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

(3x−5y) dx + (x+y) dy = 0

(3x−5y)dx+(x+y)dy=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)

(3x−5y)dx+(x+y)dy=0(1)Puty=xt⇒dy=xdt+tdx(1)⇔(3x−5xt)dx+(x+xt)(xdt+tdx)⇔[3−5t+t(1+t)]dx+(1+t)xdt=0⇔(t2−4t+3)dx+(1+t)xdt=0⇔−dxx=(1+t)dtt2−4t+3⇔−dxx=(1+t)dt(t−1)(t−3)Integratingtwosidesweget−ln∣x∣+lnC=∫2t−3dt−∫1t−1dt=ln∣(t−3)2t−1∣⇔lnC∣x∣=ln∣(t−3)2t−1∣⇔C∣x∣=(t−3)2t−1⇔C∣x∣=(yx−3)2∣(yx−1)∣⇔C=(y−3x)2x2∣y−x∣(C>0−constant)

Answered by benjo_mathlover last updated on 21/Oct/20

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