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Question Number 212435 by Ar Brandon last updated on 13/Oct/24

Answered by mr W last updated on 13/Oct/24

y′=−((2y^2 +xy)/(x^2 +xy+y^2 )) let y=px p+x(dp/dx)=−((2p^2 +p)/(p^2 +p+1)) x(dp/dx)=−((p(p+1)(p+2))/(p^2 +p+1)) ((p^2 +p+1)/(p(p+1)(p+2)))dp=−xdx ∫((1/(2p))−(1/(p+1))+(3/(2(p+2))))dp=−∫xdx ln p−2ln (p+1)+3ln (p+2)=−x^2 −C_1 ln ((p(p+2)^3 )/((p+1)^2 ))=−(x^2 +C_1 ) (((y/x)((y/x)+2)^3 )/(((y/x)+1)^2 ))=(C/e^x^2 ) ⇒((y(y+2x)^3 )/((y+x)^2 ))=((Cx^2 )/e^x^2 )

y′=−2y2+xyx2+xy+y2lety=pxp+xdpdx=−2p2+pp2+p+1xdpdx=−p(p+1)(p+2)p2+p+1p2+p+1p(p+1)(p+2)dp=−xdx∫(12p−1p+1+32(p+2))dp=−∫xdxlnp−2ln(p+1)+3ln(p+2)=−x2−C1lnp(p+2)3(p+1)2=−(x2+C1)yx(yx+2)3(yx+1)2=Cex2⇒y(y+2x)3(y+x)2=Cx2ex2

Commented by Ar Brandon last updated on 13/Oct/24

Thanks, sir!

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