{{dx}} \\ $$$${t}+{x}\frac{{dt}}{{dx}}=\frac{\mathrm{1}+{t}−\mathrm{3}{t}^{\mathrm{2}} }{\mathrm{1}+{t}} \\ $$$${x}\frac{{dt}}{{dx}}=\frac{\mathrm{1}+{t}−\mathrm{3}{t}^{\mathrm{2}} }{\mathrm{1}+{t}}−{t} \\ $$$${x}\frac{{dt}}{{dx}}=\frac{\mathrm{1}+{t}−\mathrm{3}{t}^{\mathrm{2}} −{t}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}} \\ $$$$\frac{\mathrm{1}+{t}}{\mathrm{1}−\mathrm{4}{t}^{\mathrm{2}} }{dt}=\frac{{dx}}{{x}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\mathrm{1}+{t}}{\frac{\mathrm{1}}{\mathrm{4}}−{t}^{\mathrm{2}} }{dt}=\int\frac{{dx}}{{x}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{dt}}{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −{t}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}×\frac{−\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left(\frac{\mathrm{1}}{\mathrm{4}}−{t}^{\mathrm{2}} \right)}{\frac{\mathrm{1}}{\mathrm{4}}−{t}^{\mathrm{2}} }=\int\frac{{dx}}{{x}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{dt}}{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −{t}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{8}}\int\frac{{d}\left(\frac{\mathrm{1}}{\mathrm{4}}−{t}^{\mathrm{2}} \right)}{\frac{\mathrm{1}}{\mathrm{4}}−{t}^{\mathrm{2}} }=\int\frac{{dx}}{{x}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{1}}{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}}{ln}\left(\frac{\frac{\mathrm{1}}{\mathrm{2}}+{t}}{\frac{\mathrm{1}}{\mathrm{2}}−{t}}\right)−\frac{\mathrm{1}}{\mathrm{8}}{ln}\left(\frac{\mathrm{1}}{\mathrm{4}}−{t}^{\mathrm{2}} \right)={lnx}+{lnc} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\frac{\mathrm{1}+\mathrm{2}{t}}{\mathrm{1}−\mathrm{2}{t}}\right)−\frac{\mathrm{1}}{\mathrm{8}}{ln}\left(\frac{\mathrm{1}}{\mathrm{4}}−{t}^{\mathrm{2}} \right)={lnx}+{lnc} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\frac{\mathrm{1}+\frac{\mathrm{2}{y}}{{x}}}{\mathrm{1}−\frac{\mathrm{2}{y}}{{x}}}\right)−\frac{\mathrm{1}}{\mathrm{8}}{ln}\left(\frac{\mathrm{1}}{\mathrm{4}}−\frac{{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\right)={lnx}+{lnc} \\ $$$${ln}\left\{\left(\frac{{x}+\mathrm{2}{y}}{{x}−\mathrm{2}{y}}\right)\right\}^{\frac{\mathrm{1}}{\mathrm{4}}} −{ln}\left(\frac{{x}^{\mathrm{2}} −\mathrm{4}{y}^{\mathrm{2}} }{\mathrm{4}{x}^{\mathrm{2}} }\right)^{\frac{\mathrm{1}}{\mathrm{8}}} ={ln}\left({xc}\right) \\ $$$$\left(\frac{{x}+\mathrm{2}{y}}{{x}−\mathrm{2}{y}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} ×\left(\frac{\mathrm{4}{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{4}{y}^{\mathrm{2}} }\right)^{\frac{\mathrm{1}}{\mathrm{8}}} ={xc} \\ $$$$\left({x}+\mathrm{2}{y}\right)^{\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{8}}} ×\frac{\mathrm{1}}{\left({x}−\mathrm{2}{y}\right)^{\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}} }×\left(\mathrm{4}{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{8}}} ={xc} \\ $$$$\left({x}+\mathrm{2}{y}\right)^{\frac{\mathrm{1}}{\mathrm{8}}} ×\frac{\mathrm{1}}{\left({x}−\mathrm{2}{y}\right)^{\frac{\mathrm{3}}{\mathrm{8}}} }×\left(\mathrm{4}{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{8}}} ={xc} \\ $$$$ \\ $$
