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Question Number 175210 by mnjuly1970 last updated on 23/Aug/22
y′x+y=y2ln(x)u=y−1⇒u′=−y′y−2−y′y−2x−y−1=−ln(x)u′x−u=−ln(x)u′−1xu=−ln(x)xu=e−∫−1xdx(∫−ln(x)xe−∫1xdxdx+C)=x(−∫ln(x)x2dx+C)ln(x)=t∫te−tdt=[−e−t.t+∫e−tdt]=−1xln(x)−1xu=−ln(x)−Cx−1y=1−ln(x)−Cx−1✓
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![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)) ✓](Q175210.png)