Question Number 175210 by mnjuly1970 last updated on 23/Aug/22
![y′x + y = y^( 2) ln(x) u=y^( −1) ⇒ u′ =−y′y^( −2) −y′y^( −2) x −y^(−1) = −ln(x) u′x −u = −ln(x) u′−(1/x) u =((−ln(x))/x) u = e^( −∫−(1/x)dx) ( ∫−((ln(x))/x)e^( −∫(1/x)dx) dx +C) = x (−∫ ((ln(x))/x^( 2) )dx +C) ln(x)=t ∫te^( −t) dt= [ −e^( −t) .t +∫e^(−t) dt] = −(1/x)ln(x) −(1/x) u= −ln(x) −Cx −1 y = (1/(−ln(x)−Cx−1)) ✓](https://www.tinkutara.com/question/Q175210.png)
$$ \\ $$$$\:\:\:{y}'{x}\:+\:{y}\:=\:{y}^{\:\mathrm{2}} {ln}\left({x}\right) \\ $$$$\:\:\:\:{u}={y}^{\:−\mathrm{1}} \:\Rightarrow\:{u}'\:=−{y}'{y}^{\:−\mathrm{2}} \\ $$$$\:\:\:\:\:−{y}'{y}^{\:−\mathrm{2}} {x}\:−{y}^{−\mathrm{1}} =\:−{ln}\left({x}\right) \\ $$$$\:\:\:\:\:\:\:{u}'{x}\:−{u}\:=\:−{ln}\left({x}\right) \\ $$$$\:\:\:\:{u}'−\frac{\mathrm{1}}{{x}}\:{u}\:=\frac{−{ln}\left({x}\right)}{{x}}\: \\ $$$$\:\:\:\:\:{u}\:=\:{e}^{\:−\int−\frac{\mathrm{1}}{{x}}{dx}} \left(\:\int−\frac{{ln}\left({x}\right)}{{x}}{e}^{\:−\int\frac{\mathrm{1}}{{x}}{dx}} {dx}\:+{C}\right) \\ $$$$\:\:=\:\:{x}\:\left(−\int\:\frac{{ln}\left({x}\right)}{{x}^{\:\mathrm{2}} }{dx}\:+{C}\right) \\ $$$$\:\:\:\:{ln}\left({x}\right)={t} \\ $$$$\:\:\:\:\:\:\int{te}^{\:−{t}} {dt}=\:\left[\:−{e}^{\:−{t}} .{t}\:+\int{e}^{−{t}} {dt}\right] \\ $$$$\:\:\:\:\:\:=\:−\frac{\mathrm{1}}{{x}}{ln}\left({x}\right)\:−\frac{\mathrm{1}}{{x}} \\ $$$$\:\:\:\:\:\:{u}=\:−{ln}\left({x}\right)\:−{Cx}\:−\mathrm{1} \\ $$$$\:\:\:\:\:\:{y}\:=\:\frac{\mathrm{1}}{−{ln}\left({x}\right)−{Cx}−\mathrm{1}}\:\checkmark \\ $$