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Question Number 118975 by benjo_mathlover last updated on 21/Oct/20

 (3x−5y) dx + (x+y) dy = 0

$$\:\left(\mathrm{3}{x}−\mathrm{5}{y}\right)\:{dx}\:+\:\left({x}+{y}\right)\:{dy}\:=\:\mathrm{0} \\ $$

Answered by 1549442205PVT last updated on 21/Oct/20

 (3x−5y) dx + (x+y) dy = 0(1)  Put y=xt⇒dy=xdt+tdx  (1)⇔(3x−5xt)dx+(x+xt)(xdt+tdx)  ⇔[3−5t+t(1+t)]dx+(1+t)xdt=0  ⇔(t^2 −4t+3)dx+(1+t)xdt=0  ⇔−(dx/x)=(((1+t)dt)/(t^2 −4t+3))⇔((−dx)/x)=(((1+t)dt)/((t−1)(t−3)))  Integrating two sides we get  −ln∣x∣+lnC=∫(2/(t−3))dt−∫(1/(t−1))dt  =ln∣(((t−3)^2 )/(t−1))∣⇔ln(C/(∣x∣))=ln∣(((t−3)^2 )/(t−1))∣  ⇔(C/(∣x∣))=(((t−3)^2 )/(t−1))⇔(C/(∣x∣))=((((y/x)−3)^2 )/(∣((y/x)−1)∣))  ⇔C=(((y−3x)^2 )/(x^2 ∣y−x∣))(C>0−constant)

$$\:\left(\mathrm{3}{x}−\mathrm{5}{y}\right)\:{dx}\:+\:\left({x}+{y}\right)\:{dy}\:=\:\mathrm{0}\left(\mathrm{1}\right) \\ $$$$\mathrm{Put}\:\mathrm{y}=\mathrm{xt}\Rightarrow\mathrm{dy}=\mathrm{xdt}+\mathrm{tdx} \\ $$$$\left(\mathrm{1}\right)\Leftrightarrow\left(\mathrm{3x}−\mathrm{5xt}\right)\mathrm{dx}+\left(\mathrm{x}+\mathrm{xt}\right)\left(\mathrm{xdt}+\mathrm{tdx}\right) \\ $$$$\Leftrightarrow\left[\mathrm{3}−\mathrm{5t}+\mathrm{t}\left(\mathrm{1}+\mathrm{t}\right)\right]\mathrm{dx}+\left(\mathrm{1}+\mathrm{t}\right)\mathrm{xdt}=\mathrm{0} \\ $$$$\Leftrightarrow\left(\mathrm{t}^{\mathrm{2}} −\mathrm{4t}+\mathrm{3}\right)\mathrm{dx}+\left(\mathrm{1}+\mathrm{t}\right)\mathrm{xdt}=\mathrm{0} \\ $$$$\Leftrightarrow−\frac{\mathrm{dx}}{\mathrm{x}}=\frac{\left(\mathrm{1}+\mathrm{t}\right)\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} −\mathrm{4t}+\mathrm{3}}\Leftrightarrow\frac{−\mathrm{dx}}{\mathrm{x}}=\frac{\left(\mathrm{1}+\mathrm{t}\right)\mathrm{dt}}{\left(\mathrm{t}−\mathrm{1}\right)\left(\mathrm{t}−\mathrm{3}\right)} \\ $$$$\mathrm{Integrating}\:\mathrm{two}\:\mathrm{sides}\:\mathrm{we}\:\mathrm{get} \\ $$$$−\mathrm{ln}\mid\mathrm{x}\mid+\mathrm{lnC}=\int\frac{\mathrm{2}}{\mathrm{t}−\mathrm{3}}\mathrm{dt}−\int\frac{\mathrm{1}}{\mathrm{t}−\mathrm{1}}\mathrm{dt} \\ $$$$=\mathrm{ln}\mid\frac{\left(\mathrm{t}−\mathrm{3}\right)^{\mathrm{2}} }{\mathrm{t}−\mathrm{1}}\mid\Leftrightarrow\mathrm{ln}\frac{\mathrm{C}}{\mid\mathrm{x}\mid}=\mathrm{ln}\mid\frac{\left(\mathrm{t}−\mathrm{3}\right)^{\mathrm{2}} }{\mathrm{t}−\mathrm{1}}\mid \\ $$$$\Leftrightarrow\frac{\mathrm{C}}{\mid\mathrm{x}\mid}=\frac{\left(\mathrm{t}−\mathrm{3}\right)^{\mathrm{2}} }{\mathrm{t}−\mathrm{1}}\Leftrightarrow\frac{\mathrm{C}}{\mid\mathrm{x}\mid}=\frac{\left(\frac{\mathrm{y}}{\mathrm{x}}−\mathrm{3}\right)^{\mathrm{2}} }{\mid\left(\frac{\mathrm{y}}{\mathrm{x}}−\mathrm{1}\right)\mid} \\ $$$$\Leftrightarrow\mathrm{C}=\frac{\left(\mathrm{y}−\mathrm{3x}\right)^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} \mid\mathrm{y}−\mathrm{x}\mid}\left(\mathrm{C}>\mathrm{0}−\mathrm{constant}\right) \\ $$

Answered by benjo_mathlover last updated on 21/Oct/20

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