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2x-y-1-dx-x-4y-3-dy-

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Question Number 91750 by jagoll last updated on 02/May/20
(2x−y+1)dx = (x−4y+3)dy
$$\left(\mathrm{2}{x}−{y}+\mathrm{1}\right){dx}\:=\:\left({x}−\mathrm{4}{y}+\mathrm{3}\right){dy} \\ $$
Commented by mr W last updated on 02/May/20
see Q90770
$${see}\:{Q}\mathrm{90770} \\ $$
Commented by john santu last updated on 03/May/20
(dy/dx) = ((2x−y+1)/(x−4y+3)) let x = w+w_o ; y = v+v_o 2x−y+1=2w+2w_o −v−v_o +1 x−4y+3=w+w_o −4v−4v_o +3 set 2w_o −v_o +1=0 set w_o −4v_o +3=0 v_o = (5/7); w_o = −(1/7) ⇒x=w−(1/7) ∧y=v+(5/7) let v = wt ⇒(dv/dw)=t+w (dt/dw) =((2−t)/(1−4t)) w(dt/(dw )) = ((4t^2 −2t+2)/(1−4t)) (((1−4t)dt)/(2t^2 −t+1)) = ((2dw)/w) ∫ (((1−4t)dt)/(2t^2 −t+1)) = 2ln(w)+c ∫ ((−d(2t^2 −t+1))/(2t^2 −t+1)) = 2ln(w) +c ln(2t^2 −t+1) = −2ln(w)+c 2t^2 −t+1 = (C/w^2 ) 2((v/w))^2 −((v/w))+1=(C/w^2 ) 2v^2 −vw+w^2 =C (2v+w)(v−w) = C (((7x+7y−4)/7))(((7y−7x−6)/7)) = C (7x+7y−4)(7y−7x−6)= 49C
$$\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{2}{x}−{y}+\mathrm{1}}{{x}−\mathrm{4}{y}+\mathrm{3}} \\ $$$${let}\:{x}\:=\:{w}+{w}_{{o}} \:;\:{y}\:=\:{v}+{v}_{{o}} \\ $$$$\mathrm{2}{x}−{y}+\mathrm{1}=\mathrm{2}{w}+\mathrm{2}{w}_{{o}} −{v}−{v}_{{o}} +\mathrm{1} \\ $$$${x}−\mathrm{4}{y}+\mathrm{3}={w}+{w}_{{o}} −\mathrm{4}{v}−\mathrm{4}{v}_{{o}} +\mathrm{3} \\ $$$${set}\:\mathrm{2}{w}_{{o}} −{v}_{{o}} +\mathrm{1}=\mathrm{0} \\ $$$${set}\:{w}_{{o}} −\mathrm{4}{v}_{{o}} +\mathrm{3}=\mathrm{0} \\ $$$${v}_{{o}} =\:\frac{\mathrm{5}}{\mathrm{7}};\:{w}_{{o}} =\:−\frac{\mathrm{1}}{\mathrm{7}} \\ $$$$\Rightarrow{x}={w}−\frac{\mathrm{1}}{\mathrm{7}}\:\wedge{y}={v}+\frac{\mathrm{5}}{\mathrm{7}} \\ $$$${let}\:{v}\:=\:{wt}\:\Rightarrow\frac{{dv}}{{dw}}={t}+{w}\:\frac{{dt}}{{dw}}\:=\frac{\mathrm{2}−{t}}{\mathrm{1}−\mathrm{4}{t}} \\ $$$${w}\frac{{dt}}{{dw}\:}\:=\:\frac{\mathrm{4}{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{2}}{\mathrm{1}−\mathrm{4}{t}} \\ $$$$\frac{\left(\mathrm{1}−\mathrm{4}{t}\right){dt}}{\mathrm{2}{t}^{\mathrm{2}} −{t}+\mathrm{1}}\:=\:\frac{\mathrm{2}{dw}}{{w}} \\ $$$$\int\:\frac{\left(\mathrm{1}−\mathrm{4}{t}\right){dt}}{\mathrm{2}{t}^{\mathrm{2}} −{t}+\mathrm{1}}\:=\:\mathrm{2ln}\left({w}\right)+{c} \\ $$$$\int\:\frac{−{d}\left(\mathrm{2}{t}^{\mathrm{2}} −{t}+\mathrm{1}\right)}{\mathrm{2}{t}^{\mathrm{2}} −{t}+\mathrm{1}}\:=\:\mathrm{2ln}\left({w}\right)\:+{c} \\ $$$$\mathrm{ln}\left(\mathrm{2}{t}^{\mathrm{2}} −{t}+\mathrm{1}\right)\:=\:−\mathrm{2ln}\left({w}\right)+{c} \\ $$$$\mathrm{2}{t}^{\mathrm{2}} −{t}+\mathrm{1}\:=\:\frac{{C}}{{w}^{\mathrm{2}} } \\ $$$$\mathrm{2}\left(\frac{{v}}{{w}}\right)^{\mathrm{2}} −\left(\frac{{v}}{{w}}\right)+\mathrm{1}=\frac{{C}}{{w}^{\mathrm{2}} } \\ $$$$\mathrm{2}{v}^{\mathrm{2}} −{vw}+{w}^{\mathrm{2}} ={C} \\ $$$$\left(\mathrm{2}{v}+{w}\right)\left({v}−{w}\right)\:=\:{C} \\ $$$$\left(\frac{\mathrm{7}{x}+\mathrm{7}{y}−\mathrm{4}}{\mathrm{7}}\right)\left(\frac{\mathrm{7}{y}−\mathrm{7}{x}−\mathrm{6}}{\mathrm{7}}\right)\:=\:{C} \\ $$$$\left(\mathrm{7}{x}+\mathrm{7}{y}−\mathrm{4}\right)\left(\mathrm{7}{y}−\mathrm{7}{x}−\mathrm{6}\right)=\:\mathrm{49}{C} \\ $$

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