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3x-2-9xy-5y-2-dx-6x-2-4xy-dy-

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Question Number 90394 by jagoll last updated on 23/Apr/20
(3x^2 +9xy+5y^2 )dx = (6x^2 +4xy)dy
$$\left(\mathrm{3x}^{\mathrm{2}} +\mathrm{9xy}+\mathrm{5y}^{\mathrm{2}} \right)\mathrm{dx}\:=\:\left(\mathrm{6x}^{\mathrm{2}} +\mathrm{4xy}\right)\mathrm{dy} \\ $$
Commented by john santu last updated on 23/Apr/20
(3+9((y/x))+5((y/x))^2 ) = (6+4((y/x))) dy [ let y=tx ⇒dy = t dx + x dt ] (3+9t+5t^2 )dx = (6+4t)(t dx+x dt ) (3+3t+t^2 ) dx = x(6+4t) dt (dx/x) = ((2(2t+3)dt)/(t^2 +3t+3)) ∫ (dx/x) = ∫ ((2 d(t^2 +3t+3))/(t^2 +3t+3)) ln Cx = 2 ln(t^2 +3t+3) (√(Cx)) = ((y/x))^2 +3((y/x))+3 ((y/x)+(3/2))^2 = (√(Cx))−(3/4) (y/x) = −(3/2) ±(1/2) (√((√(16Cx))−3)) y = x ( ((−3 ± (√((√(16Cx))−3)))/2))
$$\left(\mathrm{3}+\mathrm{9}\left(\frac{{y}}{{x}}\right)+\mathrm{5}\left(\frac{{y}}{{x}}\right)^{\mathrm{2}} \right)\:=\:\left(\mathrm{6}+\mathrm{4}\left(\frac{{y}}{{x}}\right)\right)\:{dy} \\ $$$$\left[\:{let}\:{y}={tx}\:\Rightarrow{dy}\:=\:{t}\:{dx}\:+\:{x}\:{dt}\:\right] \\ $$$$\left(\mathrm{3}+\mathrm{9}{t}+\mathrm{5}{t}^{\mathrm{2}} \right){dx}\:=\:\left(\mathrm{6}+\mathrm{4}{t}\right)\left({t}\:{dx}+{x}\:{dt}\:\right) \\ $$$$\left(\mathrm{3}+\mathrm{3}{t}+{t}^{\mathrm{2}} \right)\:{dx}\:=\:{x}\left(\mathrm{6}+\mathrm{4}{t}\right)\:{dt}\: \\ $$$$\frac{{dx}}{{x}}\:=\:\frac{\mathrm{2}\left(\mathrm{2}{t}+\mathrm{3}\right){dt}}{{t}^{\mathrm{2}} +\mathrm{3}{t}+\mathrm{3}} \\ $$$$\int\:\frac{{dx}}{{x}}\:=\:\int\:\frac{\mathrm{2}\:{d}\left({t}^{\mathrm{2}} +\mathrm{3}{t}+\mathrm{3}\right)}{{t}^{\mathrm{2}} +\mathrm{3}{t}+\mathrm{3}} \\ $$$$\mathrm{ln}\:\mathrm{C}{x}\:=\:\mathrm{2}\:\mathrm{ln}\left(\mathrm{t}^{\mathrm{2}} +\mathrm{3t}+\mathrm{3}\right) \\ $$$$\sqrt{\mathrm{C}{x}}\:=\:\left(\frac{{y}}{{x}}\right)^{\mathrm{2}} +\mathrm{3}\left(\frac{{y}}{{x}}\right)+\mathrm{3}\: \\ $$$$\left(\frac{{y}}{{x}}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} =\:\sqrt{{Cx}}−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\frac{{y}}{{x}}\:=\:−\frac{\mathrm{3}}{\mathrm{2}}\:\pm\frac{\mathrm{1}}{\mathrm{2}}\:\sqrt{\sqrt{\mathrm{16}{Cx}}−\mathrm{3}} \\ $$$${y}\:=\:{x}\:\left(\:\frac{−\mathrm{3}\:\pm\:\sqrt{\sqrt{\mathrm{16}{Cx}}−\mathrm{3}}}{\mathrm{2}}\right)\: \\ $$
Commented by jagoll last updated on 23/Apr/20
thank you both
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{both} \\ $$
Answered by ajfour last updated on 23/Apr/20
let y=tx (dy/dx)=t+x(dt/dx) ⇒ t+x(dt/dx)=((3+9t+5t^2 )/(6+4t)) 2x(dt/dx)=((t^2 +3t+3)/(2t+3)) ∫(((2t+3))/(t^2 +3t+3))=(1/2)∫ (dx/x) ln ∣t^2 +3t+3∣=ln (√(cx)) cx^3 =y^2 +3xy+3x^2
$${let}\:\:{y}={tx} \\ $$$$\frac{{dy}}{{dx}}={t}+{x}\frac{{dt}}{{dx}} \\ $$$$\Rightarrow\:\:{t}+{x}\frac{{dt}}{{dx}}=\frac{\mathrm{3}+\mathrm{9}{t}+\mathrm{5}{t}^{\mathrm{2}} }{\mathrm{6}+\mathrm{4}{t}} \\ $$$$\mathrm{2}{x}\frac{{dt}}{{dx}}=\frac{{t}^{\mathrm{2}} +\mathrm{3}{t}+\mathrm{3}}{\mathrm{2}{t}+\mathrm{3}} \\ $$$$\int\frac{\left(\mathrm{2}{t}+\mathrm{3}\right)}{{t}^{\mathrm{2}} +\mathrm{3}{t}+\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{{dx}}{{x}} \\ $$$$\mathrm{ln}\:\mid{t}^{\mathrm{2}} +\mathrm{3}{t}+\mathrm{3}\mid=\mathrm{ln}\:\sqrt{{cx}} \\ $$$${cx}^{\mathrm{3}} ={y}^{\mathrm{2}} +\mathrm{3}{xy}+\mathrm{3}{x}^{\mathrm{2}} \\ $$

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