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x-2-dy-dx-3xy-2y-2-0-

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Question Number 105569 by bemath last updated on 30/Jul/20
x^2 (dy/dx) −3xy−2y^2 = 0
$${x}^{\mathrm{2}} \:\frac{{dy}}{{dx}}\:−\mathrm{3}{xy}−\mathrm{2}{y}^{\mathrm{2}} \:=\:\mathrm{0}\: \\ $$
Answered by john santu last updated on 30/Jul/20
(dy/dx) = ((3xy+2y^2 )/x^2 ) { ((y=υx ⇒(dy/dx)=υ+x (dυ/dx))),((υ+x (dυ/dx)=((3υx^2 +2υ^2 x^2 )/x^2 ))) :} ⇒υ+x (dυ/dx) = 3υ+2υ^2 x (dυ/dx) = 2υ+2υ^2 →(dυ/(υ(v+1)))=2(dx/x) ∫ ((1/v)−(1/(v+1)))dv = ln Cx^2 ln ((v/(v+1))) = ln Cx^2 (y/(y+x))= Cx^2 ⇒y = (y+x)Cx^2 (JS ♠⧫)
$$\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{3}{xy}+\mathrm{2}{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} } \\ $$$$\begin{cases}{{y}=\upsilon{x}\:\Rightarrow\frac{{dy}}{{dx}}=\upsilon+{x}\:\frac{{d}\upsilon}{{dx}}}\\{\upsilon+{x}\:\frac{{d}\upsilon}{{dx}}=\frac{\mathrm{3}\upsilon{x}^{\mathrm{2}} +\mathrm{2}\upsilon^{\mathrm{2}} {x}^{\mathrm{2}} }{{x}^{\mathrm{2}} }}\end{cases} \\ $$$$\Rightarrow\upsilon+{x}\:\frac{{d}\upsilon}{{dx}}\:=\:\mathrm{3}\upsilon+\mathrm{2}\upsilon^{\mathrm{2}} \\ $$$${x}\:\frac{{d}\upsilon}{{dx}}\:=\:\mathrm{2}\upsilon+\mathrm{2}\upsilon^{\mathrm{2}} \rightarrow\frac{{d}\upsilon}{\upsilon\left({v}+\mathrm{1}\right)}=\mathrm{2}\frac{{dx}}{{x}} \\ $$$$\int\:\left(\frac{\mathrm{1}}{{v}}−\frac{\mathrm{1}}{{v}+\mathrm{1}}\right){dv}\:=\:\mathrm{ln}\:{Cx}^{\mathrm{2}} \\ $$$$\mathrm{ln}\:\left(\frac{{v}}{{v}+\mathrm{1}}\right)\:=\:\mathrm{ln}\:{Cx}^{\mathrm{2}} \\ $$$$\frac{{y}}{{y}+{x}}=\:{Cx}^{\mathrm{2}} \:\Rightarrow{y}\:=\:\left({y}+{x}\right){Cx}^{\mathrm{2}} \\ $$$$\left({JS}\:\spadesuit\blacklozenge\right) \\ $$
Commented by bubugne last updated on 30/Jul/20
correction ln ((v/(v+1))) = ln Cx^2 ln ((y/(y+x)))=ln Cx^2 ⇒y = (y+x) Cx^2 cy −x^2 y = x^3 (c = (1/C)) y = (x^3 /(c−x^2 )) verification : x^2 (dy/dx) −3xy−2y^2 = x^(2 ) ((3cx^2 −3x^4 +2x^4 )/((c−x^2 )^2 ))−((3x^4 )/(c−x^2 ))−((2x^6 )/((c−x^2 )^2 )) x^2 (dy/dx) −3xy−2y^2 = ((3cx^4 −x^6 )/((c−x^2 )^2 ))−((3cx^4 −3x^6 )/((c−x^2 )^2 ))−((2x^6 )/((c−x^2 )^2 )) x^2 (dy/dx) −3xy−2y^2 = ((3cx^4 −x^6 −3cx^4 +3x^6 −2x^6 )/((c−x^2 )^2 )) x^2 (dy/dx) −3xy−2y^2 = 0
$$\mathrm{correction} \\ $$$$\mathrm{ln}\:\left(\frac{{v}}{{v}+\mathrm{1}}\right)\:=\:\mathrm{ln}\:{Cx}^{\mathrm{2}} \\ $$$$\mathrm{ln}\:\left(\frac{{y}}{{y}+{x}}\right)=\mathrm{ln}\:{Cx}^{\mathrm{2}} \:\Rightarrow{y}\:=\:\left({y}+{x}\right)\:{Cx}^{\mathrm{2}} \\ $$$${cy}\:−{x}^{\mathrm{2}} {y}\:=\:{x}^{\mathrm{3}} \:\:\left({c}\:=\:\frac{\mathrm{1}}{{C}}\right) \\ $$$$\boldsymbol{{y}}\:=\:\frac{\boldsymbol{{x}}^{\mathrm{3}} }{\boldsymbol{{c}}−\boldsymbol{{x}}^{\mathrm{2}} } \\ $$$$ \\ $$$$\mathrm{verification}\::\: \\ $$$$\:{x}^{\mathrm{2}} \:\frac{{dy}}{{dx}}\:−\mathrm{3}{xy}−\mathrm{2}{y}^{\mathrm{2}} \:=\:\:{x}^{\mathrm{2}\:} \:\frac{\mathrm{3}{cx}^{\mathrm{2}} −\mathrm{3}{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{4}} }{\left({c}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }−\frac{\mathrm{3}{x}^{\mathrm{4}} }{{c}−{x}^{\mathrm{2}} }−\frac{\mathrm{2}{x}^{\mathrm{6}} }{\left({c}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\:{x}^{\mathrm{2}} \:\frac{{dy}}{{dx}}\:−\mathrm{3}{xy}−\mathrm{2}{y}^{\mathrm{2}} \:=\:\frac{\mathrm{3}{cx}^{\mathrm{4}} −{x}^{\mathrm{6}} }{\left({c}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }−\frac{\mathrm{3}{cx}^{\mathrm{4}} −\mathrm{3}{x}^{\mathrm{6}} }{\left({c}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }−\frac{\mathrm{2}{x}^{\mathrm{6}} }{\left({c}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\:{x}^{\mathrm{2}} \:\frac{{dy}}{{dx}}\:−\mathrm{3}{xy}−\mathrm{2}{y}^{\mathrm{2}} \:=\:\frac{\mathrm{3}{cx}^{\mathrm{4}} −{x}^{\mathrm{6}} −\mathrm{3}{cx}^{\mathrm{4}} +\mathrm{3}{x}^{\mathrm{6}} −\mathrm{2}{x}^{\mathrm{6}} }{\left({c}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\:{x}^{\mathrm{2}} \:\frac{{dy}}{{dx}}\:−\mathrm{3}{xy}−\mathrm{2}{y}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$ \\ $$

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