Commented by peter frank last updated on 14/Dec/18
$${am}\:{getting}\: \\ $$$${x}+\mathrm{2}{y}={Bx}^{\mathrm{6}} \left({x}−\mathrm{2}{y}\right)^{\mathrm{3}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 14/Dec/18
$$\left.{b}\right){y}+{xy}^{\mathrm{2}} −{x}\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$${y}−{x}\frac{{dy}}{{dx}}=−{xy}^{\mathrm{2}} \\ $$$$\frac{{ydx}−{xdy}}{{y}^{\mathrm{2}} }=−{xdx} \\ $$$${d}\left(\frac{{x}}{{y}}\right)=−{xdx} \\ $$$$\int{d}\left(\frac{{x}}{{y}}\right)=−\int{xdx} \\ $$$$\frac{{x}}{{y}}=\frac{−{x}^{\mathrm{2}} }{\mathrm{2}}+{c} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 14/Dec/18
$$\left.{c}\right){x}={e}^{{t}} \:\:\:\frac{{dx}}{{dt}}={e}^{{t}} \\ $$$$\frac{{dy}}{{dx}}=\frac{{dy}}{{dt}}×\frac{{dt}}{{dx}}=\frac{\mathrm{1}}{{e}^{{t}} }\frac{{dy}}{{dt}} \\ $$$${e}^{{t}} \frac{{dy}}{{dx}}=\frac{{dy}}{{dt}} \\ $$$${x}\frac{{dy}}{{dx}}=\frac{{dy}}{{dt}} \\ $$$$\frac{{d}}{{dx}}\left(\frac{{dy}}{{dx}}\right)=\frac{{d}}{{dt}}\left(\frac{{dy}}{{dx}}\right)×\frac{{dt}}{{dx}} \\ $$$$=\frac{{d}}{{dt}}\left(\frac{\mathrm{1}}{{e}^{{t}} }×\frac{{dy}}{{dt}}\right)×\frac{\mathrm{1}}{{e}^{{t}} } \\ $$$$=\frac{\mathrm{1}}{{e}^{{t}} }\left[\frac{\mathrm{1}}{{e}^{{t}} }×\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }+\frac{{dy}}{{dt}}×{e}^{−{t}} ×−\mathrm{1}\right] \\ $$$$=\left(\frac{\mathrm{1}}{{e}^{{t}} }\right)^{\mathrm{2}} ×\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }−\left(\frac{\mathrm{1}}{{e}^{{t}} }\right)^{\mathrm{2}} ×\frac{{dy}}{{dt}} \\ $$$$\left({e}^{{t}} \right)^{\mathrm{2}} \frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }−\frac{{dy}}{{dt}} \\ $$$${let}\:\frac{{d}}{{dt}}=\theta\: \\ $$$${x}^{\mathrm{2}} \frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\left(\theta^{\mathrm{2}} −\theta\right){y} \\ $$$${wait}\:{busy}… \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 14/Dec/18
$${pls}\:{check}\:{the}\:{question}… \\ $$
Commented by peter frank last updated on 14/Dec/18
$$\mathrm{yes}\:\mathrm{sir}\:\mathrm{am}\:\mathrm{arleady}\:\mathrm{check}\: \\ $$
Answered by peter frank last updated on 14/Dec/18
$$\left.\mathrm{c}\right)\:{y}={vx} \\ $$$$\frac{{dy}}{{dx}}={v}+{x}\frac{{dv}}{{dx}}\:……\left({i}\right) \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{2}\frac{{dv}}{{dx}}\:\:+{x}\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }….\left({ii}\right) \\ $$$$\:{x}^{\mathrm{2}} \frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}\:} }−\mathrm{2}{x}\frac{{dy}}{{dx}}+\mathrm{2}\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}\right){y}=\mathrm{24}{x}^{\mathrm{3}} \\ $$$${x}^{\mathrm{2}} \left(\mathrm{2}\frac{{dv}}{{dx}}+{x}\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\right)−\mathrm{2}{x}\left({v}+{x}\frac{{dv}}{{dx}}\right)+\mathrm{2}{vx}\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{24}{x}^{\mathrm{3}} \\ $$$${x}^{\mathrm{3}} \frac{{d}^{\mathrm{2}} {v}}{{dx}^{\mathrm{2}} }+\mathrm{4}{vx}^{\mathrm{3}} =\mathrm{24}{x}^{\mathrm{3}} \\ $$$$\frac{{d}^{\mathrm{2}} {v}}{{dx}^{\mathrm{2}} }+\mathrm{4}{v}=\mathrm{24} \\ $$$$\frac{{d}^{\mathrm{2}} {v}}{{dx}^{\mathrm{2}} }+\mathrm{4}\left({v}−\mathrm{6}\right)=\mathrm{0} \\ $$$${t}={u}−\mathrm{6} \\ $$$$\frac{{dt}}{{dx}}=\frac{{du}}{{dx}} \\ $$$$\frac{{d}^{\mathrm{2}} {t}}{{dx}^{\mathrm{2}} }=\frac{{d}^{\mathrm{2}} {u}}{{dx}^{\mathrm{2}} } \\ $$$${t}^{\mathrm{2}} +\mathrm{4}{t}=\mathrm{0} \\ $$$${t}=\mathrm{0}\pm\mathrm{2}{i} \\ $$$${t}={e}^{{px}} \left({A}\mathrm{cos}\:\mathrm{qx}+{B}\mathrm{sin}\:{qx}\right)= \\ $$$${t}={A}\mathrm{cos}\:\mathrm{2}{x}+{B}\mathrm{sin}\:\mathrm{2}{x} \\ $$$${u}−\mathrm{6}={A}\mathrm{cos}\:\mathrm{2}{x}+{B}\mathrm{sin}\:\mathrm{2}{x} \\ $$$$\frac{{y}}{{x}}−\mathrm{6}={A}\mathrm{cos}\:\mathrm{2}{x}+{B}\mathrm{sin}\:\mathrm{2}{x} \\ $$$${y}={A}\mathrm{cos}\:\mathrm{2}{x}+{B}\mathrm{sin}\:\mathrm{2}{x}+\mathrm{6}{x}\:…..\left(\mathrm{i}\right) \\ $$$$\frac{{dy}}{{dx}}=−\mathrm{2}{Ax}\mathrm{sin}\:\mathrm{2}{x}+{A}\mathrm{cos}\:\mathrm{2}{x}+\mathrm{2}{Bx}\mathrm{cos}\:\mathrm{2}{x}+{B}\mathrm{sin}\:\mathrm{2}{x}…..\left(\mathrm{ii}\right) \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=−\mathrm{4}{A}\mathrm{xcos2x}−\mathrm{2Asin}\:\mathrm{2x}−\mathrm{2Asin}\:\mathrm{2x}+\mathrm{2Bcos}\:\mathrm{2x}−\mathrm{4Bxsin}\:\mathrm{2x}…..\left(\mathrm{iii}\right) \\ $$$$\left.{from}\:{ii}\right) \\ $$$$\frac{{dy}}{{dx}}=−\mathrm{2}{Ax}\mathrm{sin}\:\mathrm{2}{x}+{A}\mathrm{cos}\:\mathrm{2}{x}+\mathrm{2}{Bx}\mathrm{cos}\:\mathrm{2}{x}+{B}\mathrm{sin}\:\mathrm{2}{x}…..\left(\mathrm{ii}\right) \\ $$$${given}\:{x}=\mathrm{0},\frac{{dy}}{{dx}}=\mathrm{6}\: \\ $$$$\mathrm{6}={A}+\mathrm{6}\:\:\mathrm{A}=\mathrm{0} \\ $$$${from}\:{iii} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=−\mathrm{4}{A}\mathrm{xcos2x}−\mathrm{2Asin}\:\mathrm{2x}−\mathrm{2Asin}\:\mathrm{2x}+\mathrm{2Bcos}\:\mathrm{2x}−\mathrm{4Bxsin}\:\mathrm{2x}…..\left(\mathrm{iii}\right) \\ $$$$\mathrm{4}=\mathrm{2}{B}+\mathrm{2}{B}\:\:\:\:\mathrm{B}=\mathrm{1} \\ $$$${from}\left(\:{i}\right) \\ $$$${y}={A}\mathrm{cos}\:\mathrm{2}{x}+{B}\mathrm{sin}\:\mathrm{2}{x}+\mathrm{6}{x}\:…..\left(\mathrm{i}\right) \\ $$$${put}\:{A}=\mathrm{0}\:\:\:\:{B}=\mathrm{1} \\ $$$${y}={x}\mathrm{sin}\:\mathrm{2}{x}+\mathrm{6}{x} \\ $$$${required}\:{solution} \\ $$$$ \\ $$$$ \\ $$
Answered by peter frank last updated on 14/Dec/18
$$\left.\mathrm{b}\right)\:{divide}\:{by}\:{x}\:{bothsides} \\ $$$$\frac{{y}}{{x}}+{y}^{\mathrm{2}} −\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$${y}={vx} \\ $$$$\frac{{dy}}{{dx}}={v}+{x}\frac{{dy}}{{dx}} \\ $$$${v}+\left({vx}\right)^{\mathrm{2}} ={v}+{x}\frac{{dy}}{{dx}} \\ $$$${xdx}=\frac{{dv}}{{v}^{\mathrm{2}} } \\ $$$$\int{xdx}=\int\frac{{dv}}{{v}^{\mathrm{2}} } \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{2}\:}=−\frac{{x}}{{y}}+{A} \\ $$$${y}=−\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} −\mathrm{2}{A}} \\ $$$${y}=\frac{\mathrm{2}{x}}{\mathrm{2}{A}−{x}^{\mathrm{2}} } \\ $$$$ \\ $$