Question

Solve

${\displaystyle y\log y \frac{dx}{dy}=\log y –x}$

Answer 1

${\displaystyle y\log y \frac{dx}{dy}=\log y –x}$

${\displaystyle y\log y\frac{dx}{dy}+x=\log y}$

∴ ${\displaystyle \frac{dx}{dy}+\frac{1}{y\log y}x=\frac{1}{y}}$

The given equation is of the form ${\displaystyle \frac{dx}{dy}+px=Q}$

where, ${\displaystyle P=\frac{1}{y\log y}\text{and}Q=\frac{1}{y}}$

∴ ${\displaystyle I.F.=e^{{\int }^{pdy}=e^{{\int }^{\frac{1}{y\log y}}dy}=e^{\log \left|\log y\right|}=\log y}}$

∴ Solution of the given equation is

${\displaystyle x(I.F.)=\int Q(I.F.)dy+c_1}$

∴ ${\displaystyle x.\log y=1y\int \log y dy+c_1}$

In R. H. S., put log y = t

Differentiating w.r.t. x, we get

${\displaystyle \frac{1}{y}dy=dt}$

∴ ${\displaystyle x\log y=tdt\int +c_1=\frac{t^2}{2}+c_1}$

∴${\displaystyle x\log y=\frac{{\left(\log y\right)}^2}{2}+c_1}$

∴ 2x log y = (log y)² + c …[2c₁ = c]

x log y = ${\displaystyle \frac{1}{2}}$ (log y)² + c