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Question Number 157457 by MathSh last updated on 23/Oct/21

$${\mathrm{xy}}^{{}^{\prime }}+\mathrm{y}={\mathrm{y}}^2\ln \mathrm{x}$$$$$$

Answered by ajfour last updated on 23/Oct/21

$$xdy+ydx=y^2\left(\ln x\right)dx$$$$\frac{xdy}{y^2}+\frac{ydx}{y^2}=\left(\ln x\right)dx$$$$\frac{d\left(xy\right)}{{\left(xy\right)}^2}=\frac{\left(\ln x\right)dx}{x^2}$$$$letxy=s;\ln x=t$$$$\Rightarrow \frac{ds}{s^2}=\frac{tdt}{e^t}$$$$-\frac{1}{s}+c=e^{-t}tdt=f\left(t\right)dt$$$$I=\int e^{-t}tdt$$$$=-{te}^{-t}-e^{-t}$$$$hence$$$$\frac{1}{xy}=\frac{\ln x+1}{x}+c$$$$\frac{1}{y}=\ln x+cx+1$$$$$$

Commented by MathSh last updated on 23/Oct/21

$$\mathrm{Thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{dear}\mathbf{Ser}$$

Answered by mnjuly1970 last updated on 23/Oct/21

$$y^{\prime }+\frac{y}{x}=\frac{y^2}{x}ln\left(x\right)$$$$y^{-1}=u\Rightarrow -y^{-2}{y}^{\prime }=u^{\prime }$$$$-y^{-2}{y}^{\prime }-y^{-1}\frac{1}{x}=-\frac{1}{x}ln\left(x\right)$$$$u^{\prime }-\frac{u}{x}=-\frac{ln\left(x\right)}{x}$$$$u=e^{\int \frac{1}{x}dx}(\int -\frac{ln\left(x\right)}{x}{e}^{-\int \frac{1}{x}dx}dx+C)$$$$=x(\int -\frac{ln\left(x\right)}{x^2}dx+C)$$$$=x[\frac{ln\left(x\right)}{x}-\int \frac{1}{x^2}dx+C]$$$$=ln\left(x\right)+1+Cx$$$$y=\frac{1}{ln\left(x\right)+1+Cx}$$

Commented by MathSh last updated on 23/Oct/21

$$\mathrm{Tthank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{dear}\mathbf{Ser}$$

Commented by mnjuly1970 last updated on 23/Oct/21

$$youarewelcomesir$$

Answered by CAIMAN last updated on 23/Oct/21

Commented by MathSh last updated on 23/Oct/21

$$\mathrm{Thank}\mathrm{you}\mathrm{Ser}$$

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