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Question Number 85835 by sahnaz last updated on 25/Mar/20

xydy=(y^2 +x)dx

$$\mathrm{xydy}=\left(\mathrm{y}^{\mathrm{2}} +\mathrm{x}\right)\mathrm{dx} \\ $$

Answered by TANMAY PANACEA. last updated on 25/Mar/20

xydy−y^2 dx=xdx  y(xdy−ydx)=xdx  ((y/x))(((xdy−ydx)/x^2 ))=(dx/x^2 )  ((y/x))d((y/x))=x^(−2) dx  intregating  ((((y/x))^2 )/2)=((−1)/x)+c

$${xydy}−{y}^{\mathrm{2}} {dx}={xdx} \\ $$$${y}\left({xdy}−{ydx}\right)={xdx} \\ $$$$\left(\frac{{y}}{{x}}\right)\left(\frac{{xdy}−{ydx}}{{x}^{\mathrm{2}} }\right)=\frac{{dx}}{{x}^{\mathrm{2}} } \\ $$$$\left(\frac{{y}}{{x}}\right){d}\left(\frac{{y}}{{x}}\right)={x}^{−\mathrm{2}} {dx} \\ $$$${intregating} \\ $$$$\frac{\left(\frac{{y}}{{x}}\right)^{\mathrm{2}} }{\mathrm{2}}=\frac{−\mathrm{1}}{{x}}+{c} \\ $$

Commented by sahnaz last updated on 25/Mar/20

thank you dear friend

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{friend} \\ $$

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