Given (lnx)dxdy=2xy3(ln x) \frac{dx}{dy}=\frac{2}{xy^3}(lnx)dydx=xy32
It can be written as,
(xlnx)dx=2y3dy(xln x){dx}=\frac{2}{y^3}dy(xlnx)dx=y32dy
2y3dy=(xlnx)dx\frac{2}{y^3}dy = (xln x){dx}y32dy=(xlnx)dx
Integrating both sides,
−1y2=[12x2ln(x)−x24]+C-\frac{1}{y^2}=[\frac{1}{2}x^2\ln \left(x\right)-\frac{x^2}{4}] + C−y21=[21x2ln(x)−4x2]+C
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