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Question Number 161241 by help last updated on 14/Dec/21

Answered by mr W last updated on 15/Dec/21

let y=xt y′=t+x(dt/dx)=((t^2 −1)/t) x(dt/dx)=((t^2 −1)/t)−t=−(1/t) tdt=−(dx/x) (t^2 /2)=−ln Cx (y^2 /(2x^2 ))=−ln Cx ⇒y^2 =−2x^2 ln Cx

lety=xty′=t+xdtdx=t2−1txdtdx=t2−1t−t=−1ttdt=−dxxt22=−lnCxy22x2=−lnCx⇒y2=−2x2lnCx

Answered by yeti123 last updated on 15/Dec/21

(dy/dx) = ((y^2 − x^2 )/(xy)) (dy/dx) −x^(−1) y = −xy^(−1) ....(1) let z = y^2 ⇒ (dz/dx) = 2y(dy/dx) (1) ⇒ 2y(dy/dx) − 2y^2 x^(−1) = −2x (dz/dx) − 2zx^(−1) = −2x x^(−2) (dz/dx) − 2zx^(−3) = −2x^(−1) (d/dx)(zx^(−2) ) = −2x^(−1) zx^(−2) = −2ln(x) + c z = x^2 {−2ln(x) + c} y^2 = x^2 {−2ln(x) + c}

dydx=y2−x2xydydx−x−1y=−xy−1....(1)letz=y2⇒dzdx=2ydydx(1)⇒2ydydx−2y2x−1=−2xdzdx−2zx−1=−2xx−2dzdx−2zx−3=−2x−1ddx(zx−2)=−2x−1zx−2=−2ln(x)+cz=x2{−2ln(x)+c}y2=x2{−2ln(x)+c}

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