Differential Equation Questions

Question Number 105569 by bemath last updated on 30/Jul/20

$${x}^{\mathrm{2}} \:\frac{{dy}}{{dx}}\:−\mathrm{3}{xy}−\mathrm{2}{y}^{\mathrm{2}} \:=\:\mathrm{0}\: \\$$

Answered by john santu last updated on 30/Jul/20

$$\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{3}{xy}+\mathrm{2}{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} } \\$$$$\begin{cases}{{y}=\upsilon{x}\:\Rightarrow\frac{{dy}}{{dx}}=\upsilon+{x}\:\frac{{d}\upsilon}{{dx}}}\\{\upsilon+{x}\:\frac{{d}\upsilon}{{dx}}=\frac{\mathrm{3}\upsilon{x}^{\mathrm{2}} +\mathrm{2}\upsilon^{\mathrm{2}} {x}^{\mathrm{2}} }{{x}^{\mathrm{2}} }}\end{cases} \\$$$$\Rightarrow\upsilon+{x}\:\frac{{d}\upsilon}{{dx}}\:=\:\mathrm{3}\upsilon+\mathrm{2}\upsilon^{\mathrm{2}} \\$$$${x}\:\frac{{d}\upsilon}{{dx}}\:=\:\mathrm{2}\upsilon+\mathrm{2}\upsilon^{\mathrm{2}} \rightarrow\frac{{d}\upsilon}{\upsilon\left({v}+\mathrm{1}\right)}=\mathrm{2}\frac{{dx}}{{x}} \\$$$$\int\:\left(\frac{\mathrm{1}}{{v}}−\frac{\mathrm{1}}{{v}+\mathrm{1}}\right){dv}\:=\:\mathrm{ln}\:{Cx}^{\mathrm{2}} \\$$$$\mathrm{ln}\:\left(\frac{{v}}{{v}+\mathrm{1}}\right)\:=\:\mathrm{ln}\:{Cx}^{\mathrm{2}} \\$$$$\frac{{y}}{{y}+{x}}=\:{Cx}^{\mathrm{2}} \:\Rightarrow{y}\:=\:\left({y}+{x}\right){Cx}^{\mathrm{2}} \\$$$$\left({JS}\:\spadesuit\blacklozenge\right) \\$$

Commented by bubugne last updated on 30/Jul/20

$$\mathrm{correction} \\$$$$\mathrm{ln}\:\left(\frac{{v}}{{v}+\mathrm{1}}\right)\:=\:\mathrm{ln}\:{Cx}^{\mathrm{2}} \\$$$$\mathrm{ln}\:\left(\frac{{y}}{{y}+{x}}\right)=\mathrm{ln}\:{Cx}^{\mathrm{2}} \:\Rightarrow{y}\:=\:\left({y}+{x}\right)\:{Cx}^{\mathrm{2}} \\$$$${cy}\:−{x}^{\mathrm{2}} {y}\:=\:{x}^{\mathrm{3}} \:\:\left({c}\:=\:\frac{\mathrm{1}}{{C}}\right) \\$$$$\boldsymbol{{y}}\:=\:\frac{\boldsymbol{{x}}^{\mathrm{3}} }{\boldsymbol{{c}}−\boldsymbol{{x}}^{\mathrm{2}} } \\$$$$\\$$$$\mathrm{verification}\::\: \\$$$$\:{x}^{\mathrm{2}} \:\frac{{dy}}{{dx}}\:−\mathrm{3}{xy}−\mathrm{2}{y}^{\mathrm{2}} \:=\:\:{x}^{\mathrm{2}\:} \:\frac{\mathrm{3}{cx}^{\mathrm{2}} −\mathrm{3}{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{4}} }{\left({c}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }−\frac{\mathrm{3}{x}^{\mathrm{4}} }{{c}−{x}^{\mathrm{2}} }−\frac{\mathrm{2}{x}^{\mathrm{6}} }{\left({c}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\$$$$\:{x}^{\mathrm{2}} \:\frac{{dy}}{{dx}}\:−\mathrm{3}{xy}−\mathrm{2}{y}^{\mathrm{2}} \:=\:\frac{\mathrm{3}{cx}^{\mathrm{4}} −{x}^{\mathrm{6}} }{\left({c}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }−\frac{\mathrm{3}{cx}^{\mathrm{4}} −\mathrm{3}{x}^{\mathrm{6}} }{\left({c}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }−\frac{\mathrm{2}{x}^{\mathrm{6}} }{\left({c}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\$$$$\:{x}^{\mathrm{2}} \:\frac{{dy}}{{dx}}\:−\mathrm{3}{xy}−\mathrm{2}{y}^{\mathrm{2}} \:=\:\frac{\mathrm{3}{cx}^{\mathrm{4}} −{x}^{\mathrm{6}} −\mathrm{3}{cx}^{\mathrm{4}} +\mathrm{3}{x}^{\mathrm{6}} −\mathrm{2}{x}^{\mathrm{6}} }{\left({c}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\$$$$\:{x}^{\mathrm{2}} \:\frac{{dy}}{{dx}}\:−\mathrm{3}{xy}−\mathrm{2}{y}^{\mathrm{2}} \:=\:\mathrm{0} \\$$$$\\$$