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Question-212435




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  )
$${y}'=−\frac{\mathrm{2}{y}^{\mathrm{2}} +{xy}}{{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} } \\ $$$${let}\:{y}={px} \\ $$$${p}+{x}\frac{{dp}}{{dx}}=−\frac{\mathrm{2}{p}^{\mathrm{2}} +{p}}{{p}^{\mathrm{2}} +{p}+\mathrm{1}} \\ $$$${x}\frac{{dp}}{{dx}}=−\frac{{p}\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right)}{{p}^{\mathrm{2}} +{p}+\mathrm{1}} \\ $$$$\frac{{p}^{\mathrm{2}} +{p}+\mathrm{1}}{{p}\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right)}{dp}=−{xdx} \\ $$$$\int\left(\frac{\mathrm{1}}{\mathrm{2}{p}}−\frac{\mathrm{1}}{{p}+\mathrm{1}}+\frac{\mathrm{3}}{\mathrm{2}\left({p}+\mathrm{2}\right)}\right){dp}=−\int{xdx} \\ $$$$\mathrm{ln}\:{p}−\mathrm{2ln}\:\left({p}+\mathrm{1}\right)+\mathrm{3ln}\:\left({p}+\mathrm{2}\right)=−{x}^{\mathrm{2}} −{C}_{\mathrm{1}} \\ $$$$\mathrm{ln}\:\frac{{p}\left({p}+\mathrm{2}\right)^{\mathrm{3}} }{\left({p}+\mathrm{1}\right)^{\mathrm{2}} }=−\left({x}^{\mathrm{2}} +{C}_{\mathrm{1}} \right) \\ $$$$\frac{\frac{{y}}{{x}}\left(\frac{{y}}{{x}}+\mathrm{2}\right)^{\mathrm{3}} }{\left(\frac{{y}}{{x}}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{{C}}{{e}^{{x}^{\mathrm{2}} } } \\ $$$$\Rightarrow\frac{{y}\left({y}+\mathrm{2}{x}\right)^{\mathrm{3}} }{\left({y}+{x}\right)^{\mathrm{2}} }=\frac{{Cx}^{\mathrm{2}} }{{e}^{{x}^{\mathrm{2}} } } \\ $$
Commented by Ar Brandon last updated on 13/Oct/24
Thanks, sir!

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