Question Number 92114 by mhmd last updated on 04/May/20
Answered by mr W last updated on 05/May/20
$${u}={y}' \\ $$$${u}\frac{{du}}{{dy}}+\mathrm{3}{y}^{\mathrm{2}} {u}=\mathrm{0} \\ $$$$\Rightarrow{u}=\mathrm{0}\:\Rightarrow\frac{{dy}}{{dx}}=\mathrm{0}\:\Rightarrow{y}={C} \\ $$$$\Rightarrow\frac{{du}}{{dy}}+\mathrm{3}{y}^{\mathrm{2}} =\mathrm{0}\:\Rightarrow{u}=−\left({y}^{\mathrm{3}} −{C}^{\mathrm{3}} \right) \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=−\left({y}^{\mathrm{3}} −{C}^{\mathrm{3}} \right) \\ $$$$\Rightarrow\frac{{dy}}{{y}^{\mathrm{3}} −{C}^{\mathrm{3}} }=−{dx} \\ $$$$\Rightarrow\int\frac{{dy}}{{y}^{\mathrm{3}} −{C}^{\mathrm{3}} }=−{x} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{3}{C}^{\mathrm{2}} }\int\left(\frac{\mathrm{1}}{{y}−{C}}−\frac{{y}+\mathrm{2}{C}}{{y}^{\mathrm{2}} +{Cy}+{C}^{\mathrm{2}} }\right){dy}=−{x} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{3}{C}^{\mathrm{2}} }\int\left(\frac{\mathrm{1}}{{y}−{C}}−\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}{y}+{C}}{{y}^{\mathrm{2}} +{Cy}+{C}^{\mathrm{2}} }−\frac{\mathrm{3}{C}}{\mathrm{2}}×\frac{\mathrm{1}}{{y}^{\mathrm{2}} +{Cy}+{C}^{\mathrm{2}} }\right){dy}=−{x} \\ $$$$...\:{this}\:{should}\:{be}\:{easy}\: \\ $$$$\Rightarrow{x}+\frac{\mathrm{1}}{\mathrm{6}{C}^{\mathrm{2}} }\mathrm{ln}\:\mid\frac{\left({y}−{C}\right)^{\mathrm{2}} }{{y}^{\mathrm{2}} +{Cy}+{C}^{\mathrm{2}} }\mid−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}{C}^{\mathrm{2}} }\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}{y}+{C}}{\sqrt{\mathrm{3}}{C}}+{C}_{\mathrm{1}} =\mathrm{0} \\ $$