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Question Number 85050 by john santu last updated on 18/Mar/20

(x−4y+3)dx = (x−5y+4)dy

$$\left(\mathrm{x}−\mathrm{4y}+\mathrm{3}\right)\mathrm{dx}\:=\:\left(\mathrm{x}−\mathrm{5y}+\mathrm{4}\right)\mathrm{dy} \\ $$

Answered by mr W last updated on 18/Mar/20

let x=u+a y=v+b x−4y+3=u−4v+(a−4b+3) x−5y+4=u−5v+(a−5b+4) set a−4b+3=0 a−5b+4=0 ⇒a=1, b=1 ⇒x=u+1 ⇒dx=du ⇒y=v+1 ⇒dy=dv (u−4v)du=(u−5v)dv ⇒(dv/du)=((u−4v)/(u−5v)) let v=ut ⇒t=(v/u) (dv/du)=t+u(dt/du) ⇒t+u(dt/du)=((1−4t)/(1−5t)) ⇒u(dt/du)=((5t^2 −5t+1)/(1−5t)) ⇒(((1−5t)dt)/(5t^2 −5t+1))=(du/u) ⇒∫(((1−5t)dt)/(5t^2 −5t+1))=∫(du/u) ⇒∫((1/(5t^2 −5t+1))−((5t)/(5t^2 −5t+1)))dt=ln (cu) ......

$${let} \\ $$$${x}={u}+{a} \\ $$$${y}={v}+{b} \\ $$$${x}−\mathrm{4}{y}+\mathrm{3}={u}−\mathrm{4}{v}+\left({a}−\mathrm{4}{b}+\mathrm{3}\right) \\ $$$${x}−\mathrm{5}{y}+\mathrm{4}={u}−\mathrm{5}{v}+\left({a}−\mathrm{5}{b}+\mathrm{4}\right) \\ $$$${set} \\ $$$${a}−\mathrm{4}{b}+\mathrm{3}=\mathrm{0} \\ $$$${a}−\mathrm{5}{b}+\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow{a}=\mathrm{1},\:{b}=\mathrm{1} \\ $$$$\Rightarrow{x}={u}+\mathrm{1}\:\Rightarrow{dx}={du} \\ $$$$\Rightarrow{y}={v}+\mathrm{1}\:\Rightarrow{dy}={dv} \\ $$$$ \\ $$$$\left({u}−\mathrm{4}{v}\right){du}=\left({u}−\mathrm{5}{v}\right){dv} \\ $$$$\Rightarrow\frac{{dv}}{{du}}=\frac{{u}−\mathrm{4}{v}}{{u}−\mathrm{5}{v}} \\ $$$${let}\:{v}={ut}\:\Rightarrow{t}=\frac{{v}}{{u}} \\ $$$$\frac{{dv}}{{du}}={t}+{u}\frac{{dt}}{{du}} \\ $$$$\Rightarrow{t}+{u}\frac{{dt}}{{du}}=\frac{\mathrm{1}−\mathrm{4}{t}}{\mathrm{1}−\mathrm{5}{t}} \\ $$$$\Rightarrow{u}\frac{{dt}}{{du}}=\frac{\mathrm{5}{t}^{\mathrm{2}} −\mathrm{5}{t}+\mathrm{1}}{\mathrm{1}−\mathrm{5}{t}} \\ $$$$\Rightarrow\frac{\left(\mathrm{1}−\mathrm{5}{t}\right){dt}}{\mathrm{5}{t}^{\mathrm{2}} −\mathrm{5}{t}+\mathrm{1}}=\frac{{du}}{{u}} \\ $$$$\Rightarrow\int\frac{\left(\mathrm{1}−\mathrm{5}{t}\right){dt}}{\mathrm{5}{t}^{\mathrm{2}} −\mathrm{5}{t}+\mathrm{1}}=\int\frac{{du}}{{u}} \\ $$$$\Rightarrow\int\left(\frac{\mathrm{1}}{\mathrm{5}{t}^{\mathrm{2}} −\mathrm{5}{t}+\mathrm{1}}−\frac{\mathrm{5}{t}}{\mathrm{5}{t}^{\mathrm{2}} −\mathrm{5}{t}+\mathrm{1}}\right){dt}=\mathrm{ln}\:\left({cu}\right) \\ $$$$...... \\ $$

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